Variables Complex Notes for Math 703. Updated Fall 2011 Anton R. Schep
CHAPTER 1
Holomorphic (or Analytic) Functions
1. Definitions and elementary properties In complex analysis we study functions f : S C, where S C. When referring to open sets in C and continuity of functions f we will always consider C (and its subsets) as a metric space with respect to the metric d(z1 , z2 ) = |z1 - z2 |, where | | denotes the complex modulus, i.e., |z| = x2 + y 2 whenever z = x + iy with x, y R. An open ball with respect this metric will be also referred to as an open disc and denoted by B(a, B(a, r) = {z C : |z - a| < r}, where a is the center and r > 0 is the radius of the open ball. The closed disc with center a and radius r is denoted by B(a, r), so B(a, r) = {z C : |z - a| r}. Recall that G C is called open if for all a G there exists r > 0 such that B(a, r) G. If z = x + iy, then the conjugate z of z is defined by z = x - iy. Now zz = |z|2 , z 1 so that z = |z|2 for z = 0. Elementary properties of complex numbers are given by: 1 (1) The real part Re z of z satisfies Re z = 2 (z + z), while the imaginary part 1 Im z of z is given by Im z = 2i (z - z). (2) For all z1 , z2 C we have z1 + z2 = z1 + z2 and z1 z2 = z1 z2 . (3) For all z1 , z2 C we have |z1 z2 | = |z1 | |z2 |. 2. Elementary transcendental functions Recall also that if z = x+iy = 0, then, using polar coordinates, we can write z = r cos + ir sin . In this case we write arg z = { + 2k : k Z}. By Arg z we will denote the principal value of the argument of z = 0, i.e. = Arg z arg z if - < . Note that if z1 = |z1 |(cos 1 + i sin 1 ) and z2 = |z2 |(cos 2 + i sin 2 ), then we have z1 z2 = |z1 ||z2 |(cos 1 cos 2 - sin 1 sin 2 + i(sin 1 cos 2 + cos 1 sin 2 )) = |z1 z2 |(cos(1 + 2 ) + i(sin(1 + 2 )). Hence we have arg (z1 z2 ) = arg z1 + arg z2 . Define now ez = ex (cos y + i sin y). Then |ez | = ex and arg ez = y + 2k. In particular e2i = 1 and the function ez is 2i-periodic, i.e., ez+2i = ez e2i = ez for all z C. We want now to define log w such that w = ez where z = log w, but we can not define it as just the inverse of ez as ez is not one-to-one. Consider therefore the equation w = ez for a given w. We must assume that w = 0 as ez = 0 (and thus log 0 is not defined). Then |w| = |ez | = ex and y = Arg w + 2k (k Z). Hence {log |w| + i(Arg w + 2k) : k Z} is the set of all solutions z of w = ez . We write log w for any w in the set {log |w| + i(Arg w + 2k) : k Z}.
3
4
1. HOLOMORPHIC (OR ANALYTIC) FUNCTIONS
Definition 2.1. Let G C be an open connected set and f : G C a continuous function such that z = ef (z) for all z G. Then f is called a branch of the logarithm on G. It is clear that if f is a branch of the logarithm on G, then 0 G and f (z) = / log |z| + i(Arg z + 2k) for some k Z, where k can depend on z. Also, if f is a branch of the logarithm on G, then for fixed k also g(z) = f (z) + 2ki is a branch of the logarithm on G. The converse also holds. Proposition 2.2. Let G C be an open connected set and f : G C a branch of the logarithm on G. Then every other branch of the logarithm on G is of the form f + 2ki for some fixed k Z. Proof. Suppose g is another branch of the logarithm on G. Then define 1 h = 2i (f - g). Then h is continuous on G, h(G) Z, and G connected implies that h(G) = {k} for some k Z. To find a branch of log z for a given open and connected set G requires finding (as log |z| is continuous on C \ {0}) a continuous selection of arg z in {Arg z + 2k}. As G is connected, the range of this continuous selection has to be an interval of length at most 2, but such a selection does not always exist! This happens e.g. in case G = C \ {0}, then G is open and connected, but there does not exist a branch of log z on G, i.e., Arg z is discontinuous on the negative x-axis. in the next examples we construct some branches of log z. Example 2.3. (i) Let G = C\{z R : z 0}. Then Arg z is continuous on G, so f (z) = log |z| + iArg z is a branch of log z on G. This branch is called the principal branch of log z and denoted by Log z. (ii) Let G = C \ {z R : z 0}. Let (z) denote the unique value of arg z such that 0 < (z) < 2. Then f (z) = log |z| + i(z) is a branch of log z on G. 3. Differentiable functions Definition 3.1. Let G C be an open set and f : G C. Then f is differentiable at z G if f (z + h) - f (z) lim h0 h exists. When this limit exists we denote it by f (z) and call it the (complex) derivative of f at z. If f (z) exists at every point of G, then we call f analytic or holomorphic on G. Notation. H(G) = {f : g C; f holomorphic in G}. If S C is any set, then we say that f is holomorphic in S if f H(G) for some open set G S. Remarks 3.2. 1. The function f is differentiable at z G, if for |h| small enough we can write f (z + h) = f (z) + f (z)h + (h)h, where (h) 0 as h 0. From this it follows directly that if f is differentiable at z, then f is continuous at z.
3. DIFFERENTIABLE FUNCTIONS
5
2. Note that f is differentiable at z0 G with derivative equal to f (z0 ) is equivalent to saying that for all > 0 there exists a > 0 such that f (z + h) - f (z) - f (z0 ) < h for all h C with 0 < |h| < . In particular we can take h = x with x real and 0 < |x| < or h = iy with y real and 0 < |y| < . This fact will be exploited in the proof of the next theorem. Theorem 3.3. (CauchyRiemann equations) Let G C be an open set and f : G C be differentiable at z = x + iy G. Let f (z) = u(x, y) + iv(x, y), where v u and v are real valued functions on G. Then the first order partials u , u , x x y v and y exist at (x, y) and satisfy the CauchyRiemann equations u v u v = and =- x y y x at the point (x, y). Proof. In the definition of the derivative we can restrict ourselves first to real valued h 0. We get then that f (z) =
h0,hR
lim
u(x + h, y) - u(x, y) v(x + h, y) - v(x, y) +i h h
=
v u +i x x
exists at z = x + iy and similarly by restricting to h = ik with k real valued and k 0, we get f (z) =
k0,kR
lim
u(x, y + k) - u(x, y) v(x, y + k) - v(x, y) +i ik ik u v u v = and =- x y y x
= -i
u v + . y y
Equating the two expressions for f (z) we get that
at the point (x, y). Example 3.4. v v (i) Let f (z) = zz = x2 + y 2 . Then u = 2x, y = 0, u = 2y and x = 0. x y Hence the CauchyRiemann equations hold if and only if (x, y) = (0, 0). At z = 0 we have f (0 + h) - f (0) =h0 h as h 0. Hence f is differentiable only at z = 0 and thus nowhere holomorphic as there exists no open set G containing 0 on which f is differentiable. (ii) Let f (z) = c, where c C is a constant. Then f (z) = 0 for all z C, so f H(C). Similarly if g(z) = z, then g (z) = 1 for all z C, so g H(C) (iii) Let f (z) = 1/z on C \ {0}. Then -1 -1 f (z + h) - f (z) = 2 h z(z + h) z for all z = 0, so that f is holomorphic on C \ {0}.
6
1. HOLOMORPHIC (OR ANALYTIC) FUNCTIONS
Definition 3.5. A function f : C C is called entire if f is holomorphic on C. The above example shows that f (z) = c and f (z) = z are entire functions. To get additional examples of holomorphic and entire functions we first observe that analogously to the rules of differentiation of real valued functions one can prove the following proposition. Proposition 3.6. Let G be a nonempty open subset of C. Then the following holds. (1) If f, g holomorphic on G and C, then so are f + g, f , and f g. (2) If f (G) G1 , where G1 is open and g H(G1 ), then h = g f is holomorphic on G and h (z) = g (f (z))f (z) for all z G. Proof. We will only prove 2. Let z G and put w = f (z). Then f being holomorphic at z implies that we can write f (z + h) - f (z) = [f (z) + where where
1 (h) 1 (h)]h,
0 as h 0. Similarly g(w + k) - g(w) = [g (w) +
2 (k)]k,
2 (k)
0 as k 0. Putting k = f (z + h) - f (z) we get
1 (h))
g(f (z + h)) - g(f (z)) = (g (f (z)) + 2 (f (z + h) - f (z)))(f (z) + h g (f (z))f (z) as h 0.
Corollary 3.7. (1) Any polynomial p(z) = a0 +a1 +. . .+an z n is entire. (2) Any rational function f (z) = p(z) , where p and q are polynomials, is q(z) holomorphic on C \ {z C : q(z) = 0}. We will now compare complex differentiability of f = u + iv with the real differentiability of the map (u, v) : R2 R2 . Recall first the definition of real differentiability of a vector valued mapping. Definition 3.8. Let G Rm an open set and F : G Rn . Then F is real differentiable at c G if there exist a linear mapping DF (c) : Rm Rn such that
h0
lim
F (c + h) - F (c) - DF (c)h = 0. h
Writing F = (F1 , , Fn ), where Fi : Rm R, then real differentiability of F at c G is equivalent with the real differentiability of each Fi and DFi (c)h = Fi (c) h, where Fi denotes the gradient of Fi and thus DF (c) is the linear map given by the Jacobian matrix of F . We now take m = n = 2 to compare complex differentiability of f = u+iv at z0 = x0 +iy0 with real differentiability of F = (u, v) at c = (x0 , y0 ). We first deal with the special case of a linear map. Lemma 3.9. Let A : R2 R2 be a real linear map, given by the matrix [ai,j ]. Then A = (u, v) where f = u + iv is a complex linear map from C to C if and only if a1,1 = a2,2 and a1,2 = -a2,1 .
