Verity of Residue theorem of [0,2pi]

Question

After I turn $$ cos\theta=\frac12(z+\frac1{z})$$and $$ d\theta=\frac1{iz}dz$$ the denominator become a mess $$ \frac{dz}{(a^2+\frac{b^2}4(z^2+2+\frac1{z^2})+\frac{ab}2(z+\frac1z))(iz)}$$ How can a find out the pole? enter image description here

You need to do the algebra. Multiply by $z^2/z^2$ and factor. — , Nov 27 '13 at 19:32
Then it become something like this $$\frac{4zdz}{i(b^2z^4+2abz^3+(2ab+4a^2)z^2+2abz+b^2)} $$ it is a mess again — user111605, Nov 27 '13 at 19:41

score 1 · Answer 1 · 2013-11-27T20:35:42.713

1

If you bring a $z^2$ in to the denominator factor without expanding the squared quantity, you can get something like

$$\frac{z}{\left( a z + \frac{b}{2}(z^2+1)\right)^2}$$

Factor the quadratic term to determine the poles (of order 2) and use the residue theorem.

edited Nov 27 '13 at 20:35

answered Nov 27 '13 at 20:20

Verity of Residue theorem of [0,2pi]

1 Answers1