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Math Help - Complex Analysis argument question

  1. #1
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    Complex Analysis argument question

    Could someone help me on this one.
    Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.
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    Quote Originally Posted by frankdent1 View Post
    Could someone help me on this one.
    Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.
    If z\not = 0 then write z = |z|e^{\theta i} where \theta = \arg (z). This means z^n = |z^n|e^{n\theta i} and that means n\theta is the \arg(z^n) (up to a factor of 2\pi).
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    Would you care to elaborate on the last part?
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    Quote Originally Posted by frankdent1 View Post
    Would you care to elaborate on the last part?
    I am saying if \theta is the argument of a non-zero complex number then we can write z = |z|e^{i\theta} (and conversely). Now if we raise both sides to the n we get z^n = |z|^n \left( e^{i\theta} \right)^n but |z|^n = |z^n| and \left( e^{i\theta} \right)^n = e^{ni\theta} by de Moivre theorem. Thus, we can write z^n as |z^n|e^{ni\theta} and so ni\theta is the argument of z^n, i.e. n\theta = \arg (z^n) thus n\arg (z) = \arg (z^n). I said "up to a factor of 2\pi" because as you know there can be many different argument differening by 2\pi.
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    Quote Originally Posted by ThePerfectHacker View Post
    If z\not = 0 then write z = |z|e^{\theta i} where \theta = \arg (z). This means z^n = |z^n|e^{n\theta i} and that means n\theta is the \arg(z^n) (up to a factor of 2\pi).
    The red word factor is probably not the word I'd use since it implies multiplication of the multiples of 2 pi, whereas you are adding the multiples of 2 pi.

    Quote Originally Posted by frankdent1 View Post
    Would you care to elaborate on the last part?
    If you think of z^n as representing some complex number (1 + i, say) and then think of an Argand diagram showing 1 + i, you should be able to that adding (or subtracting) multiples of 2 pi 'keeps you' at 1 + i. Capisce?
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