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Math 60370, Fall ‘ Assignment 3 Due Friday, September 18 (by noon)
Textbook problems:
p66: # Solution. One way to handle this is to do a partial fractions decompo- sition: 1 (ζ − 1)(ζ − 2 i)
a ζ − 1
a ζ − 2 i
where one can of course compute the value of a, but all that matters here is that a is the same in both fractions on the right. Since both 1 and 2i lie inside γ, we obtain from Lemma 2.4.1 in the textbook that ∫
γ
dζ (ζ − 1)(ζ − 2 i)
= 2πi(a − a) = 0.
Problem 1. (From Ahlfors) Compute (integrating in the counterclockwise direction) ∫
|z|=
|z − 1 | |dz|.
Solution. Let γ(t) = eit, t ∈ [0, 2 π] be the usual parametrization of the unit circle. Then |γ′(t)| = |ieit| = 1, whereas
|γ(t) − 1 | =
(cos t − 1)^2 + sin^2 t =
2 − 2 cos t =
2 sin^2 (t/2) = 2| sin(t/2)|.
Hence ∫
|z|=
|z − 1 | |dz| = 2
∫ (^2) π
0
| sin(t/2)| dt = 2
∫ (^2) π
0
sin(t/2) dt = 2 · 4 = 8.
Problem 2. (Also from Ahlfors) Given a ∈ C, R ∈ R and an analytic polynomial P (z), show that (integrating once counterclockwise again) ∫
|z−a|=R
P (z) dz = 2πiR^2 P ′(a).
Solution. Recall that a polynomial P (z) can be written out in terms of powers of (z − a) for any a (not just 0). That is,
P (z) = P ((z − a) + a) =
∑^ d
j=
cj (z − a)j^ ,
where d is the degree of P and cj ∈ C. In particular, P ′(a) = c 1.
Now if we take the parametrization γ : [0, 2 π] → C, γ(t) = a + Reit, then we get γ′(t) = iReit^ and P (γ(t)) =
∑d j=0 cj^ R
j (^) eijt. Hence
∫
|z−a|=R
P (z) dz =
∑^ d
j=
i¯cj Rj+
∫ (^2) π
0
ei(1−j)t^ dt = 2πic¯ 1 R^2 = 2πiR^2 P ′(a).
The second equality comes from the fact that (as one can verify)
∫ (^2) π 0 e
ikt (^) dt = 0 unless k = 0, in which case the integral is 2π.
Problem 3. (And again, from Ahlfors) Given a, b > 0, let γ parametrize the line segment from 0 to a + bi. Show that ∣ ∣ ∣ ∣
γ
cos(z^2 ) dz
a^2 + b^2 2 ab
sinh(2ab).
Solution. Let λ = a + bi and γ : [0, 1] → C, γ(t) = λt be the parametrization for our path. Then |γ′(t)| = |λ| =
a^2 + b^2. Moreover,
| cos(z^2 )| =
|eiz 2
|eiz 2 | + |eiz 2 | 2
e−Im^ z 2
2
= cosh Im z^2.
Hence ∣ ∣ ∣ ∣
γ
cos(z^2 ) dz
γ
| cos(z^2 )| |dz| =
0
|λ|| cos(λ^2 t^2 )| dt
≤ |λ|
0
cosh Im (t^2 λ^2 ) dt ≤ |λ|
0
cosh(t^2 Im (λ^2 )) dt.
Finally, since cosh x is an increasing function of |x| and that |t^2 Im (λ^2 )| < |tIm (λ^2 )| for t ∈ [0, 1], I have cosh(t^2 Im (λ^2 )) < cosh(tIm (λ^2 )) for t ∈ [0, 1]. Thus I continue the previous display obtaining ∣ ∣ ∣ ∣
γ
cos(z^2 ) dz
∣ ≤ |λ|
0
cosh(t(Im λ^2 )) dt =
|λ| Im (λ^2 )
sinh(2ab) =
a^2 + b^2 2 ab
sinh Im (λ^2 ),
as desired.
Problem 4. Let U ⊂ C be an open set with C^1 boundary and f : V → C be a C^1 function on an open set V ⊃ U^. Recall Green’s Theorem: if^ ω^ =^ a dx^ +^ b dy^ is a^ C^1 1-form on^ V^ and
we define dω :=
∂b ∂x −^
∂a ∂y
dx dy, then ∫
bU
ω =
U
dω,
where the boundary bU is parametrized so that rotating γ′^ counterclockwise by π/4 (i.e. multiplying γ′^ by i) gives the normal vector pointing into U.
(a) Use Green’s Theorem to show that
∂U
f (z) dz = 2i
U
∂f ∂ z ¯
dx dy.
