**Unformatted text preview: **MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS (1) State the Cauchy-Goursat theorem. (10 points). If a function f is analytic at all points of a domain
and its interior, then C f (z)dz = 0. Theorem. contour C D containing a simple closed (2) State Taylor's theorem, being careful to include the conditions on the function f (z), and
where the series converges. (10 points)
Let f be analytic on a domain D , and let z0 ∈ D . Then, for any r > 0 so that
B(z0 , r) := { z ∈ C| |z − z0 | < r} is contained in D, Theorem. the disk ∞ f (z) =
n=0 f (n) (z0 )
(z − z0 )n .
n! In particular, the series will converge on any such disk. (3) Let u(x, y) be real-valued harmonic function on a domain D. Prove that h(z) = ux − iuy is
analytic on D. (10 points)
Solution: If u is harmonic, then by denition uxx , uxy = uyx , and uyy exist, and uxx +
uyy = 0. If h(z) = ux − iuy := U + iV , then
Ux = uxx
= −uyy
= +Vy , and
Uy = uxy
= uyx
= −Vx , which are the Cauchy-Riemann equations for h(z), so h(z) is indeed analytic.
(4) Show that, if an analytic function f (z) has constant modulus |f (z)| on a domain D, then
f (z) is constant. (10 points)
Solution: By far the easiest way to prove this is by using the maximum modulus theorem.
This is sort of circular because this fact is used to prove the maximum modulus theorem.
However, I would count that as OK. That theorem states that the maximum of |f (z)| for
a non-constant function can only occur on the boundary of D, but if |f (z)| is constant
then its maximum occurs everywhere, so f must be constant.
The easiest direct proof is to note that, since f (z) is deemed to be analytic, then so
would be logα (f (z)) for some α, in some disk, except at any point where f (z) = 0. But
unless |f (z)| = 0, the fact that the modulus is constant would say that f (z) = 0 in D.
On the other hand, if |f (z)| = 0, then f is always 0 since |f (z)| is constant. Now,
g(z) := logα (f (z)) = ln (|f (z)|) + iargα (z), so that analytic function has constant real part. If g = u+iv , then u would be constant.
But then the Cauchy-Riemann equations say that vx = −uy = 0 and vy = ux = 0, so
v is also constant, so g has to be constant, which then means that f is constant. Date : May 3, 2013.
1 2 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS You can also work directly with the condition. If |f (z)| is constant, then, for now
f = u + iv , u2 + v 2 will be constant. So, using the Cauchy-Riemann equations,
∂
u2 + v 2
∂x
2uux + 2vvx
2uux − 2vuy , and
∂
u2 + v 2
∂y
2uuy + 2vvy
2uuy + 2vux
2vux + 2uuy . 0 =
=
=
0 =
=
=
= In matrix form,
2u −2v
2v 2u ux
uy = 0
0 , and that matrix has rank 2 unless its determinant is 0, u2 + v 2 = 0, so except, again,
where |f (z)| = 0, we would have u = 0, so u is constant. The same would hold for v .
But if |f (z)| = 0, since |f (z)| is constant, then again f ≡ 0 and we are done.
(5) Find the contour integral
z 2 − 3z dz,
C where C is the circle { z| |z − 1| < 1|}. (10 points)
Solution: Since the integrand is analytic everywhere, and C is a simple closed curve, the
Cauchy-Goursat theorem above says that
z 2 − 3z dz = 0.
C (6) Find the contour integral
C (z + 1)
dz
(z 2 + 1) over the unit circle |z| = 2, with the positive orientation. (10 points)
(z+1)
has two simple poles, at z = i and z = −i, with
Solution: Since f (z) =
(z 2 +1)
Resz=i f (z) =
=
= (z + 1)
(z − i)
(z 2 + 1)
(z + 1)
(z + i) z=i
i+1
2i z=i and
Resz=−i f (z) =
=
= (z + 1)
(z + i)
(z 2 + 1)
(z + 1)
(z − i)
−i + 1
,
−2i z=−i z=−i MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS 3 the Residue Theorem implies that
(z + 1)
dz = 2πi (Resz=i f (z) + Resz=−i f (z))
(z 2 + 1)
i+1 i−1
= 2πi
+
2i
2i
= 2πi. C (7) Find the following contour integral C ez dz
,
(z − 1)2 where C is the circle |z| = 2, with the positive orientation.