3. DIFFERENTIABLE FUNCTIONS
7
Proof. Assume first that f (z) = Cz for some C = c1 + ic2 . Then u(x, y) = (c1 x - c2 y) and v(x, y) = (c2 x + c1 y), which implies immediately that A = (u, v) is a linear map with matrix [ai,j ], where a1,1 = a2,2 = c1 and a1,2 = -a2,1 = -c2 . Conversely, if a1,1 = a2,2 = c1 and a1,2 = -a2,1 = -c2 , then it is straightforward to check that f (z) = Cz with C = c1 + ic2 . Remark 3.10. Note that the condition on the matrix A are the ones imposed by the Cauchy-Riemann equations for f (z) = Cz = u + iv. As the real derivative DF (c) of a linear map F : C C is F (c) this says that a linear map from R2 R2 corresponds to a complex differntiable map from C to C if and only if it is complex linear. An immediate consequence of of the Lemma is the following theorem. Theorem 3.11. Let G C be an open set and f : G C, where f (z) = u(x, y) + iv(x, y). Let z0 = x0 + iy0 G. then the following are equivalent. (1) f is complex differentiable at z0 . (2) F = (u, v) is real differentiable at (x0 , y0 ) and the derivative DF (x0 , y0 ) is complex linear. (3) F = (u, v) is real differentiable at (x0 , y0 ) and the Cauchy-Riemann equations hold at (x0 , y0 ). To prove a theorem about complex differentiability when the Cauchy-Riemann equations hold, we need first a result from vector calculus. Lemma 3.12. Let G be an open subset of R2 and u : G R2 a function which has partial derivatives on G, which are continuous at (x0 , y0 ) G. Then there exist 1 (h), and 2 (h) in a neighborhood of (0, 0) with 1 (h) 0 and 2 (h) 0 as h = (h1 , h2 ) (0, 0) such that u u (x0 , y0 )h1 + (x0 , y0 )h2 + 1 (h)h1 + 2 (h)h2 . u(x0 + h1 , y0 + h2 ) = u(x0 , y0 ) + x y Proof. Let r > 0 such that for h = (h1 , h2 ) with h < r we have that (x0 + h1 , y0 + h2 ) G. Let h < r. Then by the Mean Value theorem there exist k1 between x0 and x0 + h1 and k2 between y0 and y0 + h2 such that u(x0 + h1 , y0 + h2 ) - u(x0 , y0 ) = u(x0 + h1 , y0 + h2 ) - u(x0 , y0 + h2 ) + u(x0 , y0 + h2 ) - u(x0 , y0 ) = The proof now follows if we put u u y (x0 , k2 ) - y (x0 , y0 ).
1 (h)
u u (k1 , y0 + h2 )h1 + (x0 , k2 )h2 . x y
u x (k1 , y0
=
+ h2 ) -
u x (x0 , y0 )
and
2 (h)
=
Theorem 3.13. Let G C be an open set and f : G C. Let f (z) = u(x, y) + iv(x, y), where u and v are real valued functions on G. Assume that the v v first order partials u , u , x and y exist on G, are continuous at (x, y) and x y satisfy the CauchyRiemann equations u v u v = and =- x y y x at the point (x, y). Then f is complex differentiable at z = x + iy.
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1. HOLOMORPHIC (OR ANALYTIC) FUNCTIONS
Proof. Identifying C with R2 we can find by the above lemma j (h) 0 as h = h1 + ih2 0 for j = 1, , 4 such that
j (h)
with
f (z + h) - f (z) u h1 u h2 h1 h2 = (x, y) + (x, y) + 1 (h) + 2 (h) h x h y h h h h1 v h1 h1 h2 v (x, y) + (x, y) + 3 (h) + 4 (h) +i x h y h h h u v h1 h2 = (x, y) + i (x, y) + 1 (h) + 2 (h) x x h h h1 h2 + i 3 (h) + i 4 (h) h h u v (x, y) + i (x, y) x x h2 h1 as h 0, since | h | 1 and | h | 1. Corollary 3.14. Let f (z) = ez . Then f is entire and f (z) = ez for all z C. Proof. If f = u + iv, then u(x, y) = ex cos y and v(x, y) = ex sin y. Now v v = ex cos y, x (x, y) = ex sin y, u (x, y) = -ex sin y, and y (x, y) = y e cos y. Hence the Cauchy-Riemann equations hold for all (x, y) and, as the partial are continuous, it follows from the above theorem that f is holomorphic at all v z C. Moreover f (z) = u (x, y) + i x (x, y) = ez . x
u x (x, y) x
Proposition 3.15. Let G1 , G2 C be open sets and let f : G1 G2 , g : G2 G1 be continuous mappings such that g(f (z)) = z for all z G1 . If g is holomorphic on G2 and g (z) = 0 for all z G2 , then f is holomorphic on G1 and f (z) = g (f1(z)) for all z G1 . Proof. Let z G1 . Then for h = 0 but small enough we have z + h G1 and f (z + h) = f (z), since g(f (z)) = z = (z + h) = g(f (z + h)). Now 1= g(f (z + h)) - g(f (z)) f (z + h) - f (z) f (z + h) - f (z) h
implies that f is differentiable at z and 1 = g (f (z))f (z). Corollary 3.16. Let G C be an open connected set and f : G C a branch 1 of the logarithm on G. Then f is holomorphic on G and f (z) = z for all z G. Proof. Take g(z) = ez in the above proposition. We conclude this section with some remarks about harmonic functions. Recall that if G R2 is open and u : G R satisfies the Laplace equation u = 2u 2v x2 (x, y) + y 2 (x, y) = 0 0n G. Let now f H(G), let u = Re f and v = Im f . Assume that u and v have continuous second order partials (an assumption which 2 2 we will show later on to be always true). Then u = xu (x, y) + yv (x, y) = 2 2 + - v (x, y) = 0. Hence u is harmonic on G. Similarly v is harmonic yx on G. Two harmonic functions u, and v are called conjugate harmonic functions, when f = u+iv is holomorphic on G. Another consequence of the Cauchy-Riemann equations is that the inner product of the gradients u and v satisfy u v = 0, i.e, the level curves u(x, y) = c1 and v(x, y) = c2 intersect orthogonally.
2v xy (x, y)
2
4. POWER SERIES
9
4. Power series In this section we will see how one can use power series to get a large class of examples of holomorphic functions. In fact, in a later chapter we will see that locally every holomorphic function can be so obtained. We start by recalling some basic facts concerning series. Recall that if an n0 is a sequence of complex numbers, then the series n=0 an converges to s C if |s - sn | 0 as n , where sn = a0 + . . . + an . The number s is then called the sum of the series. The series is said to diverge, if it does not converge to any s C. As in the real variable case we have: (1) If n=0 an converges, then an 0 as n . (2) If n=0 |an | converges, then n=0 an converges. A power series is a series of the form n=0 cn (z - a)n . Usually we will treat z as a variable and the cn 's and a as constants in this expression. Example 4.1. Consider the geometric series n=0 z n . The partial sums sn n+1 for all z = 1. Hence for |z| < 1 are in this case given by sn = 1 + . . . + z n = 1-z 1-z 1 the series n=0 z n converges and has sum equal to 1-z , while if |z| 1 the series n diverges, since in that case it is not true that z 0 as n . The following simple result turns out to be a useful tool in studying the convergence of power series. Theorem 4.2. (Weierstrass Mtest) Let G C and un : G C such that |un (z)| Mn on G, where 0 Mn < . Then 0 un (z) converges uniformly on G. Proof. For fixed z G we have that 0 |un (z)| 0 Mn < . Hence the series 0 un (z) converges for all z G. Let f (z) = 0 un (z) for z G denote the sum of the series and let > 0. Then there exists N such that k=N +1 Mk < . Then we have for all z G and all n N that
n
f (z) -
k=0
un (z) =
k=n+1
un (z)
k=n+1
|un (z)|
k=n+1 0
Mk <
for all n N and all z G and thus the series f (z) on G.
un (z) converges uniformly to
For a given power series n=0 cn (z - a)n we define the radius of convergence 1 R, 0 R , by R = lim n |cn |. The circle {z C : |z - a| = R} is called the circle of convergence of the power series. Theorem 4.3. (Cauchy Root test) Let n=0 cn (z - a)n be a power series with radius of convergence R. Then the following holds. n (1) n=0 cn (z - a) converges absolutely for |z - a| < R. n (2) n=0 cn (z - a) diverges for |z - a| > R. (3) If 0 < r < R, then n=0 cn (z - a)n converges uniformly on |z - a| r. Proof. Let |z - a| < r < R. Then that |cn |
1 n
1 r
>
1 R
implies that there exists N such
|z-a| r n
<
1 r
for all n N . It follows that |cn (z - a)n |
for all n N .
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1. HOLOMORPHIC (OR ANALYTIC) FUNCTIONS
Since |z-a| < 1, it follows that n=0 cn (z - a)n converges absolutely for |z - a| < r r for any r < R and thus 1. holds. Let now |z - a| > r > R. Then there exist infinitely many n such that |cn | n >
1
1 r.
Hence |cn (z - a)n |
n
|z-a| r
n
> 1 for
infinitely many n, i.e., the series - a) diverges for |z - a| > r for any r > R and thus 2. holds. To prove 3. let 0 < r < s < R. Then as above there 1 n exists N such that |cn | n < 1 for all n N . It follows that |cn (z - a)n | r s s for all n N and all |z - a| < r. Since r < 1, it follows that n=0 cn (z - a)n s converges uniformly on |z - a| r by the Weierstrass Mtest. In dealing with power series with coefficients involving factorials, it is often easier to use the following result. Theorem 4.4. (Ratio test) Let n=0 cn (z - a)n be a power series with radius of convergence R. Assume cn = 0 for all n. Then cn+1 1 cn+1 lim . lim cn R cn In particular, if limn Proof. Exercise A power series can converge or diverge at any point of its circle of convergence as can be seen from the following examples. Example 4.5. (i) The series
(z+1)n 1 1 n n=0 2n+1 has R = 2, as lim 2n+1 = 2 . Note that the sum 1 1 1 of series equals 1-z for all |z + 1| < 2, since 1-z = 2-(z+1) = 1 1-1 = z+1 2 2 |z+1| 1 z+1 n for 2 < 1. n=0 2 2 n The series n=1 z 2 has R = 1 (e.g. by the Ratio test), and the series n 1 converges absolutely for any z on the circle of convergence as n=1 n2 < cn+1 cn
n=0 cn (z
exists, then
1 R
= limn
cn+1 cn
.
