Proving that $z$ is Purely Imaginary I'm stuck on a question in relation with complex numbers:
If $z\neq0$, and that
$$\left\vert{\frac{z+1}{z-1}}\right\vert=1,$$
prove that $z$ is purely imaginary.
I tried breaking the modulus up into two separate parts, and then multiplying both sides. Then I squared both sides and used the formula $\vert{z}\vert^2=z\bar{z}$, and replaced $z$ with $a+bi$. But it seems that no matter what manipulation I do, it either turns real or doesn't make sense.
Any help is much appreciated!! Thanks!
 A: If $|z-1|=|z+1|$, then writing $z=x+iy$ and squaring both sides we get
$$ (x-1)^2+y^2=(x+1)^2+y^2$$
which implies that $x=0$. If $z\neq 0$, this means that $z$ is purely imaginary.
There is also a nice geometric interpretation: $|z-1|$ is the distance from $z$ to $1$, and $|z+1|$ is the distance to $-1$. If these are equal, then $z$ must lie on the perpendicular bisector of the line segment connecting $1$ and $-1$, which is the line $x=0$.
A: The easiest way to see this is to note that this equation is equivalent to $|z-1|=|z+1|$. 
This is just the locus of points whose distance from $1$, (precisely $|z-1|$) is the same as they're distance from $-1$ (precisely $|z+1|$).
The set of points equidistant from $1$ and $-1$ in the complex plane is clearly just the imaginary axis!
Hope this helps!!!
A: Doing it OP's way:

Then I squared both sides and used the formula $|z|^2 =z \bar z$

$$\require{cancel}
\begin{align}
|z+1|^2=|z-1|^2 &\iff (z+1)(\bar z + 1) = (z-1)(\bar z - 1) \\ 
 &\iff \cancel{z \bar z}+ z + \bar z + \bcancel{1} = \cancel{z \bar z}- z - \bar z + \bcancel{1} \\
 &\iff 2(z+\bar z) = 0 \\[5px]
 &\iff \operatorname{Re}(z)=0
\end{align}
$$
