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Thread: Modulus of the square of a complex number ??!?

  1. #1
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    Modulus of the square of a complex number ??!?

    Hi, I've been asked to show the following for a complex number z
    Modulus of the square of a complex number ??!?-screenshot-2019-03-28-17.46.51.png

    Attempt:

    If I let z = a +bi and do the algebra I end up with

    a2 - b2 +2abi / a2 + b2 = 1

    And I can't see how this could always be true?

    If I follow this through I get ai = b ?

    Help would be greatly appreciated this is driving me nuts! thanks in advance :-)
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  2. #2
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    Re: Modulus of the square of a complex number ??!?

    in general this isn't true which should be fairly obvious as $z^2$ can have an imaginary part whereas $|z^2|$ is always real (and non-negative)
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  3. #3
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    Re: Modulus of the square of a complex number ??!?

    I suspect that the problem was supposed to be to show that $\displaystyle \frac{|z|^2}{|z^2|}= 1$.
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  4. #4
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    Re: Modulus of the square of a complex number ??!?

    This has sat for a couple of days now so: writing z= a+ bi, $\displaystyle |z|= \sqrt{a^2+ b^2}$ so $\displaystyle |z|^2= a^2+ b^2$. On the other hand $\displaystyle z^2= (a+ bi)^2= a^2- b^2+ (2ab)i$ so $\displaystyle |z^2= \sqrt{(a^2- b^2)^2+ 4a^2b^2}= \sqrt{a^4- 2a^2b^2+ b^4+4a^2b^2}$$\displaystyle = \sqrt{a^4+ 2a^2b^2+ b^4}= \sqrt{(a^2+ b^2)^2}= a^2+ b^2$ also.

    Almost trivial would be to write z in polar form: $\displaystyle z= re^{i\theta}$. Then |z|= r so that $\displaystyle |z|^2= r^2$ and $\displaystyle z^2= r^2e^{2i\theta}$ and $\displaystyle |z^2|= r^2$ also.
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