If $|z+w|=|z-w|$, prove that $\arg z-\arg w =\pm\pi/2$ If $|z+w|=|z-w|$, how to prove that 
$$\arg z-\arg w =\pm\frac{\pi}{2}$$
I squared the modulus so that I can use properties of conjugates to simplify it yielding $zw*=-wz*$.
 A: lenght of diagonals equal of a parallelogram whose 
vertex at origin and othe two vertex are z and w.
since diagonal equal that implies square or rectangle.
hence angle is right angle
A: 
Looking at the above image, we see that if $|z+w|=|z-w|$, the two triangles $z,w,z+w$ and $z,-w,z-w$ are congruent, which means that the two marked angles are the same. This can only happen if those angles are right, or if $z,w$ are at right angles to each other, or $\arg z-\arg w=\pm\pi/2$.
A: A bit of geometry.
Correct me if wrong.
A rotation does not change distances, and angles between 2 vectors, i.e.
$|z-w|=|z+w|$ implies $|z'-w'|=|z'+w'|$, where $z',w'$ are the vectors in the rotated coordinate system. 
Given  $w, -w$, rotate the complex plane by $\phi$, where 
$w=|w| e^{i\phi}.$
$w \rightarrow w'$ , where $w' =|w|$, is a point on the real axis in the rotated complex plane.
$|z'-|w|| =|z'+|w||$ .
$z'$ is the locus of points equidistant from $|w|$ and $-|w|$, i .e.
$z'$ is on the perpendicular bisector, the complex axis.
Finally $\arg z - \arg w = \arg z' - \arg w' =$
$ \pm π/2 -0=\pm π/2.$
(Why $\pm$ ?)
A: So after squaring etc, you got $z\bar{w}+\bar{z}w=0$. This means real part of $z\bar{w}$ is $0$. If $z=re^{i\alpha}$ and $w=se^{i\beta}$. Then $z\bar{w}=rse^{i(\alpha -\beta)}$. So real part zero gives you $\cos(\alpha -\beta)=0$.
A: You can visualise these complex numbers in the form of vectors. No what is given is the difference and sum of these vectors, which you know form the diagonals of a parallelogram. Now if diagonals of a parallelogram are equal, then it's a rectangle. So difference of arguments is 90 degrees. 
