1. ## Compex integral evaluation

Evaluate the integral around C of [(e^3z)/(z+pi/2)] dz if C is the circle |z| = 5.

2. Originally Posted by kiwijoey
Evaluate the integral around C of [(e^3z)/(z+pi/2)] dz if C is the circle |z| = 5.

$\displaystyle f(z)$ has a simple pole at $\displaystyle -\frac{pi}{2}$

so the residue is $\displaystyle \lim_{z}^{-\frac{\pi}{2}}\left( z+\frac{\pi}{2}\right)f(z)=e^{-\frac{3\pi}{2}}$

so

$\displaystyle \oint_{|z|=5} \frac{e^{3z}}{z+\frac{\pi}{2}}dz=2\pi i e^{-\frac{3\pi}{2}}=2\pi i (\cos\left(\frac{3\pi}{2} \right)+i\sin\left(\frac{3\pi}{2} \right))=2\pi$

3. $\displaystyle e^{-\frac{3\pi}{2}} \neq \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})$

4. Originally Posted by TheEmptySet

$\displaystyle \oint_{|z|=5} \frac{e^{3z}}{z+\frac{\pi}{2}}dz=2\pi i e^{-\frac{3\pi}{2}}=2\pi i (\cos\left(\frac{3\pi}{2} \right)+i\sin\left(\frac{3\pi}{2} \right))=2\pi$
You can also use Cauchy's theorem:
Let $\displaystyle f(z) = e^{3z}$ then $\displaystyle \frac{1}{2\pi i}\oint_{|z|=5} \frac{f(z)}{z+\frac{\pi}{2}} dz = f ' \left( -\frac{\pi}{2} \right)$

5. Is this Cauchy's int.theorem - where alpha is -pi/2 - oh Sorry new to this and as I type this I see you already have an answer! Glad to see I was on the right track. Good luck! Ruth