Let $z_1,z_2,z_3$ be any three complex numbers, such that $|z_1|=|z_2|=2, |z_3|=1$ and $\left|\arg\left(\frac{z_1-z_3}{z_2-z_3}\right)\right|=\frac{\pi}{2}$. Then, the difference between the maximum and minimum values of $|z_1+z_2|$ is equal to
My Approach:
All I can think about is angle between vector $z_1-z_3$ and $z_2-z_3$ is $\frac{\pm\pi}{2}$
Then I tried to plot them on argand plane but can't think further.
Best Answer
For simplicity, I use same notation of points and their complex coordinates.
$z_1, z_2$ lie on a circle with radius $2,$ centered at $0.$
$z_3$ lies on the unit circle ${C}$ centered at $0,$ and on circle $C'$ with diameter $z_1z_2.$
WLOG, set $z_1=2.$
$|z_1+z_2|$ is a diagonal length of the rhombus with vertices $0, z_1, z_1+z_2, z_2.$ It is grey one in the picture.
$|z_1+z_2|$ is maximum if $z_1, z_2$ are as close as possible to each other, e.g. circles $C, C'$ touch externally in a single point $z_3.$
For minimum, I chose the labeling $w_1, w_2$ instead of $z_1, z_2.$
To better see the relation between lengths, set $w_1=z_2.$
$|w_1+w_2|$ is minimum if $w_1, w_2$ are as far apart as possible, and $C$ touches $C'$ internally.
The difference between the maximum and minimum is twice the radius of unit circle. $$|z_1+z_2|-|w_1+w_2|=2$$