# Thread: Complex Analysis - Partial Fractions and Entire fucntion

1. ## Complex Analysis - Partial Fractions and Entire fucntion

Hi. I'm a bit stuck on this problem.

a) Find the partial fraction decomposition of $\displaystyle \frac{1}{(z-a)^2(z-b)}$

Let $\displaystyle \frac{1}{(z-a)^2(z-b)} = \frac{A}{(z-a)} +\frac{B}{(z-a)^2}+\frac{C}{(z-b)}$

After some working and expanding and simplifying i get the following:
$\displaystyle A=-\frac{1}{(b-a)^2}=-\frac{1}{(a-b)^2}$
$\displaystyle B=\frac{1}{a-b}$
$\displaystyle C=\frac{1}{(b-a)^2}=\frac{1}{(a-b)^2}$

b) Hence show that, if $\displaystyle f$ is an entire function and $\displaystyle R> |a|,|b|$ then
$\displaystyle$
\int_{|z|=R} \frac{f(z)}{(z-a)^2(z-b)}\,dz= 2\pi i(\frac{f(b)-f(a)}{(a-b)^2} +\frac{f'(a)}{a-b})$I can show this. Not hard. Just applying Cauchy's Integral Forumlae. c) Calculate$\displaystyle $\int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz$

$\displaystyle f(z)=exp(sin\frac{\pi z}{2})$
$\displaystyle f(1)=exp(sin\frac{\pi}{2})=exp(1)$
$\displaystyle f(2)=exp(sin\frac{2\pi}{2})=0$
$\displaystyle f'(z)=\frac{1}{2}cos(\frac{\pi z}{2})exp(sin\frac{\pi z}{2})$
$\displaystyle f'(1)=0$

So, $\displaystyle$ \int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz=

2\pi i(\frac{0-exp(1)}{(exp(1)-0)^2} +\frac{0}{exp(1)-0})
=-\frac{2\pi *i*exp(1)}{exp(2)} $d) Calculate$\displaystyle $\int_{|z|=2} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-3)}\,dz$
Does this just equal to zero because one of the roots is outside |z|=2? Did I put it in correct termiology?

2. edit: I solved it. don't worry. and d) doesn't equal to zero. thanks.