Finding number of solutions when condition is given.

Question

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.

Answer 1

Let $z=re^{i\theta}$. Then

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

Answer 2

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:

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

$a^2-b^2=0$

Answer 3

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.

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