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

Question

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.

Answer 1

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 $-1$.

Answer 2

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

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

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

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

Answer 3

$|z_1 +z_2 |= |z_{1} - z_{2}| Rightarrow|

if z_{2} ne0. In other words, z_{1} / z_{2} is equidistant from 1 and -1.

|z_{1}/z_{2} + 1| = |z_{1}/z_{2} - 1|

As geometrically we know that if a point Is equidistant from 2 fixed points it lie on the perpendicular bisector of it. In this case the imaginary axis will act like the perpendicular bisector of of line joining (-1,0) & (1,0) Therefore, Z1/z2 it will line on the imaginary axis except (0,0) as it is complex no it's y coordinate can't be zero