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Patrick Corn Math 185, 7/18/ Solution Set 6 Problem 1: Notice that Log(w) is defined and holomorphic in the disc |w − 1 | < 1. Let α(z) = Log(f (z)); then α is defined and holomorphic in G, by the Chain Rule. We have already seen that (^) dwd Log(w) = 1/w, so then α′(z) = f ′(z)/f (z) by the Chain Rule, and thus we are done by the Fundamental Theorem of Calculus.
Problem 2: If we expand the binomial, we’ll get a sum of various (positive and negative) powers of z. The only one that will integrate to something nonzero over γ is z−^1. The coefficient of z−^1 in the expansion is pretty clearly
( 2 n n
. So we have
2 πi
2 n n
γ
z
z +
z
) 2 n dz =
∫ (^2) π
0
e−it^
eit^ + e−it
) 2 n ieit^ dt = 2^2 ni
∫ (^2) π
0
cos^2 n^ t dt
and so it follows that
1 2 π
∫ (^2) π
0
cos^2 n^ t dt =
22 n
2 n n
(2n)! (2 · 4 · · · (2n))^2
1 · 3 · · · (2n − 1) 2 · 4 · · · (2n)
Problem 3: For any ǫ > 0, we can pick δ > 0 such that if |z − z 0 | < δ, |f (z) − f (z 0 )| < ǫ/ 2 π. Then if r < δ, (^) ∣ ∣ ∣ ∣
Cr
f (z) − f (z 0 ) z − z 0
dz
ǫ/ 2 π r
2 πr = ǫ,
so we see that
lim r→ 0
Cr
f (z) z − z 0
dz − 2 πif (z 0 ) = lim r→ 0
Cr
f (z) − f (z 0 ) z − z 0
dz = 0.
Problem 4: One way to do this is to look at the proof of Cauchy’s theorem (beta). Let Dr be the interior of Cr. If you let f = u + iv, then looking at your notes gives the first equality of the following series of equalities:
1 r^2
Cr
f (z) dz =
r^2
Dr
[(−vx − uy ) + i(ux − vy )] dA
πr^2
Dr
2 πi
∂f ∂z
(x, y) dA
which equals the average value of the integrand on Dr. As r → 0, this clearly goes to 2πi ∂f ∂z (z 0 ),
because ∂f ∂z is continuous. Another way to do this, which is a lot more elegant, is to show that if f is C^1 , then
lim z→z 0
f (z) − f (z 0 ) − ∂f ∂z (z 0 )(z − z 0 ) − ∂f ∂z (z − z 0 ) z − z 0
but I do not know if there is an easy way to prove this rigorously without writing everything out in terms of u, v, x, y. If you look at it for a minute, you should be convinced that it is true, if you think it’s reasonable to view complex functions as functions of z and z instead of x and y. (You can change variables back and forth pretty easily.) Anyway, once you establish this, consider what happens when you integrate this fraction divided by z − z 0 over Cr, and use the previous exercise. I’ll leave the details to you. For the second part, simply check that ∂f ∂z (z 0 ) = 0 if and only if f satisfies the Cauchy-Riemann equations at z 0 ; since f is C^1 , this is equivalent to f being differentiable at z 0.
Problem 5: Let γ 1 be the edge of the triangle on the real axis. Let γ 2 be the edge of the triangle not touching the origin. Let γ 3 be the remaining edge. Orient them so that the triangle is oriented counterclockwise. Then immediately we see that limR→∞
γ 1 e
−z^2 dz = √π/2.
A parameterization for γ 2 is γ 2 (t) = R(1 + αt), t ∈ [0, 1], where α = eπi/^8 − 1. So ∣∣ ∣ ∣
γ 2
e−z
2 dz
0
e−R
(^2) (1+αt) 2 Rα dt
∣ ≤^ e
−R^2 aR|α|,
where a = min{Re((1 + αt)^2 ) : t ∈ [0, 1]}. The only thing that matters is that a is positive. (The argument of 1 + αt is between 0 and π/8, so its square has argument between 0 and π/4.) At any rate, it is clear that this integral is bounded above in absolute value by something approaching 0 as R → ∞, so it must approach 0 as R → ∞. Now a parameterization for −γ 3 is simply (−γ 3 )(t) = teπ/^8 , t ∈ [0, R], and by Cauchy’s theorem
and what we just did, we must have that limR→∞
γ 1 e
−z^2 dz = limR→∞^ ∫ −γ 3 e
−z^2 dz. So we get √ π 2
0
e−t
(^2) eπi/ 4 eπi/^8 dt
and, putting the constant on the other side and taking real parts, we get
cos(π/8)
π 2
0
e−t
(^2) /√ 2 cos(t^2 /
- dt
Letting u = t/ 21 /^4 , we get
2 −^1 /^4 cos(π/8)
π/2 =
0
e−u
2 cos(u^2 ) du
and I will let you discover why cos(π/8) = 2−^3 /^4
2 (hint: double-angle formula). Problem 6: As I mentioned in class, it’s silly not to integrate over a sector instead of a triangle, so this is what I’ll do. Again, make γ 1 , γ 2 , γ 3 just as in the previous exercise (except now γ 2 is an
arc and not a line). Again we have limR→∞
γ 1 e
−z^2 dz = √π/2. Now ∫
γ 2
e−z
2 dz =
∫ (^) π/ 4
0
e−R
(^2) cos 2θ−iR (^2) sin 2θ ieiθ^ dθ,
so we get that ∣ ∣∣ ∣
γ 2
e−z
2 dz
∫ (^) π/ 4
0
e−R
(^2) cos 2θ dθ ≤
∫ (^) π/ 4
0
e−R
(^2) (1− 4 θ/π) dθ =
π 4 R^2
(1 − e−R
2 )
which goes to 0 as R → ∞. Note that I used the inequality cos 2θ ≥ 1 − 4 θ/π, θ ∈ [0, π/4]. If you want to see why this is true, try drawing the graph of the left and right sides. Just as in the last problem, then, Cauchy’s theorem gives that √ π 2
0
e−(e
πi/ (^4) t) 2 eπi/^4 dt = eπi/^4
0
e−it
2 dt.
So we can equate real and imaginary parts of the resulting equation
e−πi/^4
π 2
0
cos(t^2 ) dt − i
0
sin(t^2 ) dt
to obtain (^) ∫ ∞
0
cos(t^2 ) dt =
0
sin(t^2 ) dt =
π 2
