Complex Integral $\int_{C(0,2)} \frac{e^z}{i\pi -2z}$

Question

Evaluate: $\int_{C(0,2)} \frac{e^z}{i \pi -2z}dz$

So using the Cauchy Integral Formula: $\int_{C} \frac{f(z)}{z-z_0} = 2\pi i f(z_0)$

If I define $f(z)= \frac{e^z}{-2}$ then the $z-z_0=\frac{-i\pi}{2} + z$

Hence $z_0=\frac{i\pi}{2}$

So $\int_{C(0,2)} \frac{e^z}{i \pi -2z} dz= 2\pi i f(z_0)$ becomes:

$2\pi i (\frac{e^\frac{i\pi}{2}}{2})$

Where $\frac{e^\frac{i\pi}{2}}{2} = \frac{1}{2i}$

Hence

$\int_{C(0,2)} \frac{e^z}{i \pi -2z}dz= \frac{2\pi i}{2i} = \pi$

Is this correct as I think it should only hold if $z_0 \in C(0,2)$ and I'm not certain this is true in this case. Any help would be great.

Answer 1

Let $f(z)=-\frac{e^z}2$. Then\begin{align}\int_{C(0,2)}\frac{e^z}{i\pi-2z}\,\mathrm dz&=\int_{C(0,2)}\frac{f(z)}{z-\pi i/2}\,\mathrm dz\\&=2\pi if\left(\frac{\pi i}2\right)\text{ (because $\left|\frac{\pi i}2\right|<2$)}\\&=-2\pi i\frac i2\\&=\pi.\end{align}