Agfdhyk

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.

Let $z=re^{itheta}$. Then

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


2. Also $|z|=asqrt{2}$ implies $r=asqrt{2} neq 0$.

From these two, we get $cos(2theta)=0$. This implies $2theta=frac{(2n+1)pi}{2}$ or $theta=frac{(2n+1)pi}{4}$.

Thus $z=asqrt{2}e^{ifrac{(2n+1)pi}{4}}$, where $n in Bbb{Z}$. Now you can count the distinct solutions out of these as:
begin{align*}
z& =asqrt{2}e^{ifrac{pi}{4}}=aleft(1+iright)\
z&=asqrt{2}e^{ifrac{3pi}{4}}=aleft(-1+iright)\
z&=asqrt{2}e^{ifrac{5pi}{4}}=aleft(-1-iright)\
z&=asqrt{2}e^{ifrac{7pi}{4}}=aleft(1-iright).
end{align*}

Let $z=a+bi$ where $i$ is the imaginary unit and $a,binmathbb{R}$.

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 $|z|=alphasqrt{2}$ for some $alpha>0$ means, by definition that

$$|z|=sqrt{a^{2}+b^{2}}=alphasqrt{2}$$

Squaring both sides will give us that

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

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 $|z|=sqrt{a^2+b^2}=alphasqrt2$

And now we know:

$$sqrt{a^2+b^2}=alphasqrt2$$
$$a^2-b^2=0$$

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

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

$a^2-b^2=0$