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Write i^(2i) in the form a + bi?

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Write the expression i2i in the form a + bi

    2. Relevant equations

    Honestly we haven't treated such subjects during the classes, but I've made some researches and found the Euler identity might help me.

    3. The attempt at a solution

    By using the Euler identity, I found that i = ei.(π/2), so i2i = e (I skip a few steps). Now this looks great! But it is not in the form a + bi, as the problem requires. Any idea?


    Thank you very much in advance for your answers.

    PS: Next question is the same for (-1 + i)(1-i) :DD


    J.
     
  2. jcsd
  3. Nov 3, 2015 #2

    Mark44

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    ##e^{-\pi}## is a real number, so you could write it as ##e^{-\pi} + 0i##

    For the other question, start by writing -1 + i in polar form; i.e., as ##r(\cos(\theta) + i \sin(\theta))##
     
  4. Nov 3, 2015 #3
    Ahaaa that didn't occur to me. Thank you very much for your help!

    For (-1 + i)(1 - i), I did that:

    (-1 + i)(1 - i) = (-1)(1 - i) + (ei(π/2))(1 - i)
    = -1 + e(1 - i)⋅i⋅(π/2)
    = -1 + ei - i2⋅(π/2)
    = -1 + ei + π/2

    Now I am failing again at bringing the i down.
     
  5. Nov 3, 2015 #4

    HallsofIvy

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    [itex]e^{a+ b}= e^ae^b[/itex] and [itex]e^{i}= cos(1)+ i sin(1)[/itex].
     
  6. Nov 3, 2015 #5

    SammyS

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    What allows you to "distribute" the exponent this way?
     
  7. Nov 3, 2015 #6

    Mark44

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    For -1 + i, which is in rectangular or Cartesian form, what is r? What is the angle ##\theta##?
    ##(a + b)^m \ne a^m + b^m##
    [/quote]
     
  8. Nov 3, 2015 #7
    I'm afraid I'm a bit clueless about your questions. I wonder why our teacher is giving us such homework before we even mentioned complex numbers, that is just frustrating...

    But I guess the angle with the horizontal axis is 135° (-1 is the real axis, and +1 is the imaginary axis right?) and r...well, maybe √((-1)2 + 12)?


    Thank you for your help in any case, I appreciate it.
     
  9. Nov 3, 2015 #8

    Mark44

    Staff: Mentor

    Or in radians, ##3\pi/4##.
    No, the real axis is the horizontal axis, and the imaginary axis is the vertical axis. Or maybe you meant that -1 is on the real axis (to the left of the origin), and +1i is on the imaginary axis (above the origin).
    Yes. So -1 + i = ##\sqrt{2}(\cos 3\pi/4 + i \sin 3\pi/4) = \sqrt{2}e^{i3\pi/4}##
     
  10. Nov 3, 2015 #9
    Thank you for your answer. I did not know (or remember from school, it's been a while) about such properties of e actually. I attempted to follow your indications, but I got stuck again with the power:

    (-1 + i)(1 - i) = [√2(cos 3π/4 + i sin 3π/4)](1 - i)
    = (√2 ei 3π/4)(1 - i)

    I randomly feel like there might be a supplementary step with √2 ei 3π/4 to introduce i, but I can't find anything on internet or in my books. I'm sorry to not have been able to solve the problem on my own yet, I keep on trying though :DD


    J.
     
  11. Nov 3, 2015 #10
    Is there a way to write √2 with an imaginary number for example?
     
  12. Nov 3, 2015 #11

    haruspex

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    You don't want to do that. In your first problem, i2i, what did you do to the first 'i' in order to make progress? Can you do something similar to the √2?
     
  13. Nov 6, 2015 #12
    I still can't figure out, i've been trying all week!

    Well I expressed i in terms of e, and that luckily dissolved the i that was stuck inside the power. I tried to "isolate" √2, but no matter what I do there is a i remaining up there. That's where I'm stuck so far:

    (-1 + i)(1 - i) = (√2)(1 - i)e(i ⋅ 3π/4 + 3π/4)
     
  14. Nov 6, 2015 #13

    haruspex

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    How can you write √2 as a power of e?
     
  15. Nov 6, 2015 #14

    Mark44

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    This doesn't work. ##(-1 + i) \ne \sqrt{2}##
    Use my hint of earlier this week
    You have found r, which is ##\sqrt{2}## and you found the angle, ##\theta##, which is ##3\pi/4##.

    Euler's formula says that ##(\cos(\theta) + i \sin(\theta)) = e^{i\theta}##, from which we can derive ##r(\cos(\theta) + i \sin(\theta)) = re^{i\theta}##.
     
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