# $|z_{1} + z_{2}| = |z_{1} - z_{2}| \implies z_{1}/z_{2}$ is Imaginary

Two complex numbers $$z_{1}$$ and $$z_{2}$$ are taken such that $$|z_{1} + z_{2}| = |z_{1} - z_{2}|$$ and $$z_{2}$$ is not $$0$$. Show that $$z_{1}/z_{2}$$ is purely imaginary, i.e. it has no real part.

So they are both absolute values so $$|z_{1} + z_{2}| = |z_{1} + z_{2}|$$ is also true so how can you ever solve it if they are equal or is my assumption false? also i have tried to turn in into $$a + bi$$ and got to $$a_{1}\cdot a_{2} = -b_{1}\cdot b_{2}$$, but i dont see how this will help me to get to prove the statement. Also what it means that $$z_{2}$$ is not $$0$$ confuses me because $$b$$ or $$a$$ can still be $$0$$ then but just not both at the same time.

Can someone give me a hint without giving the whole answer? So i can still figure the rest out myself. Just a hint to be able to continue.

• Hint: Think geometrically. What does $|z_1-z_2|$ and $|z_1+z_2|$ mean geometrically? What does $z_1/z_2$ being purely imaginary mean for the geometric relationship between numbers ("vectors", if that suits you better) $z_1$ and $z_2$? Sep 22, 2018 at 22:37
• It is certainly true that $\lvert z_1+z_2\rvert=\lvert z_1+z_2\rvert$, but what does the absolute value have to do with that?
– user562983
Sep 22, 2018 at 22:58
• Hint: Look at $|z_1+z_2|^2=(z_1+z_2)(\bar {z_1}+\bar {z_2})$ and $|z_1-z_2|^2$. Sep 22, 2018 at 23:39
• Thanks for the answer think geometrically was indeed really usefull. Thank you. I should try to do that more often. Sep 23, 2018 at 19:55

Hint: $$|z_1 + z_2| = |z_1 - z_2| \implies \left|\frac{z_1}{z_2}+1\right| = \left|\frac{z_1}{z_2}-1\right| \tag{1}$$ if $$z_2 \neq 0$$. In other words, $$z_1/z_2$$ is equidistant from $$1$$ and $$-1$$.

We have that

$$|z_1 + z_2| = |z_1 - z_2| \iff \left|\frac{z_1}{z_2}+1\right| = \left|\frac{z_1}{z_2}-1\right|\iff \left|\frac{z_1}{z_2}+1\right|^2 = \left|\frac{z_1}{z_2}-1\right|^2$$

and by $$w=\frac{z_1}{z_2}$$

$$|w+1|^2=|w-1|^2$$

$$(w+1)(\bar w+1)=(w-1)(\bar w-1)$$ $$|w|^2+w+\bar w+1=|w|^2-w-\bar w+1$$

$$2(w+\bar w)=0 \iff \Re(w)=0$$