(ii)
. n (iii) The series n=1 zn has R = 1 (e.g. by the Ratio test), but it does not 1 converge absolutely for any z on the circle of convergence as n=1 n = . In particular it diverges for z = 1. One can show however (but this is not completely trivial) that it converges for any z = 1 with |z| = 1 (for z = -1 this follows e.g. from the so-called alternating series test). n (iv) The series n=0 z has R = (e.g. by the Ratio test). We will see after n! the next theorem that ez equals the sum of this series. n has R = 0 (e.g. by the Ratio test). Hence it (v) The series n=1 n!z converges only for z = 0.
n Proposition 4.6. Let R be the radius of convergence of n=0 cn (z - a) . n-1 Then R is also the radius of convergence of the power series n=1 ncn (z - a) = (n + 1)cn+1 (z - a)n . n=0 Proof. From calculus we know that limn n n = 1. Hence
lim
n
(n + 1) |cn+1 | = lim
n+1
(n + 1) |cn+1 |
n+1 n
=
1 . R
4. POWER SERIES
11
Note, if we apply the above proposition twice, we get that n=2 n(n - 1)z n-2 converges absolutely for |z - a| < R. The following theorem says that inside the circle of convergence the sum of the power series is a holomorphic function.
n Theorem 4.7. Let n=0 cn (z - a) have radius of convergence R = 0 and n define f (z) = n=0 cn (z - a) for |z - a| < R. Then f H(B(a, R)) and f (z) = n-1 for |z - a| < R. n=1 ncn (z - a)
Proof. It follows from the above corollary that g(z) = n=1 ncn (z - a)n-1 also converges in B(a, R). Remains to show that f (z) = g(z) on |z - a| < R. W.l.o.g. we can assume that a = 0. In the argument below we will use that n (z + h)n - z n = h k=1 (z + h)k-1 z n-k . Let z, z + h B(0, r), where 0 < r < R. Then we have f (z + h) - f (z) - g(z) = h =
n=2
cn
n=1
(z + h)n - z n - nz n-1 h
n
cn
k=1 n
(z + h)k-1 z n-k - z n-1 z n-k (z + h)k-1 - z k-1
k=2 n k-1
n=2
|cn |
|h|
n=2
|cn |
k=2 n
z n-k
l=1
|z + h|l-1 |z|k-1-l
|h|
n=2
|cn |
(k - 1)rn-k rk-2
k=2
= |h|
1 |cn | n(n - 1)rn-2 0 2 n=2
1 as h 0, since, by the above proposition, n=2 |cn | 2 n(n - 1)rn-2 < as r < R. Hence f (z) = g(z) on |z| < r for any r < R and the proof is complete.
Corollary 4.8. Let f (z) = n=0 cn (z-a)n have radius of convergence R = 0. Then f (k) (z) exists on B(a, R) for all k 1 and thus f (k) H(B(a, R)) and
f (k) (z) =
n=k
cn n(n - 1) . . . (n - k + 1)(z - a)n-k
for all k 1 and all |z - a| < R. In particular k!ck = f (k) (a) and thus the coefficients ck of the power series are unique. Example 4.9. Let f (z) =
zn n=0 n! .
Then by the above theorem
z n-1 nz n-1 f (z) = = = f (z) n! (n - 1)! n=1 n=1 for all z C. Let h(z) = e-z f (z). Then h (z) = -e-z f (z) + e-z f (z) = 0 for all z C. From the next proposition it follows that h(z) = h(0) = 1 for all z, i.e., f (z) = ez for all z.
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1. HOLOMORPHIC (OR ANALYTIC) FUNCTIONS
Proposition 4.10. Let G C be an open and connected set. Assume f H(G) such that f (z) = 0 for all z G. Then f is constant on G. Proof. Let z0 G and put A = {z G : f (z) = f (z0 )}. Then the continuity of f implies that A is closed. Let now a A. Then there exists > 0 such that B(a, ) G. Let z B(a, ) and put g(t) = f (tz + (1 - t)a) for 0 t 1. Then by the chain rule g (t) = f (tz + (1 - t)a)(z - a) = 0 for 0 < t < 1, so g is constant on 0 t 1. Hence f (z) = g(0) = g(1) = f (z0 ), and thus B(a, ) A. It follows that A is nonempty open and closed subset of G, thus A = G.
CHAPTER 2
Integration over contours
1. Curves and Contours A curve is a continuous map : [a, b] C. We call (a) the initial point and (b) the end point of the curve , and [a, b] is called the parameter interval of . If (a) = (b), then is called a closed curve. Denote by the range of . The curve induces an orientation of , namely the direction in which (t) traces as t increases from a to b. Often we will specify a curve by its range together with an orientation indicating how (and possibly how often) the range is traversed. Given a curve we can find an oriented curve -, with identical range, but with opposite orientation, e.g., (-)(t) = (a + b - t) where a t b as a parametrization of the curve -. If 1 and 2 are two curves with with parameter intervals [a1 , b1 ], [a2 , b2 ] respectively such that 1 (b1 ) = 2 (a2 ), then we can join the two curves to get the curve = 1 2 by taking (t) = 1 (t) a1 t b1 2 (t + a2 - b1 ) b1 t b1 + b2 - a2 .
A curve is called smooth, if (t) exists and is continuous for all a t b ( with one-sided derivatives at a and b). Note if we write (t) = x(t) + iy(t), then (t) exists if and only if x (t) and y (t) exist. From multi-variable calculus we know that (t) represents a tangent vector to the curve . A path or contour is a piecewise smooth curve, i.e., : [a, b] C such that there exist a = t0 < t1 < . . . < tn = b where restricted to [ti-1 , ti ] is smooth for i = 1, . . . , n. Note that can have corners at the points (ti ), i.e., the right and left hand derivatives of (t) at ti can differ. A path is called simple if : [a, b] C is such that (s) = (t) for all a s < t b, except possibly for s = a and t = b. The path is closed if (a) = (b). Example 1.1. (i) The directed line segment C from z1 to z2 is the range of a smooth curve. A parametrization of C is given : [0, 1] C defined by (t) = (1-t)z1 +tz2 . We will denote this curve by [z1 , z2 ]. (ii) A circular arc oriented counterclockwise is the range of an curve. Suppose the arc is part of the circle with center z0 and radius r, then (t) = z0 +reit with 1 t 2 will trace a circular arc counterclockwise. If 2 - 1 = 2 the curve will be the complete circle. Note the curve is simple if and only if 2 - 1 2.
13
14
2. INTEGRATION OVER CONTOURS
1.1. Conformal mappings. Let f be a holomorphic function on an open set G C. Let z0 G be a fixed point and let : [a, b] C be a smooth curve in G passing through z0 with non-zero tangent, i.e., (t0 ) = z0 for some t0 (a, b) and (t0 ) = 0. Then 1 = f is a curve passing through f (z0 ) and 1 (t0 ) = f (z0 ) (t0 ). If now f (z0 ) = 0, we see that arg 1 (t0 ) = arg f (z0 ) + arg (t0 ) and |1 (t0 )| = |f (z0 )|| (t0 )|. Thus the tangent vector (t0 ) to the curve at z0 is under the mapping f rotated over an angle arg f (z0 ) and stretched by a factor |f (z0 )|. Applying this to two curves passing through z0 we see that under the mapping f the angle between the two curves is preserved (including the direction they are measured), while their tangent vectors are stretched by the same amount. Mappings which preserve angles (including the direction they are measured) between smooth curves are called conformal. Thus we have proved: Theorem 1.2. Let f be a holomorphic function on an open set G C.Assume f (z) = 0 for all z G. Then f is conformal on G. We will now see that in fact the converse is true too, To do so we will introduce some additional notation. Let f = u + iv as usual. Then we define f = f - i f z x y and
f z
=
f x
+ i f . It is now a routine calculation to show that u and v satisfy y
f z
the Cauchy-Riemann equations if and only if
= 0.
Theorem 1.3. Let f = u + iv be a function on an open set G C with continuous partials. Assume f is conformal on G. Then f is holomorphic on G and f (z) = 0 for all z G. Proof. Let be a smooth curve with non-zero tangent passing through z0 G. Let 1 (t) = f ((t)). Write (t) = x(t) + iy(t). Then 1 = u x + u y + i v v + x y v
v i y y = f x x
+
f y y
=
f z
+
f z
. Let (t0 ) = z0 . Then
1 (t0 ) f f (t0 ) = + . (t0 ) z z (t0 ) Now f conformal implies that the argument of the left hand side of this equation is constant modulo 2. This implies that f (z0 ) = 0, since the argument of (t0 ) is z (t0 ) not constant modulo 2, when we take e.g. (t) = z0 + tei . Hence u and v satisfy the Cauchy-Riemann equations at z0 and thus f (z0 ) exists and f (z0 ) = f (z0 ) = z
1 (t0 ) (t0 )
= 0.
2. Contour integrals Definition 2.1. A curve : [a, b] C is called rectifiable if is of bounded variation, i.e., if
n
() = sup
i=1
|(ti ) - (ti-1 )| : a = t0 < . . . < tn = b
< .
In this case the length of is defined to be (). Given a continuous : [a, b] b b b C we define a (t) dt = a Re (t) dt + i a Im (t) dt. In case is (piecewise)
2. CONTOUR INTEGRALS
15
smooth we have by the Fundamental Theorem of Calculus for real integrals that b (t) dt = (b) - (a). a Lemma 2.2. Let f : [a, b] C be a continuous function. Then
b b
f (t) dt
a a
|f (t)| dt.
Proof. Let =
b a
f (t) dt. If = 0, then the inequality is trivial. Assume
b a
= 0. Then we can write = rei , where r = |
b
f (t) dt|. Now we have
b
r = Re (e-i ) =
a
Re (e-i f (t)) dt
a
|f (t)| dt.
Theorem 2.3. Let : [a, b] C be a piecewise smooth curve. Then is rectifiable and
b
() =
a
| (t)| dt.
Proof. Without loss of generality we can assume that is smooth. Let a = t0 < . . . < tn = b be a partition of [a, b]. Then by the Fundamental Theorem of Calculus and the above lemma we have
ti ti
|(ti ) - (ti-1 )| =
ti-1
(t) dt
ti-1 b
| (t)| dt.