Solution: This uses the corollary of the CIF,
n!
2πi f (n) (z0 ) = (10 points) f (z)dz
(z − z0 )n+1 C with f (z) = ez and z0 = 1, which is inside the contour, and n = 1. So
e1 = ez dz
2 , or
C (z − 1)
ez dz
.
(z − 1)2 1
2πi 2πie =
C You can also show this using the Residue Theorem, since there is only one singularity
inside the contour, at z = 1. That is a pole of order 2, so C ez dz
(z − 1)2 = 2πi Resz=1
= 2πi ez
(z − 1)2 ez
(z − 1)2
(z − 1)2 z=1 z = 2πi (e )
= 2πie. z=1 (8) Find the Taylor series, centered at the point z0 indicated, for the following functions. Don't
worry about the radius of convergence. (10 points each)
(a) f (z) = z+2
, z0 = 1.
z−3 Solution: The easiest way to do this, as we have seen, is to rst divide out:
z+2
z−3 = 1+ 5
,
z−3 4 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS and then convert the second term into a geometric series centered at 1.
z+2
z−3 = 1+ 5
(z − 1) − 2 = 1− 5
2 = 1− 5
2 1
∞ 3
= − −
2 (b) f (z) = z−1
2 1− 1
(z − 1)n
2n n=0
∞
n=1 5 2n+1 1
, z0 = 2.
z2 Solution: This is a derivative: f (z) = −
1
z = 1
2 = 1
2 1
z , so, since 1
2 + (z − 2) = (z − 1)n . 1
1 − − z−2
2
∞ n=0
∞ =
n=0 (−1)n
(z − 2)n
2n (−1)n
(z − 2)n ,
2n+1 then
f (z) = − 1
z
∞ = −
n=0
∞ =
n=1
∞ =
m=0 (−1)n
(z − 2)n
2n+1 (−1)n+1 n
(z − 2)n−1 , now set m = n − 1
2n+1
(−1)m (m + 1)
(z − 2)m .
m+2
2 (c) f (z) = Log(z), z0 = 2.
1
Solution: Since f (z) = , and
z
1
z = 1
2 + (z − 2) = 1
2 = 1 (z−2)
2
∞
n
(−1) n=0 1+ 2n+1 (z − 2)n , MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS 5 then
∞ (−1)n
(z − 2)n dz
2n+1 f (z) = n=0
∞ = C+
n=0 (−1)n
(z − 2)n+1 .
(n + 1)2n+1 But that constant is not 0.
Log(2) = f (2)
∞ (−1)n
(2 − 2)n+1
(n + 1)2n+1 ln(2) = C +
n=0 = C, so
∞ (−1)n
(z − 2)n+1 , or
(n + 1)2n+1 Log(z) = ln(2) +
n=0
∞ = ln(2) +
m=1 (−1)m−1
(z − 2)m .
m
m2 (9) Find the Laurent series of the following functions, in a punctured disk 0 < |z−z0 | < R, at the
indicated point z0 . Also nd the largest R so that the series converges on 0 < |z − z0 | < R.
(10 points each) (a) f (z) = z2 1
, z0 = 1.
−1 1
1
First separate the pieces, f (z) = z−1 z+1 . The rst piece is already
in terms of powers of (z − 1). For the second, again use the geometric series: Solution: f (z) =
= = = 1
z−1
1
z−1
1
z−1 =
n=0 = 2 1
z−1
∞ = 1
z+1
1
2 + (z − 1) 1
1 2 1 + (z−1)
∞ 1
2 n=0 (−1)n
(z − 1)n
2n n (−1)
(z − 1)n−1
2n+1 1 1
+
2 (z − 1)
1 1
+
2 (z − 1) ∞
n=1
∞
m=0 (−1)n
(z − 1)n−1
2n+1
(−1)m−1
(z − 1)m .
2m+2 6 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS This converges for 0 < |z − 1| < 2, since 2 is the distance to the next singularity,
so a punctured disk of that radius, center at 1, ts in the domain where f is
analytic.
z+2
, z0 = 2.
