Pfthb

Q - Find number of solutions when it is given that Re(z²) = 0 and |z| = a $sqrt2$ , 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.

Let $z=re^itheta$. Then

1. 
   $Re(z^2)=0$, implies $r^2cos(2theta)=0$.


2. Also $|z|=asqrt2$ implies $r=asqrt2 neq 0$.

From these two, we get $cos(2theta)=0$. This implies $2theta=frac(2n+1)pi2$ or $theta=frac(2n+1)pi4$.

Thus $z=asqrt2e^ifrac(2n+1)pi4$, where $n in BbbZ$. Now you can count the distinct solutions out of these as:
beginalign*
z& =asqrt2e^ifracpi4=aleft(1+iright)\
z&=asqrt2e^ifrac3pi4=aleft(-1+iright)\
z&=asqrt2e^ifrac5pi4=aleft(-1-iright)\
z&=asqrt2e^ifrac7pi4=aleft(1-iright).
endalign*

Let $z=a+bi$ where $i$ is the imaginary unit and $a,binmathbbR$.

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 $|z|=alphasqrt2$ for some $alpha>0$ means, by definition that

$$|z|=sqrta^2+b^2=alphasqrt2$$

Squaring both sides will give us that

$$a^2+b^2=alpha^22$$

We could just say that $a^2+b^2=2beta$ 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 $a=pmalpha$ and $b=pmalpha$ will suffice. To see this note that

$$a^2-b^2=(pmalpha)^2-(pmalpha)^2=alpha^2-alpha^2=0$$
$$a^2+b^2=(pmalpha)^2+(pmalpha)^2=alpha^2+alpha^2=2alpha^2=2beta$$

So, as Anurag A showed, for a given $alpha>0$ there are exactly four solutions.

$$alpha+alpha i$$
$$alpha-alpha i$$
$$-alpha+alpha i$$
$$-alpha-alpha i$$

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 $|z|=sqrta^2+b^2=alphasqrt2$

And now we know:

$$sqrta^2+b^2=alphasqrt2$$
$$a^2-b^2=0$$

Since it's given that $alpha>0$, you can square both sides of the first equation without changing anything:

$a^2+b^2=2alpha^2$

$a^2-b^2=0$