Let be two complex numbers $z$, $w$ satisfying $|z| = |w| = 2,$ and $ (1 + i z)(\overline{w} + i)$ is a pure imaginary complex numbers.
Find the least and the greatest value of the expression $|z-w|.$
I tried. Put $$z = 2( \cos x + i \sin x),$$ and $$w = 2( \cos y + i \sin y).$$ From the condition $ (1 + i z)(\overline{w} + i)$ is a pure imaginary complex numbers, we have $$-2 \cos x+2 \cos y-4 \cos y \sin x+4 \cos x \sin y=0,$$ or $$\cos y -\cos x - 2 (\cos y \sin x - \cos x \sin y) = 0. $$ Therefore $$\cos y -\cos x - 2\sin (x-y)=0.$$ and then $$2\sin \left (\dfrac{x-y}{2} \right )\sin \left (\dfrac{x+y}{2} \right ) - 4\sin \left (\dfrac{x-y}{2} \right )\cos \left (\dfrac{x-y}{2} \right )=0$$
First of all realize, that the points with abscisses $z$ or $w$ lie on the circle with radius $2,$ centered at origin. Thus the distance of the points satisfies $$0\le|z-w|\le4.$$ It is straightforward to see that $\;w=z\;$ fulfil the given condition (below), and $|z-w|=0.$
Now, the maximum. With $z=a+ib$ and $w=c+id$ is the condition $$(1 + i z)(\overline{w} + i)=ip,\; p\in\mathbb{R}\tag 1$$ fulfilled iff $$c-a+ad-bc=0,$$ or equivalently $$a(d-1)=c(b-1).\tag 2$$ If $a=0,$ then $b=\pm 2.$ The relation $(2)$ forces $c=0,$ thus $d=2$ or $d=-2.$
If $w=\overline{z}=0\pm2i,$ the greatest value is attained, as $$|z-w|=|z-\overline{z}|=4.$$