(b) f (z) =
2
z(z − 2) As before, separate out the (z − 2)2 factor from the denominator, f (z) =
. Then do the long division of the remaining part before you use the
geometric series: Solution: 1
(z−2)2 f (z) =
=
= = = z+2
z 1
(z − 2)2
1
(z − 2)2 z+2
z
1+ 2
z
1 1
(z − 2)2 1+ 1
(z − 2)2 1+ 1
(z − 2)2 1+1+ 1+
∞
n=0 (z−2)
2 (−1)n
(z − 2)n
2n
∞
n=1 (−1)n
(z − 2)n
2n
∞ 1
2 = (−1)n
2
+
(z − 2)n−2 , now set m = n − 2
−
2
(z − 2)
2n
(z − 2)
n=2 = 1
2
(−1)m
2
+
(z − 2)m .
−
2
m+2
(z − 2)
2
(z − 2)
m=0 ∞ This converges on the largest annulus missing any other singularities of f , for all
z so that 0 < |z − 2| < 2.
1
(c) g(z) = e z , z0 = 0.
w and replace w = 1/z :
Solution: Just use the Taylor series of e
∞ g(z) =
n=0 (10) Compute the following residues: (10 points each) (a) Resz=−1 f (z), where f (z) =
Solution: 1 −n
z .
n! z 3 + 2z
.
(z + 1)2 Since −1 is a pole of order 2,
Resz=−1 z 3 + 2z
(z + 1)2 = z 3 + 2z
2 = 3(−1) + 2
= 5 (b) Resz=0 f (z), if f (z) = sin 1
.
z z=−1 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS 7 For this you really need the Laurent series, since z = 0 is not a pole of
f (z), but an essential singularity. But, similarly to 9c, this is easy to nd: Solution: ∞ 1
(−1)n
,
(2n+1)
(2n + 1)! z f (z) =
n=0 so the z −1 coecient, the residue, is (setting n = 0), 1, thus
Resz=0 f (z) = 1. (11) Evaluate the following real integrals. Be sure to indicate which contours you are using, and
the integrals over each part of the contours. Indicate why each integral is as you claim.
Show your work. (10 points each)
(a) ∞ dx −∞ (x2 + 1)2 Take the contour CR to be the real line segment from −R to R, followed
by the semicircle |z| = R, (z) > 0. Then, there is one singularity of f (z) =
1
inside the contour, which is a pole of order 2 at i. So, the residue is
(z 2 +1)2 Solution: Resz=i f (z) =
=
=
= 1
(z 2 + 1) 2 (z − i)2
z=i 1
(z + i)2
−2
(z + i)3 z=i
1
4i z=i Now, also, since
dz
|z| = R (z 2 + 1)2
(z) > 0 ≤ πR
(R2 − 1)2 , the limit as R → ∞ will be 0, and so
∞
−∞ (x2 dx
+ 1) 2 = dz lim R→∞ CR (z 2 = 2πiResz=i
= 2πi
=
∞ (b) Find
0 + 1)2
1 (z 2 + 1)2 1
4i π
.
2 cos(x)
dx.
(x2 + 4) You need to work with the function f (z) = (ze+4) , and use the same
2
contour CR as for the previous problem. Since f (z) has only one singularity Solution: iz 8 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS inside the region, a simple pole at z = 2i, for any R > 2, CR eiz
eiz
dz = 2πi Resz=2i 2
(z 2 + 4)
(z + 4)
= 2πi
= 2πi
= 2πi eiz
(z − 2i)
(z 2 + 4) z=2i eiz
(z + 2i) z=2i e−2
4i π
.
2e2 = But, since cos(x)/(x2 + 4) is an even function
lim R→∞ CR eiz
dz =
(z 2 + 4) lim R→∞ |z|=R, (z)>0 eiz
dz + 2
(z 2 + 4) ∞
0 cos(x)
dx,
(x2 + 4) and, with eiz = e−y+ix = e−y (z)>0 lim R→∞ CR ≤ πRe−y
R2 − 4 ≤ |z|=R, eiz
dz
(z 2 + 4) πR
,
R2 − 4 eiz
dz = 0,
(z 2 + 4) so
∞
0
2π (c)
0 cos(x)
dx =
(x2 + 4) π
.
