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Thread: Find solutions of z bar = z^2

  1. #1
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    Find solutions of z bar = z^2

    Im trying find all solutions for: $\displaystyle \bar{z}=z^2$

    Should I convert z into $\displaystyle x +iy$ or $\displaystyle r(cos(\theta) +isin(\theta))$ if I want to interpret my result geometrically?

    What I have done so far is:

    $\displaystyle x-iy = (x+iy)^2$

    $\displaystyle x-iy = x^2+i2xy-y^2$

    $\displaystyle x - iy = x^2 - y^2 + i2xy$

    Therefore $\displaystyle x = x^2 -y^2$ and $\displaystyle y = 2xy$

    Not sure what to do with my results. I'm a little lost with the general values for z; I am fine with the definite values such as $\displaystyle 6 + i\sqrt3$ etc. Not sure which direction to head to solve for this equation so I can interpret geometrically.

    Thanks.
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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: $\displaystyle \bar{z}=z^2$

    Should I convert z into $\displaystyle x +iy$ or $\displaystyle r(cos(\theta) +isin(\theta))$ if I want to interpret my result geometrically?

    What I have done so far is:

    $\displaystyle x-iy = (x+iy)^2$

    $\displaystyle x-iy = x^2+i2xy-y^2$

    $\displaystyle x - iy = x^2 - y^2 + i2xy$

    Therefore $\displaystyle x = x^2 -y^2$ and $\displaystyle \color{red}y = 2xy$
    You have a mistake. It should be:
    $\displaystyle -y=2xy$
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  3. #3
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    Re: Find solutions of z bar = z^2

    So I get $\displaystyle x = -\frac{1}{2}$ and $\displaystyle x = x^2 -y^2$

    I'm not sure what to do with these results.
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    So I get $\displaystyle x = -\frac{1}{2}$ and $\displaystyle x = x^2 -y^2$

    I'm not sure what to do with these results.
    Can you find y from the second equation?

    Also, don't forget that y = 0 is also a solution of -y = 2xy.
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by alexmahone View Post
    Can you find y from the second equation?
    Also, don't forget that y = 0 is also a solution of -y = 2xy.
    There are only two solutions to this question:
    $\displaystyle z=0\text{ or }z=1$.

    Note that $\displaystyle z=\bar{z}$ if and only if $\displaystyle \text{Im}(z)=0$ or $\displaystyle z\in\mathbb{R}$.

    What real numbers have the property that $\displaystyle x=x^2~?$
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by Plato View Post
    There are only two solutions to this question:
    $\displaystyle z=0\text{ or }z=1$.
    How about $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$?
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by alexmahone View Post
    Can you find y from the second equation?

    Also, don't forget that y = 0 is also a solution of -y = 2xy.
    $\displaystyle y = \sqrt{x^2-x}$

    $\displaystyle \therefore$ y = 0 when x = 0

    and

    $\displaystyle y = \sqrt{\frac{3}{4}} = \frac{\sqrt3}{2}$ when $\displaystyle x = -\frac{1}{2}$
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by alexmahone View Post
    How about $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$?
    You are right. Good catch.
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by Plato View Post
    You are right. Good catch.
    I threw you off with my initial mistake, obviously
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    ...so I can interpret geometrically.
    Since you mentioned this, here's another approach:

    Take absolute values of both sides of $\displaystyle \bar{z}=z^2$, getting $\displaystyle |\bar{z}|=|z|=|z|^2$, so |z|=0 or |z|=1. Since 0 is easily seen to be a solution, we're focused on possible solutions on the unit circle.

    Now multiply your original equation by z, getting $\displaystyle z\bar{z}=z^3$. Do you know an identity relating $\displaystyle z\bar{z}$ and |z|? This should lead you to consider cube roots of unity, which have a nice geometric interpretation. You might get some extraneous solutions here though (possibly, too tired to trust myself now lol).
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by LoblawsLawBlog View Post
    Since you mentioned this, here's another approach:

    Take absolute values of both sides of $\displaystyle \bar{z}=z^2$, getting $\displaystyle |\bar{z}|=|z|=|z|^2$, so |z|=0 or |z|=1. Since 0 is easily seen to be a solution, we're focused on possible solutions on the unit circle.

    Now multiply your original equation by z, getting $\displaystyle z\bar{z}=z^3$. Do you know an identity relating $\displaystyle z\bar{z}$ and |z|? This should lead you to consider cube roots of unity, which have a nice geometric interpretation. You might get some extraneous solutions here though (possibly, too tired to trust myself now lol).
    Thanks to you, I just realized that I had missed a solution: $\displaystyle z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
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    Re: Find solutions of z bar = z^2

    I know that $\displaystyle \bar{z}z$ gives you $\displaystyle |z|^2$ right? the positive real part.
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    Re: Find solutions of z bar = z^2

    Quote Originally Posted by terrorsquid View Post
    I know that $\displaystyle \bar{z}z$ gives you $\displaystyle |z|^2$ right? the positive real part.
    Yes. And $\displaystyle |z|=1$, so $\displaystyle z^3=1$.
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    Re: Find solutions of z bar = z^2

    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?
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  15. #15
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    Re: Find solutions of z bar = z^2

    |z|=0 (so z=0) was one of the solutions to |z|=|z|^2. Once you verify that 0 is a solution, then you can focus on the solutions where |z|=1, which you just listed, so we have all 4 solutions.
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