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Thread: Find solutions of z bar = z^2

  1. #16
    Grand Panjandrum
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    Im trying find all solutions for: $\displaystyle \bar{z}=z^2$
    Work in polar form.

    $\displaystyle \overline{z}=z^2$,

    so:

    $\displaystyle |\overline{z}|=|z|=|z^2|=|z|^2$

    so for non-zeros solutions $\displaystyle |z|=1$, and $\displaystyle z=e^{i\theta}$ for some $\displaystyle \theta \in [0,\2\pi)$

    Then:

    $\displaystyle \overline{z}=e^{-i\theta}=z^2=e^{2i\theta} $

    etc

    CB
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  2. #17
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    Re: Find solutions of z bar = z^2

    Can someone explain how $\displaystyle |\bar{z}|$ gives you $\displaystyle |z|^2$ ? I can see that $\displaystyle z\bar{z} =|z|^2$ but how does $\displaystyle |z| = |z|^2$ ?
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  3. #18
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    Can someone explain how $\displaystyle |\bar{z}|$ gives you $\displaystyle |z|^2$ ? I can see that $\displaystyle z\bar{z} =|z|^2$ but how does $\displaystyle |z| = |z|^2$ ?

    You are looking for solutions to:

    $\displaystyle \overline{z}=z^2$

    now just take absolute values and use $\displaystyle |z^2|=|z|^2.$

    CB
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  4. #19
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    Re: Find solutions of z bar = z^2

    Oh, right. I think I misread an equation earlier. $\displaystyle \bar{z} = |z|$ so the new equation is:

    $\displaystyle |z| = |z|^2$

    I read a comment as $\displaystyle |z| = |z|^2$ as an identity separate to my equation - nevermind :S

    Thanks.

    Out of curiosity how would you derive these solutions if I converted the original equation into:

    $\displaystyle r(cos(\theta) - isin(\theta)) = [r(cos(\theta)+isin(\theta))]^2$

    can you?
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  5. #20
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    So applying this geometrically I would have a solution set of $\displaystyle cos(0+\frac{2\pi k}{3})$ where $\displaystyle k = 0, 1, 2$

    Which is the same as the $\displaystyle z = 1$, $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\displaystyle z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

    But where does the z = 0 fit in?
    These are the solutions assuming z was NOT 0. Since you have already determined that z= 0 satisfies the original equation, it has four solutions.
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