Finding the modulus and argument for z bar = z^2

**terrorsquid** · September 4th 2011, 04:37 AM

How do I find $r$ and $\theta$ for $\bar{z} = z^2$ ?

If I convert to polar form, I get

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

$\equiv cos\theta - isin\theta = r(cos(2\theta)+isin(2\theta))$

Am I trying to equate the real and imaginary parts of the two sides of the equation? If so, how does the r on the RHS affect this? or can I assume it is 1 and say

$cos\theta = cos(2\theta)$ and $-sin\theta = sin(2\theta)$ ?

How do I proceed?

**Siron** · September 4th 2011, 04:53 AM

I think that's a good way to do it and I think you can make a system of equations in two variables $r$ and $\theta$ (and $r\neq 0$ ):
If:
$\cos(\theta)-i\sin(\theta)=r\cos(2\theta)+r\cdot i\sin(2\theta)$
then:
$\cos(\theta)=r\cos(2\theta)$ (1) and $-i\sin(\theta)=r\cdot i\sin(2\theta)$ (2)

If we reform equation (2):
$-i\sin(\theta)=r\cdot 2i\sin(\theta)\cos(\theta) \Leftrightarrow -1=2r\cos(\theta) \Leftrightarrow r= \frac{-1}{2\cos(2\theta)}$

If you substitute this given in (1) then you've an equation in one variable $\theta$ and afterwards you can find $r$ .

**Siron** · September 4th 2011, 05:55 AM

Did you find the answers?

**terrorsquid** · September 4th 2011, 07:50 AM

I think so, it seems a bit convaluted how I got their though. I remember my professor doing this and it was done in 2 or 3 steps quite quickly - it seemed simpler. Maybe the problem was just simpler. I will confirm with him but here is what I did:

$r= \frac{-1}{2\cos(2\theta)}$

sub r into (1)

$cos\(\theta) = \frac{-cos(2\theta)}{2\cos(2\theta)}$

$cos(\theta) = -\frac{1}{2}$

$\theta = arcos(-\frac{1}{2}) = \frac{2\pi}{3}$

sub theta in to solve for r

$r= \frac{-1}{-\frac{2}{2}} = 1$

**Siron** · September 4th 2011, 10:50 AM

Yes, that's correct. You can check the answer:
$\cos\left(\frac{2\pi}{3}\right)-i\sin \left(\frac{2\pi}{3}\right)=\cos\left(\frac{4\pi}{ 3}\right)+i\sin\left(\frac{4\pi}{3}\right)$
$\frac{-1}{2}-i\frac{\sqrt{3}}{2}=\frac{-1}{2}-i\frac{\sqrt{3}}{2}$

So LHS=RHS.

**LoblawsLawBlog** · September 5th 2011, 10:34 AM

Didn't we solve this just a few days ago? There are 4 solutions and you've only found 1.

**HallsofIvy** · September 5th 2011, 05:21 PM

I would NOT immediately convert to polar coordinates. Taking z= a+ bi, $\overline{z}= a- bi$ and $z^2= (a^2- b^2)+ 2bi$ . Thus, $\overline{z}= z^2$ gives $a- bi= (a^2- b^2)+ 2abi$

That is, $a= a^2- b^2$ and $-b= 2ab$ . One obvious solution to the second equation is b= 0. If b is NOT 0, we can divide by it to get -1= 2a or a= -1/2.

If b= 0, then $a= a^2- 0^2$ gives $a= a^2$ so a= 0 or a= 1. Two solutions are z= 0 and z= 1.

If a= -1/2 then $-1/2= 1/4- b^2$ so $b^2= 3/4$ and $b= \sqrt{3}/2$ .

Two more solutions are $z= -\frac{1}{2}+ \frac{\sqrt{3}}{2}$ and $z= -\frac{1}{2}- \frac{\sqrt{3}}{2}$ .

It is easy to get the modulus and argument of those 4 values.

**terrorsquid** · September 6th 2011, 08:27 PM

Originally Posted by LoblawsLawBlog

Didn't we solve this just a few days ago? There are 4 solutions and you've only found 1.

I wasn't trying to solve for z here. First and foremost, I wanted to see the process of utilising polar form as it has helped me with other problems as well.

Thanks all.

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