Integral calculations using Cauchy's Integral formula

[Susanne217]

Homework Statement

Calculate the integral

Given $$\int_{C} \frac{e^z}{\pi i - 2z} dz = \int_{C} \frac{e^z}{z-\frac{\pi i}{2}} dz}$$

using Cauchy integral formula.

Homework Equations

What I know

$$\frac{1}{2\pi i} \int_{C} \frac{f(z)}{z-\zeta} = 2\pi i f(\zeta)$$

The Attempt at a Solution

This in my little girly mind amounts to

$$2\pi i f(\frac{\pi i}{2}) = \int_{C} \frac{e^z}{z-\frac{\pi}{2}i} dz \Rightarrow \int_{C} \frac{e^z}{z-\frac{\pi}{2}i} dz = 2 \pi \cdot (i) \cdot (i) = -2\pi$$

But people who are wiser than me says to me "Susanne your result is wrong!". Could someone please point out my mistake?

thanks Susanne

 

[Dick]

e^z/(i*pi-2z) isn't equal to e^z/(z-i*pi/2). It's equal to (-1/2)*e^z/(z-i*pi/2). You can't just drop the (-1/2) constant.

 

[Susanne217]

e^z/(i*pi-2z) isn't equal to e^z/(z-i*pi/2). It's equal to (-1/2)*e^z/(z-i*pi/2). You can't just drop the (-1/2) constant.

Dick,

You are a genius my man. I simply couldn't see that :D