Calculate the integral $\oint_{C(0;2)}\frac{e^{2\pi z}-1}{z(z-i)}dz$

Question

Calculate the integral $$\oint_{C(0;2)}\frac{e^{2\pi z}-1}{z(z-i)}dz$$ where $C(0;2)$ is the circle with center $0$, radius $2$ and with positive direction of rotation.

$$$$

Do we have to write the function as a sum of two functions?

$$\frac{e^{2\pi z}-1}{z(z-i)}=\frac{A}{z} + \frac{B}{z-i}\Rightarrow e^{2\pi z}-1=A(z-1)+Bz=(A+B)z-A$$

But how can we continue from here?

Or is this approach wrong?

$$$$

EDIT :

We have that $$\frac{e^{2\pi z}-1}{z(z-i)}=\frac{\frac{e^{2\pi z}-1}{z}}{z-i}$$ There is a singularity at $z=i$.

Let $f(z)=\frac{e^{2\pi z}-1}{z}$.

Then we calculate $$\text{Res}_{z=i}2\pi i(f'(z))\mid_{z=i}$$

Is that correct so far?

Answer 1

You cannot do partial fraction decomposition here because of the term $e^{2\pi z}$. You only do that for rational functions.

Let $\displaystyle f(z)=\frac{e^{2\pi z}-1}{z}$. Note that $z=0$ is a removable singularity of $f$. By the Cauchy integral formula, $$ \oint_{C(0;2)}\frac{f(z)}{z-i}\;dz = 2\pi i f(i)=0 $$

Alternatively, you can consider the function $\displaystyle g(z)=\frac{e^{2\pi z}-1}{z(z-i)}$.

$\textrm{Rez}(g;0)=0$ since $0$ is a removable singularity of $g$.

$\textrm{Rez}(g;i)=\lim_{z\to i}(z-i)g(z)=0$.

By the Residue Theorem, $$ \oint_{C(0;2)} g(z)\; dz=0 $$