Consider $ f(z) = \frac{\sqrt z}{z^2 + 2z +5}$
$z^2 + 2z +5 \implies $z= -1+2i or -1-i2.
Only z=-1+2i is up.
Res(f,-1+2i) = $ \frac{\sqrt {-1+2i}}{(-1+2i)-(-1-2i)} $= $\frac{\sqrt {-1+2i}}{4i}$
We compute $ \sqrt{-1+2i}$ = a+ib. There are two solutions, but we must choose only the one whose argument is hafl of the grgument of -1+2i.
$a^2$ - $b^2$ = -1 ab = 1
a = $ \sqrt{\frac{\sqrt5 -1}{2}} $ = $ \frac{1}{b} $
I = re( $2 \pi i \frac{a+ib}{4i})$ = $ \frac{\pi}{2}$ $ \sqrt{\frac{\sqrt5 -1}{2}}$
