# Finding poles of complex functions

1. May 18, 2015

### Kitten

I am trying to calculate a pole of f(z)=http://www4b.wolframalpha.com/Calculate/MSP/MSP86721gicihdh283d613000033ch4ae4eh37cbd4?MSPStoreType=image/gif&s=35&w=44.&h=40. [Broken] . The answer in the text book is:

Simple pole at http://www4f.wolframalpha.com/Calculate/MSP/MSP2485217c56eb3b13h612000056di60dga07378cd?MSPStoreType=image/gif&s=14&w=52.&h=26. [Broken] and a simple pole at http://www4f.wolframalpha.com/Calculate/MSP/MSP2292204iia97b59e96db00001591c4be13e5ge82?MSPStoreType=image/gif&s=19&w=66.&h=24. [Broken]

Trying to get to the same solution:

So far I have that z=±√ i

I need to find this pole in terms of z=re^iθ so I’m trying to plot this on an argand diagram but I don’t know how.

Since z=x+iy, I assumed that the real part is 0 and the imaginary part is ±1 but plotting this gives me
θ/2 and not over 4 which is required.

Last edited by a moderator: May 7, 2017
2. May 18, 2015

### fzero

Complex numbers are kind of funny. What is the real part of $(1+i)^2$?

3. May 18, 2015

### Kitten

the real part would be 0? Since (1+I)^2= 1+2i+i^2= 0+2i so the real part is 0?

4. May 18, 2015

### fzero

Correct. So the square root of an imaginary number can have a nonzero real part. Can you figure out any way to use this in your problem?

5. May 19, 2015

### hunt_mat

Note that $-1=e^{i\pi}$

So poles occur when the denominator is zero: $z^{4}-e^{i\pi}=0$

6. May 21, 2015

### HallsofIvy

Staff Emeritus
You say that you have $z= \pm\sqrt{i}$. Surely you know that the simplest way to find square roots (or other roots) or complex numbers is to put them in polar form- if $z= re^{i\theta}$, then $\sqrt{z}= \sqrt{r}e^{i\theta/2}$? "i" has absolute value 1 and argument $\pi/2$ so that $i=.(1)e^{i\pi/2}$. So the square root of i is $\sqrt{1}e^{i\pi/4}= e^{i\pi/4}= cos(\pi/4)+ i sin(\pi/4)= \frac{\sqrt{2}}{2}(1+ i)$. Of course, we can always add $2\pi$ to the argument without changing the number, $e^{i\pi/2}= e^{i\pi/2+ 2i\pi}= e^{3i\pi/2}$, and so another square root is $e^{i3\pi/4}= cos(3\pi/4)+ i sin(3\pi/4)= -\frac{\sqrt{2}}{2}(1+ i)$

Last edited by a moderator: May 21, 2015