# Solving Complex Number Equation

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.

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

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.
@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