# series expansion

• November 30th 2009, 04:22 AM
bigdoggy
series expansion
how would i write $f(z)=\frac {(z^2 -1)^ {1 \over 2}}{z^2 + 1}$ as a series expansion?
• November 30th 2009, 10:29 PM
redsoxfan325
Quote:

Originally Posted by bigdoggy
how would i write $f(z)=\frac {(z^2 -1)^ {1 \over 2}}{z^2 + 1}$ as a series expansion?

$\frac{1}{1+z^2}=1-(z^2)+(z^2)^2-(z^2)^3+...$

so

$f(z)=\frac{(z^2 -1)^{1/2}}{z^2 + 1}=\sqrt{z^2-1}\bigg(1-z^2+z^4-z^6+...\bigg)$

Is that what you're looking for?
• December 1st 2009, 01:34 AM
bigdoggy
thanks. i'm trying to find the value of the integral of that function around a circle of large radius. The function is complex and has a branch cut in the section [-1,1] of the real axis.
So I need to figure out how to express that function as a series expansion to begin with...