**Unformatted text preview: **MATH 208, SPRING, 2013, PRACTICE EXAM # 1 SOLUTIONS
(1) Find all points z ∈ C so that z 3 = 8.
Us the polar form of z = reiθ , where we will use the principal branch of the
argument, −π < θ ≤ π . If z 3 = 8, then r3 ei3θ = 8, so 3θ = 2nπ , or θ = 2nπ and
3
r = 2. Within the range of the principal branch of the argument for z , θ = 0, 2π , − 2π ,
3
3
√
√
so z = 2, −1 + 3i, or − 1 − 3i.
(2) Show that the set S := {z ∈ C |1 < |z| < 2 } is an open set.
For any z0 ∈ S , 1 < |z0 | < 2. Let δ > 0 be small enough so that 1 +
δ < |z0 | < 2 − δ , which is always possible because of the strict inequalities. Then
{ z ∈ C| |z − z0 | < δ} will be contained in S , so S is open since every point is an interior
point.
(3) Show, by an − δ argument, that limz→2 z 2 − iz = 4 − 2i.
Let > 0 be given. Then, set δ := min 1, 6 . Since |z − 2| < 1, |z| < 3, and
so by the triangle inequality, Solution: Solution: Solution: |z − (−2 + i)| ≤ |z| + |2| + |i|
< 3 + 2 + 1 = 6. So, whenever 0 < |z − 2| < δ , then (now, using both possibilities for δ )
z 2 − iz − (4 − 2i) = |z − 2| |z − (−2 + i)|
< 6 ·6= . Remark. Note that I have hidden here the "scratch" work. For that, you would look at
z 2 − iz − (4 − 2i) . But remember, since that limit is correct (you can see that by plugging z = 2 into the function), (z − 2) must be a factor:
z 2 − iz − (4 − 2i) = |z − 2| |z + (2 − i)| . Then, in order to make that right side small, < , you need to make sure that |z + (2 − i)| is
not too big. You can only control how close z is to 2, so take a standard distance, |z − 2| < 1.
In that range,
|z + (2 − i)| ≤ |z| + 2 + |i|
= 3 + 2 + 1 = 6, by the triangle inequality, since when |z − 2| < 1, |z| < 3 (here, you can just look at the
picture; the disk of radius 1 centered at 2 is contained inside the disk of radius 3 centered
at the origin). Then, you take δ to be so small that, even when multiplied by as big as
|z + (2 − i)| can get (which is 6), the product will still be less than .
You can nd a bound on |z + (2 − i)| another way:
|z + (2 − i)| = |(z − 2) + 2 + (2 − i)|
≤ |z − 2| + |4 − i|
√
< 1 + 17
< 6, and in fact, that way you can get a slightly better bound (since 4 <
doesn't matter, any reasonable bound will do. 1 √ 17 < 5). But it 2 MATH 208, SPRING, 2013, PRACTICE EXAM # 1 SOLUTIONS
(4) Dene the function f (z) = iz in terms of elementary functions (just deal with the principal
branch), and nd its derivative.
π
iπ
The denition we had was that iz = ezLog(i) = ez(0+i 2 ) = e 2 z . That is the
principal branch. By the chain rule, Solution: (iz ) = iπ e2z iπ iπ z
e2
2
iπ
= iz · .
2
(5) For which values of z1 , z2 ∈ C is it true that Arg(z1 ) − Arg(z2 ) = Arg(z1 /z2 ), where Arg(z)
is the principal branch of the argument of z ? (15 points. 5 points for nding a domain D
for which the statement is true, 5 points for showing that, on that domain D, the statement
is true, and 5 points if D is a largest possible such domain for which the statement is true
for all z1 , z2 ∈ D. )
I will take D = { z ∈ C| (z) > 0}, the "right" half-plane. For any z1 , z2 ∈ D,
r1
then z1 = r1 eiθ1 , z2 = r2 eiθ2 , with θ1 , θ2 ∈ − π , π . Then, z1 /z2 = r2 ei(θ1 −θ2 ) , so the
2 2
only question is whether θ1 − θ2 = Arg(z1 /z2 ). But, since z1 , z2 ∈ D, θ1 − θ2 ∈ (−π, π)
(this is right; check the possibilities). and so θ1 − θ2 = Arg(z1 /z2 ), and so the formula
is true on D. D is indeed a maximal such domain, and for the principal branch of the
= Solution: argument, it is the only maximal domain (not contained in a larger domain).