This implies that is rectifiable and () a | (t)| dt. For the reverse inequality, let > 0. Then is uniformly continuous on [a, b], so there exists > 0 such that | (t) - (s)| < whenever |t - s| < . Now there exists a partition a = t0 < . . . < tn = b with ti = ti - ti-1 < such that
b n
| (t)| dt -
a i=1
| (ti )|ti < .
For 1 i n we have now that ||(ti ) - (ti-1 )| - | (ti )|ti | |(ti ) - (ti-1 ) - (ti )ti |
ti
=
ti-1 ti
(t) - (ti ) dt | (t) - (ti )| dt < ti .
ti-1
16
2. INTEGRATION OVER CONTOURS
Combining the last two estimates we get
b n
| (t)| dt
a i=1 n
| (ti )|ti + (|(ti ) - (ti-1 )| + ti ) +
i=1
() + (b - a) + for all > 0. Hence
b a
| (t)| dt ().
Let : [a, b] C be a piecewise smooth curve and let f : C be continuous. b Then we define f (z)dz = a f ((t)) (t) dt. Example 2.4. Let : [a, b] C be a piecewise smooth curve. Then 1 dz = (b) - (a). This is immediate from the definition and the Fundamental Theorem of Calculus. Proposition 2.5. Let : [a, b] C be a piecewise smooth curve and let f : C be a continuous function. Then the following hold. (i) - f (z) dz = - f (z) dz, where -(t) = (a + b - t). (ii) If = 1 2 , then f (z) dz =
1
f (z) dz +
2
f (z) dz.
(iii) If |f (z)| M on , then | f (z) dz| M (). (iv) ("Independence of parametrization") Let : [a1 , b1 ] [a, b] be a smooth onto function with > 0. Then for 1 = we have f (z) dz =
1
f (z) dz. f (z) + g(z) dz =
(v) If also g : C continuous and , C, then f (z) dz + g(z) dz.
Proof. Let -(t) = (a + b - t). Then - : [a, b] C is piecewise smooth and (-) (t) = - (a + b - t) except possibly finitely many points, from which (i) follows directly. Part (ii) is an immediate consequence of the definition. Part (iii) follows from
b b
f (z) dz
a
|f ((t)) (t)| dt M
a
| (t)| dt = M ().
Part (iv) follows from the chain rule 1 (t) = ( (t)) (t) and the change of variable rules for real integrals
b1 b1
f (z) dz =
1 a1
f (1 (t))1 (t) dt =
a1 (b1 )
f (( (t))) ( (t)) (t) dt f (z) dz.
=
(a1 )
f ((s)) (s) ds =
2. CONTOUR INTEGRALS
17
Part (v) is immediate from the definition and the corresponding property of real integrals. Corollary 2.6. Let : [a, b] C be a piecewise smooth curve and let fn : C be continuous functions which converge uniformly to f on . Then fn (z) dz
f (z) dz
as n . Proof. Note first that f is also continuous on as it is the uniform limit of a sequence of continuous functions. Let Mn = supz |fn (z) - f (z)|. Then by assumption Mn 0 as n . From (iii) and (v) above we have now fn (z) dz -
f (z) dz =
fn (z) - f (z) dz Mn () 0
as n . The following example is important for the development of the theory. Example 2.7. Let : [0, 2] C be given by (t) = a + reit , i.e., is the circle with center a and radius r traversed counterclockwise. We will show that (2.1)
(z - a)n dz =
0 2i
if n Z \ {-1} if n = -1
Since is smooth we can write
2
(z - a)n dz =
0
(reit )n ireit dt
2
= irn+1
0
ei(n+1)t dt
1 i(n+1)t 2 |0 i(n+1) e
= ir
n+1
=0
if n Z \ {-1} if n = -1,
= 2i
which proves the formula. Note that this integral does not depend on r. The following Theorem will allow us to extend this example, in case n = -1, to arbitrary closed contours with a . / Theorem 2.8. Let : [a, b] C be a piecewise smooth curve and assume F is holomorphic on (an open set containing) with F continuous on . Then F (z) dz = F ((b)) - F ((a)).
In particular, if is a closed contour, then
F (z) dz = 0.
Proof. Assume first that is smooth. Then by the chainrule (F ) (t) = b F ((t)) (t) for all a t b. Hence F (z) dz = a F ((t)) (t) dt = (F ) (t) dz = F ((b)) - F ((a)), which proves the theorem for the special case of a smooth curve. In the general case, choose a = s0 < s1 < < sn = b such that
18
2. INTEGRATION OVER CONTOURS n i=1 i
i = |[si-1 ,si ] is smooth. Then F (z) dz = F ((si-1 )) = F ((b)) - F ((a)).
F (z) dz =
n i=1
F ((si )) -
Corollary 2.9. Let be any closed contour. Then (z - a)n dz = 0 for all n 0 and if in addition a , then also (z - a)n dz = 0 for all n -2. / Proof. Take F (z) =
1 n+1 (z
- a)n+1 in the above theorem.
Let now {a, b, c} be an ordered triple of complex numbers. Then = (a, b, c) denotes the triangle with vertices a, b, and c. By we denote curve obtained by joining the line segments [a, b], [b, c] and [c, a], i.e., denotes the boundary of (a, b, c) traversed counterclockwise. Hence f (z) dz =
[a,b]
f (z) dz +
[b,c]
f (z) dz +
[c,a]
f (z) dz
for any continuous f on . Theorem 2.10. (Cauchy's Theorem for a Triangle) Let G C be an open set and assume = (a, b, c) G. Let p G and f : G C such that f is continuous on G and holomorphic on G \ {p}. Then f (z) dz = 0.
Remark. If f satisfies the above hypotheses, then we shall see later that f is actually holomorphic on G. Proof. Assume first that p = (a, b, c). Let {a1 , b1 , c1 } be the midpoints / of [b, c], [c, a], and [a, b] respectively. Consider the four triangles 1 , 2 , 3 , and 4 formed by the triples {a, c1 , b1 }, {c1 , b, a1 }, {a1 , b1 , c1 } and {a1 , c, b1 } (see Figure 1). Put I = f (z) dz. Then
4
I=
j=1 j
f (z) dz.
Now |
j
f (z) dz|
|I| 4
for at least one j. By relabeling we can assume that
c
!4 b1 !1 !3 !2 b c1 a1
a
Figure 1. = 1 2 3 4
2. CONTOUR INTEGRALS
19
f (z) dz
1
|I| . 4
Dividing similarly 1 into four triangles by means of the midpoints of the edges and repeating this process, we get a sequence of triangles 1 2 such that (n ) = 21 L, where L = (), and such that n (2.2)
n
f (z) dz
|I| . 4n
Since is compact and {n } has the finite intersection property, it follows that there exists z0 n n . As p , we have that z0 = p and thus f is differentiable / at z0 . Let > 0. Then there exists r > 0 such that |f (z) - f (z0 ) - f (z0 )(z - z0 )| |z - z0 | for all z with |z - z0 | < r. Now (n ) 0 implies that there exists N such that N B(z0 , r). This implies that |z - z0 | < (N ) = 21 L for all z N . By N Corollary 2.9 we know that f (z) dz =
N N
f (z) - f (z0 ) - f (z0 )(z - z0 ) dz.
This implies that f (z) dz
N
2-N L
2-N L =
2-N
2
L2 .
From the inequality 2.2 it follows that |I| L2 for all > 0 and thus I = 0. This completes the proof in case p . Assume next that p is a vertex of the /
c
y a=p x b
Figure 2. The case a = p triangle (a, b, c), say p = a. Then pick x [a, b] and y [a, c]. Then by the above f (z) dz = (y,b,c) f (z) dz = 0 and thus (x,b,y) f (z) dz =
(a,x,y)
f (z) dz 0
as x, y a, since ((a, x, y)) 0 and f is bounded on (a, x, y). Hence f (z) dz = 0 also in the case that p is a vertex of . It remains the case that p \ {a, b, c}. In that case apply the above to the triangles (a, b, p), (b, c, p) and (c, a, p) to get the desired result.
20
2. INTEGRATION OVER CONTOURS
Definition 2.11. A set S C is called starlike if the exists a S such that the line segment [a, z] S for all z S. The point a is called a star center of S in this case. Recall that a set S C is called convex if for z1 , z2 S we have that [z1 , z2 ] S, i.e., a convex set is a starlike set such that every point of S is a star center of S. Theorem 2.12. (Cauchy's Theorem for starlike sets) Let G C be an open starlike set. Let p G and f : G C such that f is continuous on G and holomorphic on G \ {p}. Then f = F for some holomorphic F on G. In particular (2.3)
f (z) dz = 0
for every closed contour in G. Proof. Let a G be a star center of G. Then the line segment [a, z] G for all z G. Now define F (z) =
[a,z]
f (w) dw.
Let z0 G. Then there exists r > 0 such that B(z0 , r) G. Now for any z B(z0 , r) the triangle (a, z0 , z) G, so by Theorem 2.10 we have f (w) dw = 0,
(a,z0 ,z)
and thus F (z) - F (z0 ) =
[a,z]
f (w) dw -
[a,z0 ]
f (w) dw =
[z0 ,z] [z0 ,z]
f (w) dw.
Fixing z0 we get for all z = z0 in G, since 1 F (z) - F (z0 ) - f (z0 ) = z - z0 z - z0 = 1 |z - z0 |
w[z0 ,z]
1 dw = z - z0 , that
f (w) - f (z0 ) dw
[z0 ,z]
sup |f (w) - f (z0 )| |z - z0 |
w[z0 ,z]
sup |f (w) - f (z0 )| 0
as z z0 , by the continuity of f at z0 . This proves that f (z0 ) = F (z0 ) for all z0 G and thus F is holomorphic on G. Now equation 2.3 follows from Theorem 2.8. Definition 2.13. Let be a closed piecewise smooth curve in C and let a G = C \ . Then dz 1 Ind (a) = 2i z - a is called the index of with respect to a or winding number of around a.
