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Math Help - Finding the modulus and argument for z bar = z^2

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

    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?
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  2. #2
    MHF Contributor Siron's Avatar
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    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 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.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Finding the modulus and argument for z bar = z^2

    Did you find the answers?
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    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:

    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
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Finding the modulus and argument for z bar = z^2

    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.
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  6. #6
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    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.
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  7. #7
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    Re: Finding the modulus and argument for z bar = z^2

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

    Quote Originally Posted by LoblawsLawBlog View Post
    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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