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Let $z = a + ib , z^* = a - ib$

I need to find all possible solutions to $2z=(z^*)^2$

$(z^*)^2 = (a^2 - b^2) -i2ab$

$2z = 2a + i2b$

$\implies 2a + i2b = a^2 -b^2 -ia2b$ $\implies a^2 - b^2 -i2ab - 2a - i2b = 0$ $\implies a(a -i2b - 2) - b(b + i2) = 0$

I found solution to the above equation $z = 0$. However, I do not know how to find the rest of the solutions from here.

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Alternatively, you could use polar coordinates. –  1015 Mar 25 '13 at 16:14
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1 Answer

up vote 1 down vote accepted

You should separate the real and imaginary parts of your equation, keeping in mind that $a,b$ are real. So from $a^2-b^2-2iab-2a-2ib=0$ you go to $$a^2-b^2-2a=0\\-2iab-2ib=0\\ab+b=0$$ and solve the first and third as a pair of simultaneous equations.

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I don't understand why I should separate the real and imaginary parts. I just started learning the Complex system. –  0xFF Mar 25 '13 at 16:24
    
I see ... I need to make both the real and imaginary parts equal to 0 (as by definition). –  0xFF Mar 25 '13 at 16:27
    
@0xFF, do you know, $a+ib=c+id\iff a=c,b=d$. Now $0=0+i0$ i.e., $c=d=0$ –  lab bhattacharjee Mar 25 '13 at 16:30
    
@oxFF: two complex numbers are equal precisely when the real and imaginary parts are both equal. In particular, the origin of the complex plane is $0+0i$. So if you wanted to solve $z^2=i$, you could write $z=a+bi, z^2=a^2+2abi-b^2=i$, so $a^2-b^2=0, 2ab=1$ and solve these two. –  Ross Millikan Mar 25 '13 at 16:30
    
Thanks, I comprehend it now. –  0xFF Mar 25 '13 at 16:44
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