# Complex integration

• Sep 19th 2009, 04:23 AM
enjam
Complex integration
Hey guys, quick question. How would I go about showing that this following equation:

$\int_{|z|=3}^{}tan(z).dz$

equals $-4\Pi i$

...I've tried breaking it down to $sin(z) / cos(z)$, then finding the residues using the singularities that occur at $\Pi /2$ and $-\Pi /2$ but that just leads me to an answer of 0.

If anyone could show me how to do this using the residue theorem, I'd really appreciate it. Thanks.
• Sep 19th 2009, 08:37 AM
Opalg
Quote:

Originally Posted by enjam
Hey guys, quick question. How would I go about showing that this following equation:

$\int_{|z|=3}^{}tan(z).dz$

equals $-4\Pi i$

...I've tried breaking it down to $sin(z) / cos(z)$, then finding the residues using the singularities that occur at $\Pi /2$ and $-\Pi /2$ but that just leads me to an answer of 0.

If a function is of the form f(z)/g(z), and the denominator has a simple zero at $z_0$, then the residue there is given by $f(z_0)/g'(z_0)$. In this case, $f(z)/g'(z) = \sin z/(-\sin z) = -1$. So the residue at both poles is –1, giving the integral as $-4\pi i$.