Hi everyone!

This is my first post in this forum. I've only just picked up math again, after more then 10 years off the bench. Keep that in mind while you're digging through my solving. I'm studying complex analysis and I've just solved, or tried to solve some integration problems. I'd really like someone to look at my solutions, however typing in LaTeX was somewhat gnarly, but I think I got the essence down.

$\displaystyle \[ \oint_{\abs{\left|z\right|} < 1} \frac{1}{z^5 + 3z + 5}\,dz\] $

This one I used Rouche's to find out roots in the denominator. The root inside unit circle can be used to calculate the Residues. However, I couldn't find any zeros within the unit circle, with leads to Res f(z) = 0.

$\displaystyle \[\int_0^{\frac{\pi}{2}} {\frac{cos(2 \theta) } {1+2cos^2 \theta} }\,d\theta\]$

With this one I changed the limits to $\displaystyle 0 -> 2\pi$ Then I used $\displaystyle z = e^{i\theta}$ and $\displaystyle 2\cos^2(\theta) = 1+\cos(2\theta)$. This with Eulers formula, specially Cosines. Gave me:

$\displaystyle \frac{1}{4i}\[ \oint_{\abs{\left|z\right|} < 1} {\frac{z^2 + 1 } {z(z^2+4z+1) }\,dz\]$

I found two simple poles inside the unit circle, z = 0 and z = 2 - 3^0.5 Now I could calculate the residus with Hostpital's

$\displaystyle \[ Res f(z) = \|{\frac{z^2 + 1 } {(3z_k^2+8z_k+1) }\|_ {_{z_k = 0}^{z_k=2-\sqrt{3}}}$

Inserting figures.

$\displaystyle \frac{(2-\sqrt3)^2 + 1 } {(3(2-\sqrt3)^2+8(2-\sqrt3)+1) } +1$

Wish lead to, with a little help for wolframalpha

$\displaystyle \frac{77 + 2 \sqrt3}{61}$

Consequentially

$\displaystyle \frac{2\pi i}{4i} \frac{77 + 2 \sqrt3}{61}$ = $\displaystyle \pi \frac{77 + 2 \sqrt3}{122}$

The numbers feels a bit strange, but this is what I got.

Please, feel free to comment!