2. CONTOUR INTEGRALS
21
Theorem 2.14. (Cauchy's Integral Formula for starlike sets) Let G C be an open starlike set and let be a closed contour in G. Let f be holomorphic on G and z0 G \ . Then f (z0 ) Ind (z0 ) = Proof. Let z G \ and define g(z) =
f (z)-f (z0 ) z-z0
1 2i
f dz. (z) z - z0
f (z0 )
if z G \ {z0 } if z = z0
Then g satisfies the hypotheses of Theorem 2.12, so 1 g(z) dz = 0. 2i Hence 1 2i
f (z) 1 dz = z - z0 2i
f (z0 ) dz z - z0 1 1 = f (z0 ) dz 2i z - z0 = f (z0 ) Ind (z0 ),
and thus the proof of the theorem is complete. Remark 2.15. The above theorem is used most often for the case that Ind (a) = 1. We will see e.g. that Ind (z0 ) = 1, when is a circle containing z0 , traversed counter clockwise once. Theorem 2.16. (Fundamental Theorem of Algebra) Let p(z) be a polynomial of degree m 1. Then p has exactly m zeros in C, counting each zero according to its multiplicity.
1 Proof. Assume p(z) = 0 for all z C. Then f (z) = p(z) is an entire function. We can assume that p(z) = z m + + a1 z + a0 . Now a0 a1 |p(z)| = |z|m 1 + + m-1 + m z z |a1 | |a0 | 1 1 |z|m |1 - - m-1 - m |z|m Rm |z| |z| 2 2
for |z| R for R large enough. Now applying Cauchy's Integral formula to f (z) and R = Reit with 0 t 2, we get
R
f (z) 2i dz = 2if (0) = = 0, z p(0)
while f (z) 1 2 dz 2 max | | 2 m 0, z R |z|=R p(z) R as R , which is a contradiction. Hence there exists z1 C such that p(z1 ) = 0. Now factor p(z) = (z - z1 )p1 (z) and repeat the above argument.
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2. INTEGRATION OVER CONTOURS
To apply the Cauchy's Integral formula, we need to be able to compute the index of a curve. We will derive a number of properties of the index, which will facilitate this. Proposition 2.17. Let be a closed contour and let G = C\ . Then Ind (a) is an integer for all a G. Proof. Let : [b, c] C be piecewise smooth such that (b) = (c). Then 1 2i Let g(t) =
b
dz 1 = z-a 2i
t
c b
(s) ds. (s) - a
(s) ds. (s) - a
Then g(b) = 0 and g (t) =
(t) (t)-a ,
except possibly on the finite set S where is not
differentiable. Now e-g(t) ((t) - a) is a continuous function such that d -g(t) e ((t) - a) = e-g(t) (t) - g (t)e-g(t) ((t) - a) dt = e-g(t) { (t) - g (t)((t) - a)} = 0, except on the finite set S. This implies that e-g(t) ((t) - a) is constant on [b, c]. Evaluating this function at t = b and t = c gives then e-g(b) ((b) - a) = (b) - a = e-g(c) ((c) - a), which implies e-g(c) = 1, since (b) = (c). Hence g(c) = 2im for some integer 1 dz m, and thus 2i z-a = m which completes the proof of the theorem. By the above proposition the index of a closed contour is an integer m. Intuitively this integer measures how many times the contour winds around the point a and in what direction. From the properties of contour integrals we have immediately that the following properties hold. (1) Ind- (a) = -Ind (a). (2) If is obtained by joining the closed contours 1 and 2 , then Ind (a) = Ind1 (a) + Ind2 (a). We shall prove that the index of a closed contour depends continuously on the point a and that therefore the index is constant on each connected component of C \ . We recall first the relevant definitions. Let S C. Then S1 is called a (connected) component of S, if S1 is a maximal connected subset of S. One can show that if S1 is a connected subset of S, then so is the relative closure of S1 . Hence connected components of a set S are always relatively closed. Proposition 2.18. Let G be an open set in C. Then every connected component of G is also open and thus G is a countable disjoint union of open and relatively closed components. Proof. Let C denote a component of G and let z0 C. Let > 0 such that B(z0 , ) G. Then CB(z0 , ) is a connected subset of G and thus C = CB(z0 , ), i.e., B(z0 , ) C. Hence C is open. In each component we can pick a different a + bi with a, b Q, so there are countably many components.
2. CONTOUR INTEGRALS
23
Remark 2.19. If G = C \ K, where K is a compact set, then G has exactly one unbounded component. In particular, when G = C \ for a closed contour , then G has one unbounded component. Theorem 2.20. Let be a closed contour and let G = C \ . Then Ind is constant on each component of G and Ind (a) = 0 for all a in the unbounded component of G. Proof. Define f (w) = Ind (w) for w G. We first show that f : G C is continuous. Let w G. Then r = dist(w, ) > 0, since is compact. Let > 0 2 r and then take 0 < < min{ 2 , r .}, where L = (). Then for |w1 - w| < we L have (w - w1 ) 1 dz . |f (w) - f (w1 )| = 2 (z - w)(z - w1 )
r For z we have |z - w| r and |z - w1 | |z - w| - |w - w1 | > 2 . Hence
L< . r2 It follows that f is continuous. If now C G is a component, then f (C) is a connected subset of C. On the other hand f (C) Z and thus f (C) consists of a single point. To see that Ind (a) = 0 for all a in the unbounded component of G, let R > 0 such that {z : |z| > R} is contained in the unbounded component of G. Then find a C with |a| > R such that |z - a| > L for all z . Then |f (w) - f (w1 )| < 1 1 L= , 2 L 2 and thus Ind (a) = 0. As Ind (a) is constant on the unbounded component it follows that this holds for all a in the unbounded component of G. |Ind (a)| Example 2.21. (i) Let : [0, 2] C be defined by (t) = z0 + Reit . Then traces the circle |z - z0 | = R once counterclockwise. In this case Ind (a) = 1 for |a - z0 | < R and Ind (a) = 0 for |a - z0 | > R, since Ind (z0 ) = 0 and the component of G \ containing z0 equals |z - z0 | < R. (ii) Let : [0, 4] C be defined by (t) = z0 + Re-it . Then traces the circle |z - z0 | = R twice clockwise. In this case Ind (a) = -2 for |a - z0 | < R and Ind (a) = 0 for |a - z0 | > R. The following proposition provides the index for practically every curve encountered in applications. Proposition 2.22. Let : [a, b] C a closed curve. Assume there exists z0 C \ , t0 (a, b) and > 0 so that the rays Rt = {z0 + s((t) - z0 ) : s 0} have the following properties. (1) Rt = {(t)} for all t (t0 - , t0 + ) (2) The part of Rt with s > 1 lies in the unbounded component of C \ and the part with 0 < s < 1 lies in a bounded component of C \ . (3) traces {(t) : t (t0 - , t0 + )} once counter clockwise. Then Ind (z0 ) = 1.
24
2. INTEGRATION OVER CONTOURS
Proof. Let f (z) = log |z - z0 | + i arg (z - z0 ) be a branch of log(z - z0 ) with domain C \ Rt0 . Denote by the part of the curve in C \ Rt0 with initial point (t0 + ) and terminal point (t0 - ). Then
1 dz = f ((t0 - )) - f ((t0 + )) 2i z - z0 1 dz z - z0 1 dz z - z0
as
0. On the other hand
as
0 and thus Ind (z0 ) = 1.
Theorem 2.23. (Power series expansion of holomorphic functions) Let G C and let f be holomorphic on G. Then for all a G and all R > 0 such that B(a, R) G there exists (unique) cn such that
f (z) =
n=0
cn (z - a)n
for all z B(a, R). Proof. Let 0 < r < R and define : [0, 2] B(a, R) by (t) = a + reit . Then Ind (z) = 1 for all z B(a, r). Hence by the Cauchy's Integral formula (applied to the open set B(a, R)) we have f (z) = Now series
z-a -a
1 2i
f () d -z
=
|z-a| r
< 1 for all z B(a, r) and all . Hence the geometric
(z - a)n 1 = n+1 ( - a) -a n=0 1 f (z) = 2i
1 1 - z-a -a
=
1 -z
converges uniformly in on for each z B(a, r). Hence f ()
(z - a)n d ( - a)n+1 n=0 f () d (z - a)n ( - a)n+1
=
n=0
1 2i
=
n=0
cn (z - a)n
where cn = 1 2i
f () d. ( - a)n+1
The uniqueness follows from Corollary 4.8 in Chapter1, where it was shown that (n) cn = f n!(a) .
2. CONTOUR INTEGRALS
25
Corollary 2.24. Let G C be an open set and assume f : G C is holomorphic. Then f is holomorphic on G and thus f (n) exists for all n 1 on G. Moreover, if B(a, R) G and |f (z)| M on B(a, R), then n!M |f (n) (a)| n (Cauchy Estimates). R Proof. The fact that f is holomorphic on G follows immediately from the above theorem and Theorem 4.7. From Corollary 4.8 we get f () n! d, f (n) (a) = 2i ( - a)n+1 where (t) = a + reit , 0 t 2, 0 < r < R and thus n! M n!M |f (n) (a)| 2r n+1 = n . 2 r r As this holds for all 0 < r < R the proof is complete. Theorem 2.25. (Morera's Theorem) Let G C be an open set and f : G C a continuous function such that f (z) dz = 0
for all triangles G. Then f is holomorphic on G. Proof. Let B(a, R) G for a G. Then as in the proof of Theorem 2.12 we can find F holomorphic on B(a, R) such that F = f on B(a, R). From the above corollary we now conclude that f is holomorphic on B(a, R). As this holds for all B(a, R) G we conclude that f is holomorphic on G. Theorem 2.26. (Liouville's Theorem) Let f be an entire function. Assume that f is bounded on C. Then f is constant. Proof. Let f (z) = n=0 an z n be the power series expansion around z = 0. Since f is entire, this series has radius of convergence equal to . Let M be such that |f (z)| M for all z C. Then for all R > 0 we have for n 1 that |f (n) (0)| n!M 0 as R . Hence f (n) (0) = 0 for all n 1, and thus also Rn an = 0 for all n 1. Therefore f (z) = a0 for all z C.
