# Thread: Finding the modulus and argument for z bar = z^2

1. ## Finding the modulus and argument for z bar = z^2

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

If I convert to polar form, I get

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

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

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

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

How do I proceed?

2. ## Re: Finding the modulus and argument for z bar = z^2

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

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

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

4. ## Re: Finding the modulus and argument for z bar = z^2

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:

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

sub r into (1)

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

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

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

sub theta in to solve for r

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

5. ## Re: Finding the modulus and argument for z bar = z^2

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

So LHS=RHS.

6. ## Re: Finding the modulus and argument for z bar = z^2

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

7. ## Re: Finding the modulus and argument for z bar = z^2

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

That is, $\displaystyle a= a^2- b^2$ and $\displaystyle -b= 2ab$. 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 $\displaystyle a= a^2- 0^2$ gives $\displaystyle a= a^2$ so a= 0 or a= 1. Two solutions are z= 0 and z= 1.

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

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

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

8. ## Re: Finding the modulus and argument for z bar = z^2

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.

,

,

,

# find the modulus of 2 cos theta isintheta

Click on a term to search for related topics.