<s>[INST] Finding number of solutions when condition is given. Q - Find number of solutions when it is given that Re(z²) = 0 and |z| = a $\sqrt{2}$ , where z is a complex number and a>0.
First I assumed $z = x + iy$ and then squared it and equated the real part to $0$.
I don't know how to approach after that. 
Please guide. 
 [/INST] Let $z=re^{i\theta}$. Then 


*

*$\Re(z^2)=0$, implies $r^2\cos(2\theta)=0$. 

*Also $|z|=a\sqrt{2}$ implies $r=a\sqrt{2} \neq 0$. 


From these two, we get $\cos(2\theta)=0$. This implies $2\theta=\frac{(2n+1)\pi}{2}$ or $\theta=\frac{(2n+1)\pi}{4}$.
Thus $z=a\sqrt{2}e^{i\frac{(2n+1)\pi}{4}}$, where $n \in \Bbb{Z}$. Now you can count the distinct solutions out of these as:
\begin{align*}
z& =a\sqrt{2}e^{i\frac{\pi}{4}}=a\left(1+i\right)\\
z&=a\sqrt{2}e^{i\frac{3\pi}{4}}=a\left(-1+i\right)\\
z&=a\sqrt{2}e^{i\frac{5\pi}{4}}=a\left(-1-i\right)\\
z&=a\sqrt{2}e^{i\frac{7\pi}{4}}=a\left(1-i\right).
\end{align*}
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