CHAPTER 3
Zeros and singularities of holomorphic functions
1. Zeros of holomorphic functions A subset G C is called a region if G is open and connected. If f : G G, then z0 G is called a zero of f if f (z0 ) = 0. Theorem 1.1. Let f be a holomorphic function on a region G. Then either every zero of f is isolated or f is identically zero on G. For each isolated zero a G there exists a unique m N such that f (z) = (z - a)m g(z), where g is holomorphic on G and g(a) = 0. Moreover if f is not identically zero on G, then f has countably many zeros in G. Proof. Let a G such that f (a) = 0. Then there exists r > 0 such that B(a, r) G. Then by Theorem 2.23 we can expand f in a power series
f (z) =
n=0
cn (z - a)n
on B(a, r). Note that c0 = 0 as f (a) = 0. There are now two possibilities: either cn = 0 for all n in which case f vanishes identically on B(a, r) and thus on G, or there exists a smallest m 1 such that cm = 0. In the latter case we define g(z) =
f (z) (z-a)m
if z G \ {a} if z = a
cm
Then clearly f (z) = g(z)(z - a)m for all z G and thus g is holomorphic on G \ {a}. But the power series of f shows that g has a power series expansion on B(a, r) and is thus holomorphic on B(a, r). This shows that g is holomorphic on G and g(a) = cm = 0. Since g is continuous it follows that g(z) = 0 in a open disk around a and thus a is an isolated zero of f in this case. We have thus shown that a zero of f is either isolated or f is identically zero on a disk around the zero. We show next that if f has a non-isolated zero, then f is identically zero on G. Assume f has a non isolated zero on G. Then the interior U of f -1 (0) = {z G : f (z) = 0 is non-empty. We next observe that U is relatively closed in G. Let zn U such that zn z0 . Then continuity of f implies that also f (z0 ) = 0. Since the zn 's are non-isolated zeros, we can assume that zn = zm . Then z0 is a non-isolated zero of f and thus z0 U . It follows that U is open and closed in G and thus U = G by connectedness of G. It remains to show that f -1 (0) is countable, in case f is not identically zero on G. In this case every a f -1 (0) is isolated, so for every a f -1 (0) there exists r > 0 such that B(a, r) f -1 (0) = {a}. Hence there exists a1 Q + iQ and ra > 0 such that a Da = B(a1 , ra ) B(a, r). The collection {Da : a f -1 (0)} is countable and if a, b f -1 (0) with a = b, then
27
28
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
Da = Db . Hence the mapping a Da is a one-to-one mapping and thus f -1 (0) is countable. Remark 1.2. The number m associated with the zero a, as in the above theorem, is called the order of the zero a. Corollary 1.3. Let f and g be a holomorphic functions on a region G and assume there exists a subset S G with limit point in G such that f (z) = g(z) on S. Then f (z) = g(z) for all z G. Proof. Let h = f - g. Then h has a non-isolated zero in G (namely the limit point of S) and thus by the above theorem h is identically zero on G. Example 1.4. 1 1 (i) Let f be an entire function such that f ( n ) = sin( n ) for all n N. Then by the above corollary f (z) = sin z for all z C. 1 (ii) Let f be a holomorphic fuction on C \ {0} such that f (z) = sin( z ) for 1 1 all z = n , n = 1, 2, , i.e., f ( n ) = 0 for all n 1. It does not follow 1 in this case that f (z) = sin z for all z = 0, since f (z) 0 also satisfies 1 f ( n ) = 0. Theorem 1.5. (Maximum Modulus Theorem) Let G C be open and connected and f : G G holomorphic. Assume |f | attains a maximum at a point a G, i.e., |f (z)| |f (a)| for all z G. Then f is constant on G. Proof. Let a G such that |f (z)| |f (a)| for all z G. Then there exists R > 0 such that B(a, R) G. Take 0 < r < R and let (t) = a+reit for 0 t 2. Then Cauchy's Integral Formula 2.14, applied to B(a, R), gives f (a) = 1 2i f (z) dz z-a
2 0 2 0 1 2 2 0
1 = 2i = Hence |f (a)| that
1 2 2 0
f (a + reit ) it rie dt reit
1 2
f (a + reit ) dt. |f (a)| dt = |f (a)|. It follows
|f (a + reit )| dt
2
|f (a)| - |f (a + reit )| dt = 0.
0
The integrand is non-negative and continuous, so it must be identically zero. Hence |f (a)| = |f (a + reit )| for all 0 < r < R and it follows that |f | is constant on B(a, R). It is now an an exercise to show, using the Cauchy-Riemann equations, that f is constant on B(a, R). Hence f (z)-f (a) = 0 on B(a, R) and from the connectedness of G it follows that f (z) - f (a) = 0 on G. Remark 1.6. If G is a bounded region and f is holomorphic on G and continuous on G, then by the above theorem |f | must attains its maximum on the boundary G of G.
2. SINGULARITIES OF HOLOMORPHIC FUNCTIONS
29
A corollary of the maximum modulus theorem in the form of the remark is the Minimum Modulus Theorem. Corollary 1.7. (Minimum Modulus Theorem) Let G C be a bounded region and let f : G C a non-constant continuous function which is holomorphic on G. If there exists z0 G such that |f (z0 )| inf{|f (z)| : z G}, then f has a zero on G.
1 Proof. Assume f has no zero on G. Then g = f is holomorphic on G and has an interior maximum on G. Hence g is constant, which contradicts that f is non-constant.
We now note that if G is region and f : G C is non-constant, then for a C there exists a closed disk B(a, r) such that f (z) = f (a) for all z B(a, r) \ {a}. This follows from the fact that z = a is an isolated zero of f (z) - f (a). Combined withe minimum modulus theorem we can use this observation to prove the Open Mapping Theorem. Recall first that a mapping f from a metroic space X into a metric space Y is called open, if f (U ) is open in Y for all open U X. Theorem 1.8. (Open Mapping Theorem) Let f be a non-constant holomorphic function on a region G. Then f is an open mapping Proof. Let U G be open and a U . We need to prove that f (a) is an interior point of f (U ). By the above remark there exists r > 0 such that B(a, r) U and f (z) = f (a) for all z with |z - a| = r. This implies that = 1 2 min{z:|z-a|=r} |f (z) - f (a)| > 0. We claim that B(f (a), ) f (U ). To see this let w B(f (a), ). Then |f (a) - w| < . For z with |z - a| = r we have that |f (z) - w| |f (z) - f (a)| - |f (a) - w| 2 - = . This implies that |f (a) - w| < min{z:|z-a|=r} |f (z) - w|. By the minimum modulus theorem there exists z B(a, r) such that f (z) - w = 0. This shows B(f (a), ) f (U ) and the proof is complete.
2. Singularities of holomorphic functions Let G C be an open set and let a G. Assume f is holomorphic on G \ {a}, then we say that f has an isolated singularity at a. If we can define f (a) in such a way that f becomes differentiable at a, then a is called a removable singularity of f. Theorem 2.1. Let G C be an open set and let a G. Assume f is holomorphic on G \ {a} and that f is bounded on B(a, r) \ {a} for some r > 0. Then f has a removable singularity at z = a. Proof. Define h(x) = (z - a)2 f (z) z = a, z G 0 z = a.
30
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
Then f bounded on B(a, r) \ {a} implies that h (a) = 0. Hence h is holomorphic on G. It follows that h has a power series expansion
h(z) =
n=2
cn (z - a)n
for all z B(a, r). Now define f (a) = c2 , then f has the power series expansion
f (z) =
n=0
cn+2 (z - a)n
for all z B(a, r), which implies that f is holomorphic on B(a, r). Thus f is holomorphic on G. Remark 2.2. Note that if f is holomorphic on G\{a} and if limza f (z) exists in C, then f is bounded on B(a, r) \ {a} for some r > 0 and thus f has a removable singularity at z = a in that case. In particular, if f is holomorphic on G \ {a} and if f is continuous at a, then f is holomorphic on G. Conversely, if f has a removable singularity at z = a, then limza f (z) exists in C. Example 2.3. Let f (z) = power series of sin z that
sin z z
on C \ {0}. For z = 0 we find by using the
z2 z4 + - . 3! 5! Now the series on the right hand side converges for all z C to a holomorphic function g, which agrees with f on C \ {0}. Thus 0 is a removable singularity of f and by defining f (0) = g(0) = 1 we extend f to an entire function. f (z) = 1 - An isolated singularity a of f is called a pole of f if limza |f (z)| = . An isolated singularity a of f which is neither a removable singularity or a a pole is called an essential singularity of f . We first characterize poles. Theorem 2.4. Let G C be an open set and let a G. Assume f is holomorphic on G \ {a}. Then the following are equivalent. (i) f has a pole at a. (ii) There exist a unique m N and a holomorphic function g on G with g(a) = 0 such that g(z) f (z) = (z - a)m for all z G \ {a}. (iii) There exist a unique m N and c-1 , c-2 , , c-m C with c-m = 0 such that m c-k f (z) - (z - a)k
k=1
has a removable singularity at a. Proof. Assume first that (i) holds, i.e., f has a pole at a. Then limza |f (z)| = implies that there exists r > 0 such that f (z) = 0 on B(a, r) \ {a}. Then define 1 h(z) = f (z) for z B(a, r) \ {a}. Then h is holomorphic on B(a, r) \ {a} and limza h(z) = 0, so a is a removable singularity of h and by defining h(a) = 0
2. SINGULARITIES OF HOLOMORPHIC FUNCTIONS
31
we have a holomorphic function on B(a, r) with its only zero zero at a. Hence by Theorem 1.1 we know that there exist m N and a holomorphic function g1 on B(a, r) with g1 (z) = 0 such that h(z) = (z - a)m g1 (z), so (ii) holds on B(a, r) with g = g11 . Now (z - a)m f (z) is holomorphic on G \ {a} and agrees with g on B(a, r), so we can extend g to a holomorphic function on G so that (ii) holds. If (ii) holds then there exists r > 0 such that g(z) = n=0 an (z - a)n for all z B(a, r) and g(a) = a0 = 0. Now
m
f (z) -
k=1
am-k = (z - a)k
ak+m (z - a)k
k=0
for all z B(a, r) \ {a}, which shows that (iii) holds if we take c-k = am-k for k = 1, , m. If (iii) holds, then (z - a)m f (z) defines a holomorphic function g with g(a) = 0 on an open disk B(a, r) for some r > 0. Hence
za
lim |f (z)| = lim
za
|g(z)| = , |z - a|m
which completes the proof of the theorem. Remark 2.5. If f has a pole at a, then the number m as in the above theorem m c-k is called the order of the pole and k=1 (z-a)k is called the principal part of f at the pole a. Note also that (iii) above implies that if f has a pole of order m at a, then there exist r > 0 and c-m , , c-1 , c0 , c1 , such that we have f (z) = for all z B(a, r) \ {a}. We now present a limit characterization of essential singularities. Theorem 2.6. (Casorati-Weierstrass Theorem) Let G C be an open set and let a G. Assume f is holomorphic on G \ {a}. Then the following are equivalent. (i) f has an essential singularity at a. (ii) If r > 0 such that B(a, r) G, then f (B(a, r)) is dense in C, i.e., for all w C there exist zn G \ {a} with zn a such that limn f (zn ) = w. (iii) There exist zn a and zn a in G \ {a} such that limn f (zn ) and limn f (zn ) exist, but are unequal. Proof. Assume a is an essential singularity of f . If (ii) does not hold, then there exist a w C such that 1 g(z) = f (z) - w is bounded in a neighborhood of a. Hence g has a removable singularity at a. This implies that f has either a removable singularity at a (in case g(a) = 0) or a pole at a, which contradicts our assumption. hence (ii) holds. Clearly (ii) implies (iii). If (iii) holds, then a can not be a removable singularity of f by Remark 2.2 and a can not a pole either, so it must be an essential singularity of f . Example 2.7. Let f (z) = e z on G = C \ {0}. Then z = 0 is an essential 1 1 singularity of f , since limn f (- n ) = 0 and limn f ( 2ni ) = 1.
