1. No category

Math 6350 Homework #5 Solutions
1. Ahlfors pg. 108 #2: Compute
Z
x dz
|z|=r
for the positive sense of the circle, in two ways: first, by use of an explicit parameterization of the circle, and second, by observing that
r2
1
1
z+
.
x = (z + z) =
2
2
z
Solution: For the first method, use z = reit for t ∈ [0, 2π]. We get x = r cos t and
dz = ireit dt, so that
Z
Z 2π
x dz =
ir2 cos teit dt
|z|=r
0
= ir
2
Z
2π
cos2 t + i sin t cos t dt
0
= ir2 (π + 0)
= ir2 π.
For the second method, we note that
Z
Z 1
r2
x dz =
z+
dz
2 C
z
C
Z
dz
r2
=
2 C z
r2
= (2πi)
2
= ir2 π,
R
using the fact that z is analytic while C dz/z = 2πi whenever C contains the origin.
2. Ahlfors pg. 108R #7: If P (z) is a polynomial and C denotes the circle |z − a| = R, what
is the value of C P (z) dz? Answer: −2πiR2 P 0 (a). Hint: adapt the trick from problem
#2.
Solution: On the given circle we have (z−a)(z−a) = R2 , so that (z−a)dz+(z−a)dz =
0. Hence we have
z−a
R2
dz = −
dz = −
dz.
z−a
(z − a)2
We therefore see that
Z
P (z) dz = −R
C
2
Z
C
1
P (z)
dz.
(z − a)2
We may express P (z) = P (a) + P 0 (a)(z − a) + N (z)(z − a)2 where N (z) is some
lower-order polynomial, and we therefore have
Z Z
P (a)
P 0 (a)
2
+ N (z) dz = −2πiR2 P 0 (a)
P (z) dz = −R
+
2
(z − a)
z−a
C
C
since the integral of 1/(z − a)2 and the analytic function N (z) are both zero around
the circle.
R
3. Evaluate the integral γ f (z) dz where f (z) = z|z|2 and γ is the closed upper unit
semicircle traversed counterclockwise (that is, rightward along the real axis from −1
to 1, then around the unit circle back to −1).
Solution: Write γ = γ1 + γ2 where γ1 is the real axis segment parametrized by z = t
on t ∈ [−1, 1] and γ2 is the semicircle parametrized by z = eit for t ∈ [0, π].
We have
Z
Z
1
Z
Z
π
t4 1
1 1
t dt =
= − = 0,
f (z) dz =
−1
4
4 4
−1
γ1
while
3
f (z) dz =
γ2
Hence coincidentally we have
0
R
γ
t=π
eit (ieit ) dt = 2i e2it t=0 = 0.
f (z) dz = 0 although f is not analytic.
4. Prove Green’s Theorem for a rectangle R = [a, b] × [c, d]:
Z
ZZ ∂p
∂q
p dx + q dy =
−
dx dy.
∂x ∂y
∂R
R
Do not use Stokes’ theorem: do the calculation directly using the Fundamental Theorem of Calculus. Since we will not be using this result at the moment, you may assume
that p and q are as smooth as needed to interchange order of integration.
Solution: It’s easier to work with the right side. We have
Z b Z d
Z d Z b
ZZ ∂p
∂p
∂q
∂q
−
(x, y) dx dy −
(x, y) dy dx
dx dy =
∂x ∂y
c
a ∂x
a
c ∂y
R
Z d
Z b
=
q(b, y) − q(a, y) dy −
p(x, d) − p(x, c) dx.
c
a
R
We now check that this is γ1 +γ2 −γ3 −γ4 p dx + q dy, where ∂R = γ1 + γ2 − γ3 − γ4 , for
γ1 the bottom side, γ2 the right side, γ3 the top side (traversed negatively), and γ4 the
left side (also traversed negatively). For γ1 we have x = t for t ∈ [a, b] and y = c, so
that dx = dt and dy = 0; we obtain
Z
Z b
Z b
Z b
p dx + q dy =
p(t, c) dt +
q(t, c) · 0 dt =
p(t, c) dt.
γ1
a
a
2
a
For γ3 we have x = t for t ∈ [a, b] and y = d, so that dx = dt and dy = 0 again, and
Z
Z
b
p(t, d) dt.
p dx + q dy =
a
γ3
The vertical paths
RR are similar; we see that the signs work out in exactly the same way
as they do in R (qx − py ) dx dy.
3