# If $|z_1 - z_2| = |z_1 + z_2|$, then $\arg z_1 - \arg z_2 = \pi/4$

### Problem

If $|z_1 - z_2| = |z_1 + z_2|$, show $$\arg(z_1) - \arg(z_2) = \frac{\pi}{4}.$$

### Progress

I have tried squaring the modulus and using double angle formula for $\tan$, but can't get to the answer.

• $\frac \pi 2$ maybe? – sas Feb 17 '14 at 5:40
• This isn't true. For example, let $z_1=1$ and $z_2=0$ – David Peterson Feb 17 '14 at 5:40
• It must be $\frac{\pi}{2}$. See my answer. Also, both numbers must be non zero for "arg" to be meaningful. – user44197 Feb 17 '14 at 5:43
• Zero doesn't have quite well-defined complex argument. – sas Feb 17 '14 at 5:46
• @sas Yes, my fault – David Peterson Feb 17 '14 at 5:49

Geometric solution.

You have quadrilateral: two sides are your numbers $r_1$ and $r_2$, two diagonals are $r_1-r_2$ and $r_1+r_2$.

Diagonals are equal — so quadrilateral is rectangle.

• Cool answer. Nicely done – user44197 Feb 17 '14 at 5:47
• Not as cool as it seems. I forgot to tell that it is not just a quadrilateral — it is parallelogram. Because it just represents vectors and translation. Other way it could be isosceles trapezium :) – sas Feb 17 '14 at 6:00
• That is just an oversight. The idea comes through. – user44197 Feb 17 '14 at 6:02

Assume both number are nonzero.

The condition is equivalent to $$(z_1-z_2)(\bar z_1 - \bar z_2)=(z_1+z_2)(\bar z_1 + \bar z_2)$$ Multiply out to get $$\frac{z_1}{\bar z_1} = -\frac{z_2}{\bar z_2}$$ or $$2 \angle z_1 = \pm\pi + 2 \angle z_2$$ which gives $$\angle z_1 -\angle z_2 = \pm\frac{\pi}{2}$$