Q1.
Solve equa for z=a+bi where a and b are real numbers.
(z-zbar)^2=2z+6i
The ans is -18-3i but do not know how they get this.
I get
$\displaystyle (a+bi-(a-bi))^2=2(a+bi)+6i$
$\displaystyle (a+bi-a+bi)^2=2a+2bi+6i$
$\displaystyle (2bi)^2=2a+2bi+6i$
$\displaystyle 4b^2(-1)=2a+2bi+6i$
$\displaystyle -4b^2=2a+2bi+6i$
$\displaystyle -4b^2-2a-2bi=0+6i$
Now equating coefficients
$\displaystyle -2b=6 , -4b^2-2a=0\implies b = -3 ,-4(-3)^2-2a=0 $
$\displaystyle \implies -4\times 9-2a=0\implies -36-2a=0, a=-18$
$\displaystyle 3i = 3\left(\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}}\right)$
$\displaystyle = 3e^{\frac{\pi i}{2}}$.
So $\displaystyle z^2 = 3i$
$\displaystyle z^2 = 3e^{\frac{\pi i}{2}}$
$\displaystyle z = \left(3e^{\frac{\pi i}{2}}\right)^{\frac{1}{2}}$
$\displaystyle z = \pm\sqrt{3}e^{\frac{\pi i}{4}}$
$\displaystyle z = \pm\sqrt{3}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$
$\displaystyle z = \pm\sqrt{3}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}i}{2}\right)$
$\displaystyle z = \frac{\sqrt{6}}{2} + \frac{\sqrt{6}i}{2}$ or $\displaystyle z = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}i}{2}$.