MA 201
COMPLEX ANALYSIS
ASSIGNMENT–3
Z
(1) Show that
γ
eaz
dz = 2πi sin a, where γ(t) = 2eit , t ∈ [0, 2π].
z2 + 1
Answer:
Z
Z
Z
Z
eaz
1
1
1
eaz
eaz
az 1
dz = e
−
dz =
dz −
dz
2
2i z − i z + i
2i γ z − i
γ z +1
γ
γ z +i
By Cauchy integral formula,
Z
Z
eaz
eaz
1
1
dz −
dz = 2πi × [eia − e−ia ] = 2πi sin a.
2i γ z − i
2i
γ z +i
Z 2π
iθ
(2) Evaluate
ee dθ.
0
. So
Answer: Put ei θ = z. Then dθ = dz
iz
Z
Z 2π
dz
eiθ
ez
e dθ =
= 2π (by Cauchy integral formula).
iz
|z|=1
0
f (z) = 0. Show that f is constant.
(3) Let f be an entire function such that lim z→∞
z
f (z) = 0. for every > 0 there exists a M > 0
Answer: Given that lim z→∞
z such that f (z)
< whenever |z| > M. i.e
z |f (z)| < |z| whenever |z| > M =⇒ f (z) = az + b
for some a, b ∈ C. But
az + b f (z) = lim = lim a +
lim z→∞
z z→∞ z z→∞ b = 0.
z
So a = 0 and hence f is constant.
(4) Let f : C → C be a function which is analytic on C \ {0} and bounded on
R
B(0, 12 ). Show that |z|=R f (z)dz = 0 for all R > 0.
R
R
Answer: By deformation theorem, |z|=R f (z)dz = |z|=r f (z)dz for every
r > 0. Take 0 < r < 21 . Given that f is bounded (by M say) on B(0, 21 ), then
by ML inequality
Z
f (z)dz ≤ M (2πr) → 0 as r → 0.
|z|=r
1
2
MA 201 COMPLEX ANALYSIS ASSIGNMENT–3
(5) Show that an entire function satisfying f (z + 1) = f (z) and f (z + i) = f (z)
for all z ∈ C is a constant.
Answer: It follows from the hypothesis that
f (z) = f (z + n) = f (z + im),
for all z ∈ C, and for all n, m ∈ Z.
Let S be the rectangle with vertices 0, 1, 1 + i and i. For any z = x + iy ∈ C,
there exits integers n and m and z0 = x0 + iy0 ∈ S such that,
z = x + iy = x0 + n + i(y0 + m) = z0 + n + im.
This implies that f (z) = f (z0 ). In particular f (C) = f (S). Since S is a
compact set and f is a continuous function then f (S) must be a bounded
set. All together implies that f is a bounded entire function. By Liouville’s
theorem we get f is a constant function.
Z
(6) Let g(z) be an analytic in B(0, 2). Compute
f (z)dz if
|z|=1
a1
ak
+ ··· +
+ a0 + g(z)
k
z
z
where ai ’s are complex constants.
f (z) =
Answer:
Z
Since a0 + g(z) is analytic by cauchy’s theorem
ak
g(z)]dz = 0. Again k has an antiderivative for k
z
Z
ha
a2 i
k
+ · · · + 2 dz = 0. Therefore
k
z
|z|=1 z
Z
Z
a1
f (z)dz =
dz = 2πi × a1 .
|z|=1 z
|z|=1
[a0 +
|z|=1
6= 1 therefore
(7) Let f be an entire function such that |f (0)| ≤ |f (z)| for all z ∈ C. Then either
f (0) = 0 or f is constant.
1 1 Answer: If f (0) = 0 then the proof is trivial. If f (0) 6= 0 then f (z) ≤ f (0) .
1
So f (z)
is entire and bounded and by Liouville’s theorem f is constant.
(8) FindX
the radius of convergence of the following power series:
(a)
z n! (R=1)
(b)
n≥0
X
2
2n z n (R=0)
n≥0
X (−1)n
z n(n+1) (R=1)
n≥0
(
X
2n
(d)
an z n where an =
3n
n≥0
(c)
n
if n is odd
(R = 31 )
if n is even.
MA 201
COMPLEX ANALYSIS
ASSIGNMENT–3
(9) Find the power series expansion of the function f (z) = cos2 z about 0.
Answer: We know that f (z) = cos2 z = (1 + cos 2z)/2 and
∞
X
z 2n
cos z =
(−1)n
2n!
n=0
for z ∈ C. Therefore,
∞
f (z) = cos2 z =
1 1X
(2z)2n
+
(−1)n
2 2 n=0
2n!
for z ∈ C .
3
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