Toronto Math Forum

MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Tunan Jia on December 01, 2018, 05:06:45 PM

Title: 2.6 Q14
Post by: Tunan Jia on December 01, 2018, 05:06:45 PM
can anyone help with this question?
Title: Re: 2.6 Q14
Post by: Zoran on December 01, 2018, 05:27:35 PM
attached is my scanned answer, please check. Hopefully it can help you
Title: Re: 2.6 Q14
Post by: hanyu Qi 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}}$