(6) Let f (z) be analytic on a domain D. If f (z) is also analytic on D, show that f is constant.
∂v
Set f = u + iv . Since f is analytic, ∂u = ∂y , but since f is also analytic,
∂x
∂v
∂u
∂u
∂x = − ∂y as well, so (add the two equations together) 2 ∂x = 0. Similarly, since f is
∂v
∂v
analytic, ∂u = − ∂x , but since f is also analytic, ∂u = + ∂x as well, so again adding
∂y
∂y
the two equations together, 2 ∂u = 0. Since both partial derivatives of u are 0, at any
∂y
z , then u is constant. Similarly, v is also constant, so f is constant.
(7) Show that the function u(x, y) = x2 − y 2 + 2y is harmonic, and nd a harmonic conjugate
of u(x, y).
2
2
Certainly ∂ u = 2 and ∂ u = −2, so u is harmonic. To nd the conjugate v ,
∂x2
∂y 2
∂v
use the fact that ∂x = − ∂u = +2y − 2, so
∂y Solution: Solution: v = (2y − 2) dx = 2xy − 2x + h(y). You determine h(y) by dierentiating with respect to y , since you know
∂v
Cauchy-Riemann equations, ∂y = ∂u = 2x:
∂x ∂v
∂y by the ∂v
∂y
∂u
=
∂x
= 2x, 2x + h (y) = so h is a constant, and v = 2xy − 2x is the harmonic conjugate. You can also see that
f (z) = u + iv = z 2 − 2iz is the holomorphic function combining the two together.
(8) Find the following contour integrals, directly from the denition.
(a) γ z 2 dz , where γ is the quarter-circle parametrized by z(t) = 2eit , t ∈ [0, π/2]. MATH 208, SPRING, 2013, PRACTICE EXAM # 1 SOLUTIONS Solution: Parametrize by z = 2e it ,
π
2 2 z dz =
γ 3 as mentioned in the problem,t ∈ [0, π/2].
(2eit )2 2ieit dt 0
π
2 = 8ie3it dt 0
π =
= (b) γ 8i 3it 2
e
3i
0
8
(−i − 1) .
3 1
dz , where γ is the circle of radius 4, center at the origin. (10 points)
z Solution: We will again use the denition of the contour integral, not any theorems.
Here the curve γ can be parametrized by z(t) = 4eit , t ∈ [0, 2π]. Note that
dz
it
dt = 4ie ; the ordinary dierentiation rules still work with complex functions.
γ 2π 1
dz :=
z 0 1
4ieit dt
4eit 2π = idt
0 = (c) γ 2πi. z 2 dz , where γ is the quarter-circle parametrized by z(t) = 2eit , t ∈ [0, π/2].
C is again parametrized by z = 2eit , t ∈ 0, π . Then
2 Solution: π
2 2 z dz =
C 2e−it 2 2ieit dt 0
π
2 = 8 i e−it dt 0 i −it
= 8
e
−i
−i π
2 = −8 e π
2 0 −1 = − (−i − 1)
= 8 (i + 1) Again, you can turn the last integral into 02π sin(t) + i cos(t) dt, and both real
and imaginary parts are 0.
(9) State the Cauchy-Goursat theorem, being careful to include the conditions on the function
f (z).
Let f be analytic on and inside a simple closed contour C . Then Solution: f (z) dz = 0.
C (10) State the Cauchy Integral Formula, being careful to include the conditions on the function
f (z). 4 MATH 208, SPRING, 2013, PRACTICE EXAM # 1 SOLUTIONS Solution: Let f be analytic on and inside a simple closed contour C , positively oriented.
If z0 is any point in the interior of C , then
f (z0 ) = 1
2πi C f (z) dz
.
z − z0 ...

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