# Find solutions of z bar = z^2

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• Aug 28th 2011, 08:22 AM
terrorsquid
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.
• Aug 28th 2011, 08:31 AM
Plato
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by terrorsquid
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:
$\displaystyle -y=2xy$
• Aug 28th 2011, 08:40 AM
terrorsquid
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.
• Aug 28th 2011, 08:42 AM
alexmahone
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by terrorsquid
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.
• Aug 28th 2011, 09:00 AM
Plato
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by alexmahone
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 $\displaystyle x=x^2~?$
• Aug 28th 2011, 09:07 AM
alexmahone
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by Plato
There are only two solutions to this question:
$\displaystyle z=0\text{ or }z=1$.

How about $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$?
• Aug 28th 2011, 09:21 AM
terrorsquid
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by alexmahone
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 $\displaystyle x = -\frac{1}{2}$
• Aug 28th 2011, 09:23 AM
Plato
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by alexmahone
How about $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$?

You are right. Good catch.
• Aug 28th 2011, 09:25 AM
terrorsquid
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by Plato
You are right. Good catch.

I threw you off with my initial mistake, obviously :D
• Aug 28th 2011, 09:37 AM
LoblawsLawBlog
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by terrorsquid
...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).
• Aug 28th 2011, 09:44 AM
alexmahone
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by LoblawsLawBlog
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: $\displaystyle z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
• Aug 28th 2011, 09:48 AM
terrorsquid
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.
• Aug 28th 2011, 09:50 AM
alexmahone
Re: Find solutions of z bar = z^2
Quote:

Originally Posted by terrorsquid
I know that $\displaystyle \bar{z}z$ gives you $\displaystyle |z|^2$ right? the positive real part.

Yes. And $\displaystyle |z|=1$, so $\displaystyle z^3=1$.
• Aug 28th 2011, 09:58 AM
terrorsquid
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?
• Aug 28th 2011, 10:09 AM
LoblawsLawBlog
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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