4e2 dθ
√
.
2 + 3 sin(θ) Solution: For this, replace the real integral with the contour integral around |z| =
1
1
1
1
dz
2 z + z , sin(z) = 2i z − z , and dθ = iz : 1, where cos(z) =
2π
0 dθ
√
2 + 3 sin(θ) dz
1
1
iz 2 + 3 2i z − z
dz
√ 1
2iz + 3 2 (z 2 − 1)
dz
.
√
√
3 2
z + 2iz − 23
2
√ =
|z|=1 =
|z|=1 =
|z|=1 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS 9 That denominator (the numbers were chosen to make it easier, later) can be
√
√
i
factored with the quadratic formula, as 23 z + √3 z + 3i . The only singui
larity inside the circle |z| = 1 is at z = − √3 , so
2π
0 dθ
√
2 + 3 sin(θ) = dz √ 3 2
2 z |z|=1 = 2πiResz=− √
i 3
2 1 √ 3 2
2 z 3 . √ + 2iz − √ + 2iz − 3
2 = 2πi 1 √ 3 2
2 z i
z+√
3 √ + 2iz − 3
2 i
z=− √ 3 = 2πi 1 √ 3
2 i
√
3 z+ z+ √ 3i i
z+√
3 i
z=− √ 3 = 2πi 1 √ 3
2 z+ √ 3i i
z=− √ 3 = 2πi √ 3
2 = π
= 1
i
− √3 + √ 3i 1
(−1 + 3) π
.
2 (12) Find a linear fractional transformation w = f (z) which maps the bottom half-plane B =
{z | (z) < 0 } onto the disk D = {z ||z| < 1 } . (10 points)
Solution: There are several correct answers to this, depending on which points you pick.
But to map B to D, you have to map the boundary of B (the real line) to the boundary
of D (the unit circle), and in the orientation so that the inside of each region is on the
left. So, for B , take points (for example) z1 = 1, z2 = 0, z3 = −1 (as you move from
one to the next, B is on your left), and take w1 = 1, w2 = i, w3 = −1, and use the
cross-ratio:
(w − w1 )(w2 − w3 )
(w − w3 )(w2 − w1 )
(w − 1)(i + 1)
(w + 1)(i − 1) =
= (z − z1 )(z2 − z3 )
(z − z3 )(z2 − z1 )
(z − 1)(0 + 1)
.
(z + 1)(0 − 1) Solving for w (OK, this is a bit messy)
w = z+i
.
iz + 1 Since the arithmetic is messy, it is worth while to double check that when z = 1, w = 1,
when z = 0, w = i, and when z = −1, w = −1, so this takes the real line to the circle
of radius 1. You might also double check to see that the region B maps to D. If any 10 MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS point of B maps to any point of D, then since this is a 1-1 map and boundary maps
to boundary, then all of B would have to map to all of D. Take z = −i, which is in B ;
then w = 0, which is in D. Done.
z−i
(13) Show that the linear fractional transformation f (z) =
maps the rst quadrant I :=
−iz + 1
{z | (z) > 0, (z) > 0 } onto the D-shaped region D = {z ||z| < 1, (z) > 0 } . (10 points)
Solution: I thought this would be the hardest problem on the exam. But the idea is
easy. You have to check two independent things, that f maps the real line in z to the
unit circle in w (and in the order so that the upper half-plane maps to the inside), and
that f maps the imaginary line in z to the imaginary line in w. But since f (−1) = −1,
f (0) = −i, and f (1) = 1, f does map the real line to the unit circle. As you move in the
z plane from −1 to 0 to 1, you move in the w-plane counterclockwise around the circle,
so that is correct. Also, f (0) = −i, f (i) = 0, and f (2i) = i/3, so f maps imaginaries
to imaginaries, so again it takes the entire half-plane (z) > 0 to (w) > 0 since the
orientation is correct (or since f (1) = 1 which we already knew). So, f maps the upper
half-plane to the unit disk, and maps the right half-plane to the right half-plane, so it
maps points z in both the upper, and right, half plane (points in I ) to points w in both
the disk and the right half-plane (points in D). Since boundary goes to boundary, the
whole of B goes to the whole of D. ...

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