What is the value of complex number $z$ when $|3 - \sqrt{2 - |z|}| = 2$.

Question

WolframAlpha’s answer is attached but I do not understand how it is obtained. Wolfram Solution

If this question is downvoted, please explain why so I can revise it.

Answer 1

$\lvert z\rvert$ is real, so $2-\lvert z\rvert$ is also real. If $2-\lvert z\rvert\geq0$, then the square root operator is returning a nonnegative real, and $3-\sqrt{2-\lvert z\rvert}$ is real. So in this case the equation says:

$$3-\sqrt{2-\lvert z\rvert}=\pm2$$ $$\sqrt{2-\lvert z\rvert}=3\pm2\in\{1,5\}$$ Then $$2-\lvert z\rvert\in\{1,25\}$$ But $\lvert z\rvert\geq0$, so it's not possible for $2-\lvert z\rvert$ to be $25$. Instead it equals $1$, and so $\lvert z\rvert=1$. In this case $z$ is a complex number on the unit circle.

Note: If $2-\lvert z\rvert<0$, then if we like, we can consider the square root operator to be returning an imaginary number with positive imaginary part: $i\sqrt{\lvert z\rvert-2}$. But then we have a contradition, because in this case the given equation says $$3^2+\sqrt{\lvert z\rvert-2}^2=2^2$$ with the left side clearly greater than the right side.

So the only solutions are from the first case, with $z$ being a unit complex number.

This all looks very different than the posted Wolfram screenshot. Either something is wrong with Wolfram's analysis of the equation, or possibly there is an issue with how the command was issued to Wolfram.

Answer 2

In the meanwhile you could considerate

$$|3 - \sqrt{2 - |z|}| = 2 \iff 3 - \sqrt{2 - |z|}=\pm 2$$

hence

$$5-\sqrt{2 - |z|}=0, \quad {\text{or}} \quad 1-\sqrt{2 - |z|}=0 $$

i.e.

$$\sqrt{2 - |z|}=5, \quad {\text{or}} \quad \sqrt{2 - |z|}=1 $$

Putting $z=a+ib$ you will have:

$$\sqrt{2 - \sqrt{a^2+b^2}}=5, \quad {\text{or}} \quad \sqrt{2 - \sqrt{a^2+b^2}}=1 $$

Solve for a $\sqrt{2-\sqrt{a^2+b^2}}=5 \implies 2-\sqrt{a^2+b^2}-2=25-2 \implies -\sqrt{a^2+b^2}=23$; but squaring $a^2+b^2=529$ with $a=\sqrt{529-b^2},\:a=-\sqrt{529-b^2}$. Verifing the solutions into $\sqrt{2 - \sqrt{a^2+b^2}}=5$ we have false conditions.

You try with the same method with $\sqrt{2 - \sqrt{a^2+b^2}}=1$ where we have two solutions $$a=\sqrt{1-b^2},\:a=-\sqrt{1-b^2}$$

Answer 3

Since $2-|z|$ is real its square root is either real, either a pure imaginary.

• $|z|>2$ the square root is pure imaginary $\sqrt{2-|z|}=iy$ with $y$ real.

But this is not possible since $|3-iy|=\sqrt{9+y^2}\ge 3$ so this cannot be $2$

• $|z|\le 2$ the square root is real $\sqrt{2-|z|}=x$ with $x$ real $\in[0,\sqrt{2}]$

We can solve in $\mathbb R,\ |3-x|=2\iff x=1$ or $5$

$x=5$ can be ruled out since it's larger than $\sqrt{2}$

So the only solution is $|z|=1\iff z=e^{i\theta}$

We can verify it works as $|3-\sqrt{2-|z|}|=|3-\sqrt{2-1}|=|3-1|=|2|=2$