Evaluate the integral over the circle

Question

Suppose $\gamma = {z: |z|=1}$. I'm interested in evaluating the integral $\int_\gamma \frac{e^z}{2z+1}dz$. The first step I do is to take out 1/2: $\int_\gamma \frac{e^z}{2z+1}dz=\frac{1}{2}\int_\gamma \frac{e^z}{z+1/2}dz$ Could someone please help me to understand why it is equal to $\frac{1}{2}2\pi ie^{-\frac{1}{2}}$?

Answer 1

Let $f(z) = e^z$, $z_0 = -1/2$ and $\gamma$ such as above. Using the Cauchy integral formula you get

$$ \operatorname{ind}_{\gamma}(z_0) f(z_0) = \frac{1}{2\pi \mathrm i} \int_\gamma \frac{f(z)}{z - z_0} dz.$$

Applying this to your situtation, you arrive at

$$ \frac{1}{2}\int_\gamma \frac{e^z}{z+1/2}dz = \frac{1}{2}\left(2\pi \mathrm i \cdot e^{-1/2}\right). $$