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. 
 A: Let $z=a+bi$ where $i$ is the imaginary unit and $a,b\in\mathbb{R}$. 
The condition $Re(z^{2})=0$ means that $a^{2}=b^{2}$ because,
$$z^{2}=(a+bi)^{2}=a^{2}+2abi-b^{2}=(a^{2}+b^{2})+2abi$$
Then $Re(z^{2})=0$ implies $a^{2}-b^{2}=0$ which gives us that $a^{2}=b^{2}$.
Now the condition that $|z|=\alpha\sqrt{2}$ for some $\alpha>0$ means, by definition that
$$|z|=\sqrt{a^{2}+b^{2}}=\alpha\sqrt{2}$$
Squaring both sides will give us that
$$a^{2}+b^{2}=\alpha^{2}2$$
We could just say that $a^{2}+b^{2}=2\beta$ for some $\beta>0$. Note then that we have a system of two linear equations with two unkowns ($a^{2}$ and $b^{2}$).
$$a^{2}-b^{2}=0$$
$$a^{2}+b^{2}=\beta2$$
Of course $\beta$ depends on $a$ and $b$, but that doesn't make things very difficult. Setting $a=b=1$ and $\beta=1$ yields one solution. Setting $a=b=2^{n}$ for some natural number $n$ yields another solution with $\beta=2^{n}$.
Using some basic linear algebra you can show that for every $\beta>0$ there is some choice of $a$ and $b$ that will solve the system.
Choosing $\beta=1$ we get that choosing $a=\pm1$ and $b=\pm$ yields a solution. In general for a specific $\beta=\alpha^{2}$ where $\alpha>0$ we have (taking into account Anurag A's solution) that any of $a=\pm\alpha$ and $b=\pm\alpha$ will suffice. To see this note that
$$a^{2}-b^{2}=(\pm\alpha)^{2}-(\pm\alpha)^{2}=\alpha^{2}-\alpha^{2}=0$$
$$a^{2}+b^{2}=(\pm\alpha)^{2}+(\pm\alpha)^{2}=\alpha^{2}+\alpha^{2}=2\alpha^{2}=2\beta$$
So, as Anurag A showed, for a given $\alpha>0$ there are exactly four solutions.
$$\alpha+\alpha i$$
$$\alpha-\alpha i$$
$$-\alpha+\alpha i$$
$$-\alpha-\alpha i$$
A: 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*}
A: It's probably assuming that $z = a+ib$;
So from here
$z^2=(a+ib)^2=a^2+2ib-b^2=a^2-b^2+i(2b)$
-Put it this way so you can see $Re(z^2)=a^2-b^2=0$
And $|z|=\sqrt{a^2+b^2}=\alpha\sqrt2$
And now we know:
$$\sqrt{a^2+b^2}=\alpha\sqrt2$$
$$a^2-b^2=0$$
Since it's given that $\alpha>0$, you can square both sides of the first equation without changing anything:
$a^2+b^2=2\alpha^2$
$a^2-b^2=0$
