2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

4
$\begingroup$

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 $-1$.

$\endgroup$
0
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.