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Finding poles of complex functions

  1. May 18, 2015 #1
    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. jcsd
  3. May 18, 2015 #2

    fzero

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    Complex numbers are kind of funny. What is the real part of ##(1+i)^2##?
     
  4. May 18, 2015 #3
    the real part would be 0? Since (1+I)^2= 1+2i+i^2= 0+2i so the real part is 0?
     
  5. May 18, 2015 #4

    fzero

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    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?
     
  6. May 19, 2015 #5

    hunt_mat

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    Note that [itex]-1=e^{i\pi}[/itex]

    So poles occur when the denominator is zero: [itex]z^{4}-e^{i\pi}=0[/itex]
     
  7. May 21, 2015 #6

    HallsofIvy

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    You say that you have [itex]z= \pm\sqrt{i}[/itex]. 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 [itex]z= re^{i\theta}[/itex], then [itex]\sqrt{z}= \sqrt{r}e^{i\theta/2}[/itex]? "i" has absolute value 1 and argument [itex]\pi/2[/itex] so that [itex]i=.(1)e^{i\pi/2}[/itex]. So the square root of i is [itex]\sqrt{1}e^{i\pi/4}= e^{i\pi/4}= cos(\pi/4)+ i sin(\pi/4)= \frac{\sqrt{2}}{2}(1+ i)[/itex]. Of course, we can always add [itex]2\pi[/itex] to the argument without changing the number, [itex]e^{i\pi/2}= e^{i\pi/2+ 2i\pi}= e^{3i\pi/2}[/itex], and so another square root is [itex]e^{i3\pi/4}= cos(3\pi/4)+ i sin(3\pi/4)= -\frac{\sqrt{2}}{2}(1+ i)[/itex]
     
    Last edited by a moderator: May 21, 2015
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