$|z_{1} + z_{2}| = |z_{1} - z_{2}| \implies z_{1}/z_{2}$ 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.
 A: 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 $-1$.
A: 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 $w=\frac{z_1}{z_2}$
$$|w+1|^2=|w-1|^2$$
$$(w+1)(\bar w+1)=(w-1)(\bar w-1)$$
$$ |w|^2+w+\bar w+1=|w|^2-w-\bar w+1$$
$$ 2(w+\bar w)=0 \iff \Re(w)=0$$
A: $|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
