Question:
Find the real part $z_1 / z_2$ if absolute value $|z_1+z_2|=|z_1-z_2|$.
I thought $z_1= a +bi$ and $z_2= c + di$ then $z_1 + z_2= (a+c) + (b+d)i $ and $z_1-z_2= (a-c)+(b-d)i$ when computing the modulus and squaring both sides I end up with: $$(a+c)^2 +(b+d)^2= (a-c)^2 +(b-d)^2$$
and I get $4ac +4bd=0$. But how do I go from there calculating the real part $z_1/z_2$?
Convince yourself that $|z_1+z_2|=|z_1-z_2|$ can only happen if $z_2$ is at right angles to $z_1$:
^ +z_2
z_1 |
------->o
|
v -z_2
So the real part of $\frac{z_1}{z_2}$ is zero.
$z_1/z_2=\frac{z_1\overline{z_2}}{|z_2|^2}$. Note that the real part of $z_1\overline{z_2}$ equals $ac+bd$ which is $0$ by your computation.
Draw on the complex plane any non zero complex number $z_2$.
Then draw $-z_2$.
The condition
$$ |z_1-z_2|=|z_1+z_2| $$
says that $z_1$ has to be equally distant from $z_2$ and $-z_2$.
All such complex numbers, lie on the straight line orthogonal to the segment $[-z_2,z_2]$ passing thru the origin.
Thus, writing $z_2=re^{i\theta}$, then all such $z_1$ (when non zero) will be of the form $$ z_1=se^{i(\theta\pm\pi/2)}, \;\;s>0\;, $$ hence
$$ \frac{z_1}{z_2}=\frac sre^{\pm i\pi/2}=\pm i\frac sr\; $$ which is a purely imaginary number.
@ParclyTaxel's diagram is instructive in proving $z_1/z_2$ is imaginary: the two moduli are the distances of $z_1$ from $\mp z_2$, equated on a linear locus. Or if you want a proof that doesn't need the geometric insight: $\left|\tfrac{z_1/z_2+1}{z_1/z_2-1}\right|=1$ iff some $\theta\in\Bbb R$ exists with $\tfrac{z_1/z_2+1}{z_1/z_2-1}=e^{i\theta}$, which rearranges (since $\tfrac{w+1}{w-1}$ is an involution) to$$\frac{z_1}{z_2}=\frac{e^{i\theta}+1}{e^{i\theta}-1}=\frac{e^{i\theta/2}+e^{i\theta/2}}{e^{i\theta/2}-e^{i\theta/2}}=\frac{2\cos\tfrac{\theta}{2}}{2i\sin\tfrac{\theta}{2}}=-i\cot\tfrac{\theta}{2}.$$
Using $|z|^2=z\bar{z}$, one has $$ (z_1+z_2)(\bar{z}_1+\bar{z}_2)=(z_1-z_2)(\bar{z}_1-\bar{z}_2)$$ or $$ |z_1|^2+|z_2|^2+z_2\bar{z}_1+z_1\bar{z}_2=|z_1|^2+|z_2|^2-z_2\bar{z}_1-z_1\bar{z}_2.$$ From this, one attains $$ z_2\bar{z}_1+z_1\bar{z}_2=0$$ or $$ \frac{z_1}{z_2}=-\frac{\bar{z}_1}{\bar{z}_2}=-\overline{\left(\frac{z_1}{z_2}\right)} $$ which implies that $\frac{z_1}{z_2}$ is purely imaginary number or $$ \Re\left(\frac{z_1}{z_2}\right)=0. $$
When we add complex numbers, we use the "parallelogram rule."
The $|z_1 + z_2|$ equals the length of the diagonal of the parallelogram
$|z_1 - z_2|$ equals the length of the other diagonal.
If these two lengths are equal, the parallelogram is, in fact, a rectangle.
$\arg\frac {z_1}{z_2} = \arg z_1 - \arg z_2 = \frac {\pi}{2}$
The real component of any complex number with this argument is $0.$
