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

How do I find and for ?

If I convert to polar form, I get

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

and ?

How do I proceed?

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 and (and ):

If:

then:

(1) and (2)

If we reform equation (2):

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

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

Did you find the answers?

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:

sub r into (1)

sub theta in to solve for r

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

Yes, that's correct. You can check the answer:

So LHS=RHS.

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.

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

I would NOT immediately convert to polar coordinates. Taking z= a+ bi, and . Thus, gives

That is, and . 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 gives so a= 0 or a= 1. Two solutions are z= 0 and z= 1.

If a= -1/2 then so and .

Two more solutions are and .

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

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

Quote:

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.