Complex Analysis - Partial Fractions and Entire fucntion

**lpd** · September 13th 2010, 10:42 PM

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

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

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

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

b) Hence show that, if $f$ is an entire function and $R> |a|,|b|$ then
$$ <br /> \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
$$ \int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz$

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

So, $$ \int_{|z|=3} \frac{exp(sin\frac{\pi z}{2})}{(z-1)^2(z-2)}\,dz=<br /> <br /> 2\pi i(\frac{0-exp(1)}{(exp(1)-0)^2} +\frac{0}{exp(1)-0})<br /> =-\frac{2\pi *i*exp(1)}{exp(2)}$

d) Calculate
$$ \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?

**lpd** · September 14th 2010, 02:41 AM

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

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