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Homework Help: Complex numbers

  1. Apr 11, 2010 #1


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    1. The problem statement, all variables and given/known data
    If [tex]z=6e^{3i}[/tex] then find the exact answer for [tex]Re(z^4)[/tex]

    3. The attempt at a solution
    What I'm having trouble with is the fact that it's not in the usual form [tex]z=re^{i\theta}[/tex] where [itex]\theta[/itex] is some multiple of [itex]\pi[/itex]. So I guess in a way I'm dealing with a not so nice answer.

    Anyway, [tex]z^4=6^4e^{12i}[/tex]

    Now restrict the radian angle between [tex]-\pi<\theta\leq \pi[/tex] we take away [itex]4\pi[/itex]. So our angle is now [itex]12-4\pi[/itex].

    For Re(z4) I suppose we take [tex]6^4cos(12-4\pi)[/tex]

    Is this the exact answer I'm looking for?

    Oh and while I was working on this, I tried to go down this road and can't figure out why it's wrong:

    [tex]e^{3i}=\left(e^{2\pi i}\right)^{\frac{3}{2\pi}}=1^{\frac{3}{2\pi}}=1[/tex]

    Of course this is not correct since the answer to the original expression is not 1. May anyone shed some light on this?
  2. jcsd
  3. Apr 11, 2010 #2
    [tex]z^4=6^4 e^{12i} [/tex]

    [tex]z^4 = 1296 (\cos 12 + i \sin 12) [/tex]

    Also, if [tex] e^{i\theta}=\cos \theta + i \sin \theta [/tex] One thing you know is there exists a value for theta whereby [tex]e^{i\theta} \ne 1 [/tex]. Seems obvious?

    Call this angle [tex] \phi [/tex]

    Using your reasoning, [tex] e^{i\phi} \ne 1 [/tex]

    [tex] (e^{2\pi i})^{\phi \over 2\pi} = e^{i\phi} \ne 1 [/tex]

    [tex] (e^{2\pi i})^{\phi \over 2\pi} = 1^{\phi \over 2\pi} = 1 [/tex]


    [tex] 1 \ne 1 [/tex] ?
    Last edited: Apr 11, 2010
  4. Apr 11, 2010 #3

    Gib Z

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    The radian angle restriction only applies when we wish to give the Argument of a complex number without ambiguity. Even though it may seem like for eg z=1, Arg z = 0, 2pi, 4pi... all seem like reasonable answers, to make the argument a function we choose the convention to always take Arguments to be in the range you stated. This is because the actual values 0, and 2pi, etc, are different.

    It is not necessary however, to make the Argument in that form inside every expression. The simplest way to express your answer is in fact 6^4 cos(12), which is numerically the same as the answer you gave but simplified a bit.

    The last equality doesn't directly follow, eg 1^(1/4) = 1, -1, i, -i. If you look at all solutions to 1^(3/2pi) you will find your original number as one of them. Also, in C where many n-th roots exist for its elements, its no longer strictly correct to write down eg 1 = sqrt 1, as 1 is just 1 solution of sqrt 1.
  5. Apr 11, 2010 #4


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    Ahh so it's just the ambiguous case of when to take n roots and when to take the principal root. Such as how you said [tex]1^{1/4}=1,-1,i,-i[/tex], it's like solving the polynomial z4=1.

    Well, obviously! :tongue:

    Thanks for the help!
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