Complex Analysis - Final exam - Answers Exercise : (0 %) 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. By the definition given in the course, f is analytic on D(0, r) if it can be written as a series f(z) = a n z n for (a n ) n N C N such that a n r n converges. In the same way, g is analytic on D(0, s) if it can be written as a series g(z) = b n z n for (b n ) n N C N such that b n s n converges. Let t = min(r, s) and consider the sequences (S N ) N N, (T N ) N N and (U N ) N N defined by S N = a n r n and T N = b n s n and U N = a n + b n t n. As t r and t s, for N N, we have U N = a n + b n t n a n r n + ( a n + b n )t n = b n s n = S N + T N. As S N + T N is increasing ( a n r n + b n s n 0), U N S N + T N a n r n + a n t n + b n s n. b n t n Finally, U N is increasing ( a n + b n t n 0) and upper bounded so it is convergent. We proved that a n + b n t n is convergent, so the following function is analytic on D(0, t) = D(0, min(r, s)): h(z) = (a n + b n )z n. As min(r, s) r and min(r, s) 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 : We want to study the limit of z n.. (0 %) Prove that the function n N f(z) = is well defined and infinitely differentiable on D(0, /). We prove that f is analytic on D(0, /). For that, let us notice that the series ( ) n converges. Indeed, it is a geometric series with common ratio / and / <. As f is analytic on D(0, /), it is defined and infinitely differentiable.. (5 %) Prove that, for z D(0, /), Let z D(0, /). The series z n f(z) = z. is a geometric series of common ratio z. Moreover, z < / < so this series converges to /( z). Finally, f(z) = /( z). 3. (0 %) Is it true that for every z C such that z z n = z? Justify your answer. It is not true. Indeed, for z =, we have /( z) =. On the other hand, if we define (S N ) N N by S N = we get S N = n+ = N+ n = z n n n + N+ = S N + N+ so S N = N+ and the series n diverges.
4. (5 %) Let a C \ {0}. Using f, express the function f a defined around 0 by as an analytic function of the form f a (z) = a z b n z n. Give an r R with r > 0 such that this series converges on D(0, r) (r can depend on a). For any z C such that z a, we have f a (z) = a z = a z a = a f ( z a So, using the previous question, if z/a D(0, /), we get f a (z) = a ( z n = a) a n+ zn. Moreover, z/a D(0, /) if and only if z/a < / if and only if z < a / so the previous series is analytic on D(0, a /). So we can choose r = a /. Note that we could in fact prove that the series converges to f a (z) for any z D(0, a ). zn an+ 5. (0 %) Deduce an expression of the function g defined around 0 by as an analytic function of the form g(z) = z + iz + c n z n. Give an r R with r > 0 such that this series converges on D(0, r ). Hint: Write g(z) = m a z + m a z for some m, m, a, a C. You can use freely the result of Exercise. We want to get ). m a z + m a z = g(z) = z + iz +.
We have easily m a z + m a z = a m + a m (m + m )z (a z)(a z) so we should have a m + a m =, m + m = 0 and a and a are the two roots of the polynomial z + iz +. The discriminant of this polynomial is = i 4 = 9 and therefore, the square roots of are 3i and 3i. Finally, the roots of the polynomial are i + 3i i 3i = i and = i. Let us fix a = i and a = i. So we have to solve the system of equations { m + m = 0 im + im = Adding i times the first one to the second one we deduce that this system is equivalent to { m + m = 0 which admits the solution 3im = { m = /3i = i/3 Finally, we get that m = i/3. g(z) = i 3 i z i 3 i z. Using the previous question, over D(0, i /) = D(0, /), /(i z) is analytic and we have i z = i n+ zn = ( i) n+ z n. In the same way, over D(0, i /) = D(0, ), /( i z) is analytic and we have i z = ( i) n+ zn = ( ) n+ i z n. Using Exercise, we get that on D(0, min(/, )) = D(0, /), /(i z) /( i z) is analytic and ( i z i z = ( ) ) n+ i ( i) n+ z n. Thus, g is analytic on D(0, /) and g(z) = i 3 ( ) i z = i z ( i ( i) n+ 3 ( ) ) n+ i z n.
Exercise 3 : Let (a n ) n N be a sequence of complex numbers and r R >0 such that a n r n converges. n N. (0 %) Prove that the following functions are well defined and infinitely differentiable on D(0, r): f(z) = a n z n ; g(z) = First of all f is analytic on D(0, r) because a n r n n N a n n zn. converges. To prove that g is also analytic, let us denote, for N N, For N N, we have S N = S N = a n r n n a n r n. n As, for any n N, a n r n 0 and r > 0, we have N S N r a n r n r N a n r n = r a n r n. a n r n. So, as S N is increasing and upper bounded, S N 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 differentiable.. (0 %) Find all the antiderivatives of f on D(0, r). First of all, as g is analytic on D(0, r), we have, for z D(0, r), g a n (z) = n nzn = a n z n = a n z n = 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 a n F (z) = z 0 + g(z) = z 0 + n zn.
Reciprocally, it is immediate that all the analytic functions F z0 for z 0 C defined D(0, r) by a n F z0 (z) = z 0 + n zn are antiderivatives of f.