Cauchy's Integral Theorem - use partial fractions to solve integral?

1. Jul 14, 2013

dla

1. The problem statement, all variables and given/known data
Find the integral and determine whether Cauchy's Theorem applies. Use partial fractions.

$\large \oint \frac{tan \frac{z}{2}}{z^{4} -16} dz$ C the boundary of the square with vertices ±1, ±i cw

2. Relevant equations

3. The attempt at a solution
I just wanted to check if approach is correct. Since tan is not analytic at ±∏ which does not lie within the square, can I just take out the tan and solve using partial then multiply it back in?

I would have four terms since z^4-16 has four solutions but how would I evaluate each term. The integral I got is

$\huge - \oint \frac{tan \frac{z}{2}}{32(z+2)} + \oint \frac{tan \frac{z}{2}}{32(z-2)} - \oint \frac{itan \frac{z}{2}}{32(z+2i)} + \oint \frac{itan \frac{z}{2}}{32(z-2i)}$

I can't seem to evaluate it.

This section is before Cauchy's Integral Formula so I can't use that yet.. It only talks about Cauchy's Integral Theorem

Last edited: Jul 14, 2013
2. Jul 14, 2013

HallsofIvy

All you need is the theorem "the integral of an analytic function around a closed path is 0".
tan(z/2) is analytic for all z in the given square and the denominators are 0 only at 2, -2, 2i, and -2i. Again, those are all outside the square you are integrating over.