MAT334-2018F > MAT334--Lectures & Home Assignments

2.6 #14

(1/1)

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

yunhao guan:
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)-(-1-2i)} = \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.

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