If $|z|=|w|=\frac{|z-w|}{\sqrt2}\neq0$ then what is the possible value of $\frac{iz}{w}$

Question

If $|z|=|w|=\frac{|z-w|}{\sqrt2}\neq0$ then what is the possible value of $\frac{iz}{w}$ I couldn’t get anywhere.

Answer 1

Divide the given equation by $\vert w\vert$ to get $$\left\vert\dfrac{z}{w}\right\vert=1=\dfrac{\left\vert\dfrac{z}{w}-1\right\vert}{\sqrt 2}$$ Hence, $\dfrac{z}{w}$ is a point in plane which lies on unit circle around origin and is at a distance $\sqrt2$ from point $(1,0)$. Therefore, $\dfrac{z}{w}$ represents the point $(0,1)$ or $(0,-1)$. So required answer is $i\times i=-1$ or $-i\times i=1$.

Answer 2

You have$$\frac{\lvert z-w\rvert^2}2=\frac{\lvert z\rvert^2+\lvert w\rvert^2-2\operatorname{Re}\left(z\overline w\right)}2=\lvert z\rvert^2-\operatorname{Re}\left(z\overline w\right).\tag1$$On the other hand,$$\frac{\lvert z-w\rvert^2}2=\left(\frac{\lvert z-w\rvert}{\sqrt2}\right)^2=\lvert z\rvert^2.\tag2$$It follows from $(1)$ and $(2)$ that $\operatorname{Re}\left(z\overline w\right)=0$. But then$$0=\frac1{\lvert w\rvert^2}\operatorname{Re}\left(z\overline w\right)=\operatorname{Re}\left(\frac{z\overline w}{w\overline w}\right)=\operatorname{Re}\left(\frac zw\right)$$and therefore$$\operatorname{Im}\left(\frac{iz}w\right)=-\operatorname{Re}\left(\frac zw\right)=0.$$

Answer 3

On an Argand plane draw vectors from the origin to $z$ and $w$. If the length of the third side representing $|z-w|$ satisfies $|z-w|=\sqrt{2}|z|=\sqrt{2}|w|$, then by the Pythagorean Theorrm the $z$ and $w$ vectors must be perpendicular. This together with $|z|=|w|$ implies $z/w\in\{i,-i\}$, therefore $iz/w\in\{1,-1\}$.