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Find all solutions of given equation and draw them

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Find all solutions od ##e^z=2+i## and draw at least three of them.


    2. Relevant equations



    3. The attempt at a solution

    ##e^z=2+i##

    ##z=log|2+i|+2\pi in+iarctan(\frac{1}{2})##

    Now how on earth do I draw any of them?

    I guess the length is ##log|2+i|##. Ok? What about everything else?
     
  2. jcsd
  3. Apr 2, 2014 #2

    Ray Vickson

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    Write ##2+i## in polar form.
     
  4. Apr 2, 2014 #3
    Ok...

    That gives me only one solution.
     
  5. Apr 2, 2014 #4

    haruspex

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    No, combine it with the other terms. You should everything now in the form x+iy, and the y term should include an unknown n.
     
  6. Apr 3, 2014 #5
    Aaargh! -.-

    ##2+i=\sqrt{5}e^{iarctan(1/2)+2\pi i n}##

    This naturally has to be wrong because for every ##n \in \mathbb{N}## the solution is the same. There has to be something I don't get it here...
     
  7. Apr 3, 2014 #6

    ehild

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    So what are the real and imaginary parts of z?


    ehild
     
  8. Apr 3, 2014 #7
    ##log|2+i|## is real and the rest is imaginary.
     
  9. Apr 3, 2014 #8

    ehild

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    What are the real and imaginary parts of z in numbers when n=0, when n=1, when n=-1 ....?
     
  10. Apr 3, 2014 #9
    ##x=log|2+i|## and
    ##y=arctan(\frac{1}{2}) +2\pi n##

    Where ##x## is real part and ##y## is imaginary part. One question here: Is ##n \in \mathbb{N}## or is ##n \in \mathbb{Z} ##?

    Ok, ##x## is not a function of ##n## therefore it stays the same for all ##n##, while ##y##:

    if ##n=0##: ##y=arctan(\frac{1}{2})##
    if ##n=1##: ##y=arctan(\frac{1}{2}) +2\pi##
    if ##n=-1##: ##y=arctan(\frac{1}{2}) -2\pi##
     
  11. Apr 3, 2014 #10

    ehild

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    What are the values? Can you draw these complex numbers which are solutions of the original equation?

    ehild
     
  12. Apr 3, 2014 #11
    I don't want to be rude, but that is why I am here, because I can't draw them ...
    What is the value of ##log|2+i|##? How does the solution for ##n=0## differ from the one for ##n=1## - to me it seems to be the same point or is ##2\pi ## suddenly not the whole circle around origin?
     
  13. Apr 3, 2014 #12

    ehild

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    Do you know how to get the absolute value of a complex number z=u+iv? ##|z|=\sqrt{u^2+v^2}## . So |2+i|=√(22+1)=√5=2.236.. . You can find its logarithm, it will be x, the real part of z.
    You can find arctan(0.5) with your calculator (shift tan 0.5) but set the calculator to radians first.
    It should be about 0.46.
    You certainly know that pi=3.14. Add it to .46 to get the other y value, and subtract from 0.46, to get the third one.
    You have got three complex numbers, with the same x(real part) and three different y (imaginary parts). Plot these points on the complex plane.

    Adding 2πi to a complex number will change it. It is not the same as adding 2πi to the exponent.

    ehild
     
    Last edited: Apr 3, 2014
  14. Apr 3, 2014 #13
    Yes, I completely get that. :)

    What confuses me is that all solutions (for arbitrary ##n##) are on the same vertical line at ##x=log(\sqrt{5})##. I don't know why, but everybody keep mentioning periodicity which to me is not so obvious. How is anything on a straight line ever periodic?

    EDIT: Perhaps, this is the question I should have started this topic with. I keept looking for some kind of polar form which would explain my interpretation of periodicity by adding ##2\pi ## in the exponent.
     
  15. Apr 3, 2014 #14

    ehild

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    I guess you have the n-th roots in your mind. Say, you have to determine the 3rd roots of z=8eiπ/3. One will be 2eiπ/9. The second root results when adding i 2π/3 to the exponent: 2ei(π/9+2π/3). The third one is 2ei(π/9+4π/3). Adding i2π/3 to the exponent again does not result in a new root: you get back the first one.

    ehild
     
  16. Apr 3, 2014 #15
    Mhm. Ok, that explains everything than.

    Thank you ehlid!
     
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