Math 6350 Homework #6 Solutions

Math 6350 Homework #6 Solutions
1. Suppose the branch of the logarithm is chosen so that 0 < arg z < 2π. Under what
circumstances is it true that log z1 z2 = log z1 + log z2 ? Under what circumstances is it
true that log z p = p log z for a real number p?
Solution: The real parts certainly work out fine since they use only the properties of
the real logarithm: we have
ln |z1 z2 | = ln |z1 ||z2 | = ln |z1 | + ln |z2 |
and
ln |z p | = ln |z|p = p ln |z|.
Thus we only need to worry about the imaginary part.
First we ask when arg z1 z2 = arg z1 + arg z2 . The argument is assumed to be in (0, 2π),
which means the left side is in (0, 2π) while the right side can be anything between 0
and 4π. We will therefore have
(
arg z1 + arg z2
arg z1 + arg z2 < 2π
arg z1 z2 =
arg z1 + arg z2 − 2π 2π < arg z1 + arg z2 < 4π.
The condition is therefore that arg z1 + arg z2 < 2π. I’m not sure of a simpler way to
express this.
As for arg z p = p arg z, suppose z = reiθ for θ ∈ (0, 2π). Then obviously arg z = θ
so that p arg z = pθ. On the other hand z p = rp eipθ and arg z p = pθ + 2nπ for some
integer n which makes pθ + 2nπ ∈ (0, 2π). So as long as 0 < p arg z < 2π we will have
arg z p = p arg z.
2. Use the method from class to evaluate the integral
Z 2π
dt
.
5 − 3 sin t
0
Solution: We express this as the integral around the closed curve γ(t) = eiπt , using
the fact that on this curve
sin t =
z − 1/z
eit − e−it
=
.
2i
2i
We have dz = ieit dt = iz dt so that dt = dz/(iz). Therefore we can turn the integral
into
Z 2π
Z
Z
dt
dz
2i dz
= − 2
=
.
3
5 − 3 sin t
3iz + 10z − 3i
0
γ iz 5 − 2i (z − 1/z)
γ
Now factor this quadratic to get
z=
−10 ±
√
100 − 36
−10 ± 8
=
6i
6i
1
which yields z = 3i and z = i/3. Only i/3 is inside the unit disc, so we may express
this integral as
Z
Z 2π
f (z)
dt
=
5 − 3 sin t
γ z − i/3
0
where
f (z) = −
2i
.
3i(z − 3i)
The integral is therefore equal to
Z 2π
dt
4π
4π
π
= 2πif (i/3) =
=
= .
5 − 3 sin t
3i(i/3 − 3i)
8
2
0
3. Ahlfors pg. 120 #2: Compute
Z
|z|=2
dz
+1
z2
by decomposing it into partial fractions.
Solution: Assuming a decomposition of the form
z2
1
A
B
=
+
,
+1
z−i z+i
we must have A(z − i) + B(z + i) = 1 for all z, and thus B = −A and −2iA = 1 so
that A = i/2. We obtain
i 1
i 1
1
=
−
.
z2 + 1
2 z−i 2 z+i
Hence the given integral is
Z
γ
4. Compute
R
|z|=1
Z
Z
dz
i
dz
i
dz
=
−
2
z +1
2 γ z−i 2 γ z+i
i
i
= (2πi) − (2πi) = 0.
2
2
cos z/z n dz where n is any integer.
Solution: Using the general formula
Z
f (z)
2πif (n−1) (0)
=
n
(n − 1)!
γ z
if n ≥ 1, we obtain
Z
|z|=1

0
n even and positive,
2πi 
cos z dz
=
1
n = 4k + 1 for k ≥ 0,
zn
(n − 1)! 

−1 n = 4k + 3 for k ≥ 0.
On the other hand if n ≤ 0 then z −n cos z is analytic at 0 and hence everywhere, so
that the integral is zero.
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5. Ahlfors pg. 123 #2: Prove that a function f which is analytic on the whole plane and
satisfies an inequality |f (z)| < |z|n for some n and all sufficiently large z reduces to a
polynomial.
Solution: We will prove that f (n+1) (z) = 0 for all z ∈ C, which implies that f is a
polynomial of degree n. Using formula (24) we have
Z
f (ζ) dζ
(n + 1)!
(n+1)
f
(z) =
n+2
2πi
C (ζ − z)
whenever f is analytic inside C. Choose C to be a circle of large radius R; then we
have
Z
(n + 1)! 2π
Rn+1 dt
1
(n + 1)!
(n+1)
|f
(z)| ≤
=
.
n+2
2π
(|R| − |z|)
R
(1 − |z/R|)n+2
0
For fixed z, let R go to infinity, and we get |f (n+1) (z)| ≤ 0, which implies f (n+1) (z) = 0
as desired.
6. From “Counterexamples in Analysis” by Gelbaum and Olmsted. Define
(
2
x3 /y 2 e−x /y y > 0
f (x, y) =
0
y ≤ 0.
(a) Show that f is smooth (infinitely differentiable) in x for fixed y.
Solution: If y > 0 then obviously the function is smooth since it’s just a rescaling
2
of x3 e−x which is smooth. If y ≤ 0 then it’s the zero function which is smooth
in x.
(b) Show that f is continuous in y for any fixed x.
Solution: It’s clearly continuous at points y > 0 and at points y < 0, so we just
have to check that it’s continuous at y = 0. We have
lim+ f (x, y) = lim+
y→0
y→0
x3 z 2
1
u2
x3
=
lim
=
lim
=0
x u→+∞ eu
y 2 ex2 /y z→+∞ ex2 z
by L’Hopital’s rule. We also have limy→0− f (x, y) = f (x, 0) = 0, and since these
all match we have continuity in y.
(c) Show that
Z 1
Z 1
d ∂ f (x, y) dy 6=
f (x, y) dy.
dx x=0 0
0 ∂x x=0
Solution: First evaluate the left side. We have
Z 1
Z 1
2
f (x, y) dy =
x3 /y 2 e−x /y dy
0
Z0 1
∂
2
=
xe−x /y dy
0 ∂y
y=1
2
= xe−x /y y=0
2
= xe−x − lim+ xe−x
y→0
= xe
3
−x2
2 /y
whenever x 6= 0. Of course if x = 0 then the integral is zero, so this formula
always works. Differentiating we get
Z 1
d 2
2 f (x, y) dy = (e−x − 2x2 e−x )x=0 = 1.
dx x=0 0
Now we work on the right side. We compute (if y > 0) that
∂
∂ 3 2 −x2 /y
2
2
f (x, y) =
x /y e
= 3x2 /y 2 e−x /y − 2x4 /y 3 e−x /y ,
∂x
∂x
∂ and evaluating at x = 0 we obtain ∂x
f (x, y) = 0. On the other hand if y ≤ 0
x=0
then the derivative is clearly zero for any x (and thus for x = 0). Thus the integral
of the zero function is zero, and the right side is always zero.
(d) Show that f is not continuous at (0, 0).
Solution: We have f (0, 0) = 0 and we want to find a sequence (xn , yn ) such that
(xn , yn ) → 0 and f (xn , yn ) does not converge to zero. Suppose yn = x2n /k for
some k; then we have
k 2 e−k
.
f (xn , yn ) =
xn
Since (xn , yn ) → (0, 0), we must have xn → 0, so f (xn , yn ) → ∞.
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