1
c-m c-1 + + + (z - a)m (z - a)
ck (z - a)k
k=0
32
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
Theorem 2.8. (Laurent Series expansion) Let G C be an open set and let a G. Assume f is holomorphic on G \ {a}. Then there exists R > 0 and cn (n = 0, 1, 2, . . .) such that for all z B(a, R) \ {a} we have
f (z) =
k=-
ck (z - a)k ,
f () 1 d 2i r ( - a)k+1 and r (t) = a + reit , 0 t 2 with 0 < r < R. Moreover the series converges uniformly on any annulus 0 < r1 |z - a| r2 < R. ck = Proof. Let R > 0 such that B(a, R) G and let 0 < r1 < |z - a| < r2 < R. Define rk (t) = a + rk eit , 0 t 2 for k = 1, 2. Now write r1 r2 as a join of two curves, each one lying in a starlike open set contained in 0 < |z - a| < R and such that z is inside exactly one of the two curves (see figure 1 below). Than g() = f () is holomorphic inside the other curve. We see by Theorems 2.12 and -z
where
Figure 1. -r1 r2 as a join of two curves. 2.14 that (3.1)
r2
f () d - -z
r1
f () d = 2if (z). -z
Now we have 1 1 1 = = -z ( - a) - (z - a) - a n=0
for r2 for all |z - a| < r2 = | - a| and
z-a -a
n
1 1 1 = =- -z ( - a) - (z - a) z - a n=0
-a z-a
n
3. THE RESIDUE THEOREM AND APPLICATIONS for r1 for all |z - a| > r1 = | - a|. Inserting these expansions in 3.1 we get -1
33
(3.2) where
f (z) =
k=0
ak (z - a)k +
k=-
bk (z - a)k ,
ak = and bk =
1 2i 1 2i
r2
f () d ( - a)k+1 f () d. ( - a)k+1
r1
1 The two series for -z converge uniformly in z - a for r1 |z - a| r2 , where r1 < r1 and r2 < r2 . Thus the series (3.2) converges uniformly on r1 |z - a| r2 . Similarly to how we established the equation (3.1), we can see that f () ck = r (-a)k+1 d does not depend on r, so that ck = ak for k 0 and ck = bk for k -1, which completes the proof of the theorem.
The following corollary follows now immediately from the previous characterizations of removable singularities and poles. Corollary 2.9. Let G C be an open set and let a G. Assume f is k holomorphic on G \ {a}. Let f (z) = k=- ck (z - a) be the Laurent series expansion of f around a. Then the following hold. (i) f has a removable singularity at z = a if and only if ck = 0 for all k -1. (ii) f has a pole at z = a of order m if and only if ck = 0 for all k -(m + 1) and c-m = 0. (iii) f has an essential singularity at a if and only ck = 0 for infinitely many k < 0. Remark 2.10. Let f be holomorphic on G \ {a} and let
f (z) =
k=-
ck (z - a)k
be the Laurent series expansion of f around a. Then the coefficient c-1 is called the residue of f at a and denoted by Res(f, a). Its importance derives from the fact that if (t) = a + reit with 0 t 2 is a curve in G, then f (z) dz = 2iRes(f, a). This follows immediately from the uniform convergence of the series, which allows us to integrate the series term by term. In case f has a pole of order m at a, we can compute Res(f, a) without using the Laurent series as follows: Res(f, a) = lim dm-1 1 [(z - a)m f (z)]. za (m - 1)! dz m-1
3. The Residue Theorem and Applications We start with an application of the Laurent expansion.
34
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
Theorem 3.1. (Residue Theorem) Let G be a starlike region. Let p1 , , pn in G and f : G \ {p1 , , pn } C be holomorphic. Let be a piecewise smooth closed curve in G \ {p1 , , pn }. Then
n
f (z) dz = 2i
[Res(f, pk )]Ind (pk ).
k=1
Proof. For each pk there exists Rk > 0 such that B(pk , Rk ) G. Then for z B(pk , Rk ) \ {pk } we have a Laurent expansion f (z) = n=- cn (z - pk )n . -1 Denote by Sk (z) the singular part n=- cn (z - pk )n . Then there exist > 0 such that Sk converges uniformly on |z - pk | and such that is a subset of each |z - pk | . In particular each Sk is holomorphic on C \ {pk }. Define g(z) = n f (z)- k=1 Sk (z) on G\{p1 , , pn }. Clearly g is holomorphic on G\{p1 , , pn }. We claim that each pk is a removable singularity of g. To see this, note that on B(pk , Rk ) we have that
n
g(z) = -
j=k
Sj (z) +
m=0
cm (z - pk )m .
Both terms on the right hand side are holomorphic on B(pk , Rk ) so that z = pk is a removable singularity for g. Hence we can extend g to a holomorphic function on G. It follows now from Cauchy's theorem 2.12 that f (z) dz = 0. Hence
n
f (z) dz =
k=1
Sk (z) dz.
Now each Sk converges uniformly on , so that
-1
Sk (z) dz =
m=-
cm
(z - pk )m dz = 2i[Res(f, pk )]Ind (pk ),
from which the conclusion follows. To apply the above theorem to the evaluation of improper teal integrals, we first recall some definitions. Assume f : R R is function which is Riemann integrable over [-R1 , R2 ] for all R1 , R2 > 0 (this holds e.g. when f is continuous). Then f 0 is (improper) Riemann integrable over R if both limits limR1 -R1 f (x) dx and limR2 0 2 f (x) dx exist and - f (x) dx is by definition the sum of these two limits. One can also define the Cauchy principal value integral of f by
R R
(P V )
-
f (x) dx = lim
R
f (x) dx.
-R
It is easy to see that if f is improper Riemann integrable, then the Cauchy principal value of the integral of f exists and equals - f (x) dx, but that the converse is false in general (take e.g. f (x) = x). There are two cases, where the two integrals coincide. The first case is when f is an even function, i.e. f (-x) = f (x) for all x. In this case we have R 1 R f (x) dx = f (x) dx. 2 -R 0
3. THE RESIDUE THEOREM AND APPLICATIONS
35
The other case is when the integral of |f (x)| has a finite Cauchy principal value. We present now some examples of applications of the residue theorem.
1 Example 3.2. Let f (x) = (1+x2 )2 . We want to compute first that f is an even function, so that by the above 0
f (x) dx. Observe
f (x) dx =
0
1 lim 2 R
R
f (x) dx.
-R
1 First extend f to C \ {i, -i} by f (z) = (1+z2 )2 . Now define R (t) = Reit for 0 t and define the closed curve CR = R [-R, R]. Then for R > 1 we have IndCR (i) = 1 and IndCR (-i) = 0. By the residue theorem we have for R > 1 that
f (z) dz = 2i Res(f (z), i).
CR 1 d Now z = i is a pole of order 2, so Res(f (z), i) = dz ( (z+i)2 )|z=i = that for R > 1 1 = . f (z) dz = 2i 4i 2 CR 1 4i .
It follows
Now
CR
f (z) dz =
R
f (z) dz +
R -R
f (x) dx and R 1 0 |(1 + z 2 )2 | (R2 - 1)2
R
R
f (z) dz R max
zR
as R . It follows that 1 1 dx = lim (1 + x2 )2 2 R 0
f (x) dx =
-R 1 0 1+x4
1 = . 2 2 4 dx. We first observe that
Example 3.3. Suppose we want to compute 1 f (x) = 1+x4 is even, so that
0
1 1 dx = lim 4 1+x 2 R
R
f (x) dx.
-R
1 Let f (z) = 1+z4 . Then f is holomorphic on C \ {z1 , z2 , z3 , z4 }, where zj are the i i i solutions of z 4 = -1, i.e., z1 = e 4 = 1 2 + 2 2, z2 = - 1 2 + 2 2, z3 = -z1 , 2 2 and z4 = -z2 . Let R = R [-R, R], where R (t) = Reit with 0 t . Then for R > 1 we have
f (z) dz = 2i (Res(f, z1 ) + Res(f, z2 )) .
