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Trouble with complex numbers

  1. Dec 23, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]z=1-i[/tex]
    [tex]e^{iz} = ?[/tex]
    I have to solve this problem and than picture it.
    2. Relevant equations



    3. The attempt at a solution
    [tex]e^{iz} =e^{i(1-i)}=e^{i+1}=e^i*e [/tex]
    I don't really understand how to picture this result. I assume their is an other way, in wich the result has a geometric meaning.
     
  2. jcsd
  3. Dec 23, 2013 #2
    Do you know how to graph a point in the polar form ##z = re^{i\theta}##?
     
  4. Dec 23, 2013 #3
    Yes I know! But there is "i+1" wich fustrate. But I know how to graph for example [tex]e^i[/tex]
     
  5. Dec 23, 2013 #4
    Does the order of resolution matter?

    If not, do the graphic first. z will be in the fourth quadrant with coordinates (1, -i)

    e^(iz)=e^[i(1-i)]=e.e^i

    edit: Think of e as the radius r.
     
    Last edited: Dec 23, 2013
  6. Dec 23, 2013 #5

    HallsofIvy

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    In the standard graph of the complex plane, the point (a, b) represents the complex number a+ bi. The complex number i+ 1 (= 1+ i, of course) is represented by the point (1, 1). That strikes me as being easier than [itex]e^i[/itex]!

    But it is not difficult to go from one to the other. The distance from (0, 0) to (a, b) is [itex]\sqrt{a^2+ b^2}[/itex] and the angle the line from (0, 0) to (a, b) makes with the x-axis is arctan(b/a). a+ bi is the same as [itex]\sqrt{a^2+ b^2}e^{arctan(b/a)}[/itex]. 1+ i has a= b= 1 so [itex]\sqrt{1^2+ 1^2}= \sqrt{2}[/itex] and [itex]arctan(1/1)= \pi/4[/itex]. [itex]1+ i= \sqrt{2}e^{\pi i/4}[/itex].
     
  7. Dec 23, 2013 #6
    Thanks!
     
  8. Dec 23, 2013 #7
    Sometimes the simplest, dumbest thing is the correct one.
     
  9. Dec 23, 2013 #8

    HallsofIvy

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    That's always been my plan!
     
  10. Dec 23, 2013 #9

    haruspex

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    Yes, but you're not suggesting it equals [itex]e^i[/itex] are you?
    terbed, do you know how to write [itex]e^{iθ}[/itex] in a + ib form? What does that give in this case?
     
  11. Dec 24, 2013 #10
    I don't really know your problem! Yes I can represent it in a+bi form.
     
  12. Dec 24, 2013 #11

    haruspex

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    I thought you were trying to get a handle on what ei looked like, so I expected at some point an answer in a+ib form. But I don't see that anywhere in the thread. What did you get for that?
     
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