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Complex Analysis - Final exam - Answers
Exercise 1 : (20 %) Let r, s ∈ R> 0. Let f be an analytic function defined on D(0, r) and g be an analytic function defined on D(0, s). Prove that f +g is analytic on D(0, min(r, s)) where min(r, s) is the minimum of r and s. Answer: By the definition given in the course, f is analytic on D(0, r) if it can be written as a series
f (z) =
∑^ ∞
n=
anzn
for (an)n∈N ∈ CN^ such that
n=0 |an|r
n (^) converges. In the same way, g is analytic on
D(0, s) if it can be written as a series
g(z) =
∑^ ∞
n=
bnzn
for (bn)n∈N ∈ CN^ such that
n=0 |bn|s
n (^) converges. Let t = min(r, s) and consider the sequences (SN )N ∈N, (TN )N ∈N and (UN )N ∈N defined by
SN =
∑^ N
n=
|an|rn^ and TN =
∑^ N
n=
|bn|sn^ and UN =
∑^ N
n=
|an + bn|tn.
As t 6 r and t 6 s, for N ∈ N, we have
UN =
∑^ N
n=
|an + bn|tn^6
∑^ N
n=
(|an| + |bn|)tn^ =
∑^ N
n=
|an|tn^ +
∑^ N
n=
|bn|tn
∑^ N
n=
|an|rn^ +
∑^ N
n=
|bn|sn^ = SN + TN.
As SN + TN is increasing (|an|rn^ + |bn|sn^ > 0),
UN 6 SN + TN 6
∑^ ∞
n=
|an|rn^ +
∑^ ∞
n=
|bn|sn.
Finally, UN is increasing (|an + bn|tn^ > 0) and upper bounded so it is convergent. We proved that ∑∞
n=
|an + bn|tn
is convergent, so the following function is analytic on D(0, t) = D(0, min(r, s)):
h(z) =
∑^ ∞
n=
(an + bn)zn.
As min(r, s) 6 r and min(r, s) 6 s, the three functions f , g and h are analytic on D(0, min(r, s)) and we get immediately the relation h = f + g so f + g is analytic on D(0, min(r, s)).
Exercise 2 : We want to study the limit of ∑
n∈N
zn.
- (10 %) Prove that the function
f (z) =
∑^ ∞
n=
zn
is well defined and infinitely differentiable on D(0, 1 /2). Answer: We prove that f is analytic on D(0, 1 /2). For that, let us notice that the series ∑^ ∞
n=
)n
converges. Indeed, it is a geometric series with common ratio 1/2 and | 1 / 2 | < 1. As f is analytic on D(0, 1 /2), it is defined and infinitely differentiable.
- (5 %) Prove that, for z ∈ D(0, 1 /2),
f (z) =
1 − z
Answer: Let z ∈ D(0, 1 /2). The series ∑∞
n=
zn
is a geometric series of common ratio z. Moreover, |z| < 1 / 2 < 1 so this series converges to 1/(1 − z). Finally, f (z) = 1/(1 − z).
- (10 %) Is it true that for every z ∈ C such that z 6 = 1 ∑^ ∞
n=
zn^ =
1 − z
Justify your answer. Answer: It is not true. Indeed, for z = 2, we have 1/(1 − z) = −1. On the other hand, if we define (SN )N ∈N by
SN =
∑^ N
n=
2 n
we get
2 × SN =
∑^ N
n=
2 n+1^ =
N∑ +
n=
2 n^ =
∑^ N
n=
2 n^ − 1 + 2N^ +1^ = SN + 2N^ +1^ − 1
so SN = 2N^ +1^ − 1 and the series
n=0 2
n (^) diverges.
We have easily
m 1 a 1 − z
m 2 a 2 − z
a 2 m 1 + a 1 m 2 − (m 1 + m 2 )z (a 1 − z)(a 2 − z)
so we should have a 2 m 1 + a 1 m 2 = 1, m 1 + m 2 = 0 and a 1 and a 2 are the two roots of the polynomial z^2 + iz + 2. The discriminant of this polynomial is
∆ = i^2 − 4 × 2 × 1 = − 9
and therefore, the square roots of ∆ are 3i and − 3 i. Finally, the roots of the polynomial are −i + 3i 2
= i and
−i − 3 i 2
= − 2 i.
Let us fix a 1 = i and a 2 = − 2 i. So we have to solve the system of equations
{ m 1 + m 2 = 0 − 2 im 1 + im 2 = 1
Adding 2i times the first one to the second one we deduce that this system is equivalent to (^) { m 1 + m 2 = 0 3 im 2 = 1
which admits the solution { m 2 = 1/ 3 i = −i/ 3 m 1 = i/ 3.
Finally, we get that
g(z) =
i 3
×
i − z
i 3
×
− 2 i − z
Using the previous question, over D(0, |i|/2) = D(0, 1 /2), 1/(i − z) is analytic and we have 1 i − z
∑^ ∞
n=
in+^
zn^ =
∑^ ∞
n=
(−i)n+1zn.
In the same way, over D(0, | − 2 i|/2) = D(0, 1), 1/(− 2 i − z) is analytic and we have
1 − 2 i − z
∑^ ∞
n=
(− 2 i)n+^
zn^ =
∑^ ∞
n=
i 2
)n+ zn.
Using Exercise 1, we get that on D(0, min(1/ 2 , 1)) = D(0, 1 /2), 1/(i−z)− 1 /(− 2 i− z) is analytic and
1 i − z
− 2 i − z
∑^ ∞
n=
(−i)n+1^ −
i 2
)n+1) zn.
Thus, g is analytic on D(0, 1 /2) and
g(z) =
i 3
i − z
− 2 i − z
∑^ ∞
n=
i 3
(−i)n+1^ −
i 2
)n+1) zn.
Exercise 3 : Let (an)n∈N be a sequence of complex numbers and r ∈ R> 0 such that ∑
n∈N
|an|rn
converges.
- (10 %) Prove that the following functions are well defined and infinitely differen- tiable on D(0, r):
f (z) =
∑^ ∞
n=
anzn^ ; g(z) =
∑^ ∞
n=
an− 1 n
zn.
Answer: First of all f is analytic on D(0, r) because ∑
n∈N
|an|rn
converges. To prove that g is also analytic, let us denote, for N ∈ N,
SN =
∑^ N
n=
an− 1 n
∣ rn.
For N ∈ N, we have
SN =
∑^ N
n=
an− 1 n
∣ rn^6
∑^ N
n=
|an− 1 | rn^ = r
N∑ − 1
n=
|an|rn.
As, for any n ∈ N, |an|rn^ > 0 and r > 0, we have
SN 6 r
N∑ − 1
n=
|an|rn^6 r
∑^ ∞
n=
|an|rn.
So, as SN is increasing and upper bounded, SN converges. It finishes to prove that g is analytic on D(0, r). Finally, as f and g are analytic on D(0, r), they are defined and infinitely differen- tiable.
- (10 %) Find all the antiderivatives of f on D(0, r). Answer: First of all, as g is analytic on D(0, r), we have, for z ∈ D(0, r),
g′(z) =
∑^ ∞
n=
an− 1 n
nzn−^1 =
∑^ ∞
n=
an− 1 zn−^1 =
∑^ ∞
n=
anzn^ = f (z).
Let F be an antiderivative of f on D(0, r). For z ∈ D(0, r), we have, by definition (F − g)′(z) = F ′(z) − g′(z) = f (z) − f (z) = 0. So F − g is constant: there exists z 0 ∈ C such that
F (z) = z 0 + g(z) = z 0 +
∑^ ∞
n=
an− 1 n
zn.