R
As the pole at z1 is simple we compute the residue by 1 1 Res(f, z1 ) = lim (z - z1 )f (z) = = (- 2 - i 2). zz1 (z1 - z2 )(z1 - z3 )(z1 - z4 ) 8 2 1 Similarly Res(f, z2 ) = 8 ( 2 - i 2). Thus R f (z) dz = 2 . Now f (z) dz R max |f (z)|
R |z|=R
R 0 R4 - 1
as R . Hence
R
R
lim
f (x) dx =
-R
2 , 2
36
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
which implies that
0
1 2 dx = . 1 + x4 4
We now apply the residue theorem to location and counting of zeros and poles of a holomorphic function. Theorem 3.4. (Principle of the Argument) Let G C be a starlike region and a closed contour in G. Let f be holomorphic on G, except for poles of order lk at z = pk G \ (1 k m). Assume f has zeros of order mj at z = aj G \ (1 j n). Then Ind1 (0) = =
j=1
1 2i
n
f (z) dz f (z)
m
mj Ind (aj ) -
k=1
lk Ind (pk )
where 1 = f .
(z) Proof. Let h(z) = f (z) . Then h is holomorphic at all z G where f (z) = 0 f and has a pole at the zeros of f . If z = aj is a zero of order mj , then f (z) = (z - aj )mj g(z), where g is holomorphic on G \ {pi , , pm } and g(a) = 0. Then mj (z) there exists r > 0 such that h(z) = f (z) = z-aj + g (z) on B(a, r)\{a}, where g (z) is f g(z) g(z) holomorphic on B(a, r). Hence h has a simple pole at z = aj and Res(h, aj ) = mj . Similarly at the point z = pk we can write f (z) = (z - pk )-lk g(z), where g is holomorphic near pk and g(pk ) = 0. As above this implies that h has a simple pole at z = pk and Res(h, pk ) = -lk . Hence we have by the residue theorem that
1 2i
f (z) dz = f (z)
n
m
mj Ind (aj ) -
j=1 k=1
lk Ind (pk ).
Let : [a, b] C. Then Ind1 (0) = = 1 2i 1 2i 1 1 dz = z 2i
b a
1 b a
1 (s) ds 1 (s) f (z) dz. f (z)
f ((s)) 1 (s) ds = f ((s)) 2i
Theorem 3.5. (General Rouch's Theorem) Let G C be a starlike region and e a closed contour in G. Let f, g be holomorphic on G, except for poles of order lk at z = pk G \ (1 k m) for f and poles of order ni at z = qi G \ for g (1 j r). Let f have zeros of order mj at z = aj G (1 j n) and g have zeros of order sj at z = bj G (1 j t). Assume |f (z) + g(z)| < |f (z)| + |g(z)| on . Then
n m t r
mj Ind (aj ) -
j=1 k=1
lk Ind (pk ) =
j=1
sj Ind (bj ) -
k=1
lk Ind (qk ).
4. THE GLOBAL CAUCHY THEOREM
37
Proof. Observe first that the strict inequality |f (z) + g(z)| < |f (z)| + |g(z)| implies that f and g can't have zeros on . By hypothesis f (z) f (z) +1 +1 < g(z) g(z) on . Observe that the strict inequality prevents that
f g f (z) g(z)
is a non-negative real
number for z . Hence maps into C \ [0, ). Let 1 = ( f ) , then 0 is in g the unbounded component of C \ 1 . Hence Ind1 (0) = 0. Hence by the lefthand part of the equality in the above theorem we have 0= 1 2i f f 1 ( ) ( )-1 dz = g g 2i (
f g 1 - ) dz = f g 2i
f 1 dz - f 2i
g dz, g
and the conclusion now follows from the righthand part of the equality in the above theorem. Remark 3.6. Note most often the above theorems are applied to curves with C \ having exactly two components, namely the "exterior" of where Ind = 0 (z) 1 and the "interior" with Ind = 1. In that case 2i 1 f (z) dz = Nf - Pf , where f Nf is the number of zeros of f inside (counting with their orders) and Pf is the number of poles of f inside (also counted according to their order). With this notation and with these hypotheses we have then as conclusion in the above theorem that Nf - Pf = Ng - Pg . The following corollary is the classical Rouch's Theorem, which has a slightly e stronger hypothesis than the above theorem. Corollary 3.7. (Rouch's Theorem) Assume f and g are holomorphic in a e neighborhood of B(a, R). Assume also that |f (z)+g(z)| < |f (z)| on {z : |z-a| = R}. Let Nf , Ng denote the number of zeros of f , respectively g inside (with orders). Then Nf = Ng . Example 3.8. Let g(z) = z 5 -12z 3 +14. Let first R = 1. Then with the choice of f (z) = -14 we get |g(z) + f (z)| |z|5 + 12|z|2 = 13 < |f (z)| on |z| = 1. Hence g has the same number of zeros on B(0, 1) as f , namely zero. Now take R = 2. Now take f (z) = 12z 3 . Then |g(z) + f (z)| 25 + 14 = 46 < 96 = |f (z)| on |z| = 2. Hence g has three zeros inside |z| = 2, since f has three zeros inside |z| = 2. For R = 4 we can take f (z) = -z 5 and see that all five zeros of g lie in the disk B(0, 4). 4. The Global Cauchy Theorem The goal of this section is to obtain versions of Theorems 2.12 and 2.14 of chapter 2 without starlike assumptions. Proposition 4.1. Let G C be an open set, f be a holomorphic function on G, and let g : G G C be defined by g(, z) =
f ()-f (z) -z
f (z)
=z = z.
Then for fixed G, the function g(, ) is holomorphic on G and if is a contour in G, the function h defined by h(z) = g(, z) d is holomorphic on G.
38
3. ZEROS AND SINGULARITIES OF HOLOMORPHIC FUNCTIONS
Proof. For fixed G, the function g(, z) is holomorphic on G \ {} and continuous at , and thus by Remark 2.2 we see that g(, z) is holomorphic on G. We claim now that g(, z) is continuous on G G. To prove continuity of g we only need to consider points (a, a) G G. Let a G and > 0. Then there exists r > 0 such that B(a : r) G and |f (w) - f (a)| < for all w B(a, r). Let now , z B(a, r). Then f () - f (z) - f (a)( - z) =
[z,]
f (w) - f (a) dw
implies that |g(, z)-g(a, a)| maxw[z.] |f (w)-f (a)| < . Hence g is continuous at (a, a) for all a G. Next we prove that h(z) = g(, z) d is continuous on G, let zn z in G. Then g(, zn ) g(, z) uniformly on . This implies that h(zn ) h(z). Now let G be a triangle. Then h(z) dz =
g(, z) dz d = 0.
Hence h is holomorphic on G by Morera's Theorem. Theorem 4.2. (Global Cauchy's Theorem) Let G C be an open set and let f be a holomorphic function on G. If 1 , . . . , m are closed contours in G such that Ind1 (z) + + Indm (z) = 0 for all z C \ G, then for all z G \ m k we have k=1 Cauchy's Integral Formula
m
f (z)
k=1
Indk (z) =
1 2i
m k
k=1
f () d -z
and Cauchy's Theorem
m
f () d = 0.
k=1 k
Proof. Let g be as in the previous proposition and let h be defined as h(z) = m k=1 k g(, z) d. Then h is holomorphic on G and Cauchy's Integral Formula is equivalent to proving that h(z) = 0 for all z G \ m k . Define H = {z k=1 m m C \ k=1 k : k=1 Indk (z) = 0}. Then H is open (since the index is a continuous integer valued function) and C = G H. Now define h1 on H by
1 2i
1 h1 (z) = 2i
m k
k=1
f () d. -z
If z G H, then h1 (z) = h(z) by the definition of h and H. As in the proof of Theorem 2.23 we can expand each integral k f () d in a power series -z around each point z H and thus h1 is holomorphic on H. Therefore we can extend h to an entire function by defining h = h1 on H. Now lim|z| |h(z)| m m 1 1 lim|z| 2 k=1 | k f () d| 2 k=1 (k ) lim|z| maxk |f ()| = 0. It -z |-z| follows from Liouville's Theorem that h(z) = 0 for all z. For z G \ m k we get k=1
m
f (z)
k=1
Indk (z) =
1 2i
m k
k=1
f (z) 1 d = -z 2i
m k
k=1
f () d. -z
4. THE GLOBAL CAUCHY THEOREM To prove Cauchy's Theorem, take a G \ m k . Then k=1 m m
39
f () d =
k=1 k k=1 k
f ()( - a) d = 2i -a
m
Indk (a) f (a)(a - a) = 0.
k=1
Corollary 4.3. Let G C be an open set and let f be a holomorphic function on G. If 1 , . . . , m and 1 , . . . , n are closed contours in G such that Ind1 (z) + + Indm (z) = Ind1 (z) + + Indn (z) for all z C \ G, then
m n
f () d =
k=1 k k=1 k
f () d.
Proof. Apply Cauchy's Theorem to the closed contours 1 , . . . , m , -1 , . . . , -n . Next we prove a theorem generalizing Theorem 3.1. Theorem 4.4. (General Residue Theorem) Let G C be an open set and let f be a holomorphic function on G except for a subset A of G of isolated singularities. If 1 , . . . , m are closed contours in G \ A such that Ind1 (z) + + Indm (z) = 0 for all z C \ G, then
m m
f () d = 2i
k=1 k aA m
Res(f, a)
k=1
Indk (a) .
Proof. Let B = {a A : k=1 Indk (a) = 0}. The unbounded component m of C \ k and G are both contained in the set {z C : k=1 Indk (z) = 0} and thus B is the intersection of A with a compact subset of G. This implies that B is finite, as every point of A is isolated. Let B = {a1 , . . . , an } and define m lj = k=1 Indk (aj ) for 1 j n. Then pick rj > 0 for j = 1, . . . , n such that B(aj , rj ) are mutually disjoint, none of them intersects any k , and are contained in G \ (A \ B). Then define j to be the boundary of B(aj , rj ) traversed lj times (clockwise when lj < 0). Let G1 = (G \ A) B. Then f is holomorphic on G1 \ B m n and k=1 Indk (z) = 0 on C \ G1 . We also have j=1 Indj (z) = 0 on C \ G1 as A \ B is outside each disk B(aj , rj ) for j = 1, . . . , n. For ai B we have n m j=1 Indj (ai ) = li = k=1 Indk (ai ). Hence by Corollary 4.3 applied to G1 \ B we have
m n
f () d =
k=1 k j=1 j n
f () d
= 2i
j=1
Res(f, aj ) lj
m
= 2i
aA
Res(f, a)
k=1
Indk (a) .