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.

  • $\begingroup$ 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 $z_2$? $\endgroup$
    – Arthur
    Sep 22, 2018 at 22:37
  • $\begingroup$ 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? $\endgroup$
    – user562983
    Sep 22, 2018 at 22:58
  • $\begingroup$ Hint: Look at $|z_1+z_2|^2=(z_1+z_2)(\bar {z_1}+\bar {z_2})$ and $|z_1-z_2|^2$. $\endgroup$
    – random
    Sep 22, 2018 at 23:39
  • $\begingroup$ Thanks for the answer think geometrically was indeed really usefull. Thank you. I should try to do that more often. $\endgroup$ Sep 23, 2018 at 19:55

2 Answers 2


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


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

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


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