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Homework Help: Three problems on Complex Analysis

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data

    1)Show that [itex](1-i)^{2}=-2i[/itex] then evaluate [itex](1-i)^{2004}+(1-i)^{2005}[/itex]

    2)Prove that every complex number with moduli 1, except z=1, can be put in the form [itex]\frac{a+i}{a-i}[/itex]

    3)Let m and n be positive integers without a common factor. Define [itex]z^{m/n}=(z^{m})^{1/n}[/itex], and show that [itex]z^{m/n}=(z^{1/n})^{m}[/itex]

    3. The attempt at a solution

    1) I solved [itex](1-i)^{2}=-2i[/itex] by expanding, and then solved in the polar form, that is, [itex](1-i)^{2}=(\sqrt{2}(cos(-pi/4)+i*sin(-pi/4))^{2}=2(cos(-pi/2)+i*sin(-pi/2)=\sqrt{2}(1-i)[/itex], hence, [itex]\sqrt{2}(1-i)=-2i[/itex]. Then I stopped because something's wrong, isn't it?

    2)Let z=x+yi. Letting [itex]x+yi=\frac{a+i}{a-i}[/itex] I can obtain the following equations: [itex]xa+y=a; ya-x=1[/itex], which, solved simultaneously, yields [itex]a=\sqrt{\frac{1+x}{1-x}}[/itex], then I got stuck.

    3)I've just used DeMoivre's with z in polar form and particular properties of real numbers to invert the powers, but idk if that's correct.

    Many Thanks
  2. jcsd
  3. Aug 31, 2012 #2


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    Yes, something is wrong.

    cos(±π/2) = 0
  4. Aug 31, 2012 #3


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    For 1 , you can look at multiplication as multiplying the respective moduli and

    adding the arguments (mod 2pi if necessary). For 2, it seems you can just solve

    for a in a single equation, to have a in terms of x,y ; after all, z seems to be given

    in terms of x and y.
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