Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Aaron on December 05, 2018, 12:18:26 AM

Can someone show me the details? I am not sure my process

Let f(z) = $\frac{z}{z^2+2z+5}\ , $then $z^2+2z+5=0$. We solve the equation, and we get $z= 1 \pm 2i$, only$z= 1+ 2i$ is the upper region.
Therefore, $Res(f, 1+2i)$ = $\dfrac{\sqrt{1+2i}}{(1+2i)(12i)} = \frac{\sqrt{1+2i}}{4i}\
$ We need to compute $\sqrt{1+2i}=a+ib$. We square both sides, and we get that $a = \sqrt{\frac{\sqrt{5}1}{2}} =b^ {1}$
Thus $I = Re(2\pi\ i$$\frac{a+ib}{4i}) = \frac{\pi\ a}{2}=\frac{\pi}{2}\sqrt{\frac{\sqrt{5}1}{2}} $
Hope it helps.

Since $x^2+2x+5$ is not even function one needs to use a "keyhole" contour