### Author Topic: 2.6 Q14  (Read 1819 times)

#### Tunan Jia

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##### 2.6 Q14
« on: December 01, 2018, 05:06:45 PM »
can anyone help with this question?

#### Zoran

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##### Re: 2.6 Q14
« Reply #1 on: December 01, 2018, 05:27:35 PM »
attached is my scanned answer, please check. Hopefully it can help you

#### hanyu Qi

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##### Re: 2.6 Q14
« Reply #2 on: December 01, 2018, 06:53:08 PM »
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( $2 \pi i \frac{a+ib}{4i})$ = $\frac{\pi}{2}$ $\sqrt{\frac{\sqrt5 -1}{2}}$
« Last Edit: December 01, 2018, 06:54:48 PM by Alex Qi »