Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 3G (Complex Analysis or Complex Methods)

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- Solution 3G

Solution

The Taylor series of the complex sine is

sin z = z - \frac{z ^{3}}{3 !} + \frac{z ^{5}}{5 !} - \dots .

Therefore, for

z \neq = 0

,

\frac{s i n z}{z} = 1 - \frac{z ^{2}}{3 !} + \frac{z ^{4}}{5 !} - \dots,

and the right side defines a holomorphic function at

z = 0

with value

1

. Thus the apparent singularity is a removable singularity.

The resulting power series is

\sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} z ^{2 n}}{( 2 n + 1 )!} .

The ratio test shows convergence for every

z \in C

, so its radius of convergence is

\infty .

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