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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.

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

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  • $\begingroup$ Cool answer. Nicely done $\endgroup$ – user44197 Feb 17 '14 at 5:47
  • $\begingroup$ 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 :) $\endgroup$ – sas Feb 17 '14 at 6:00
  • $\begingroup$ That is just an oversight. The idea comes through. $\endgroup$ – user44197 Feb 17 '14 at 6:02
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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 $$ \angle z_1 -\angle z_2 = \pm\frac{\pi}{2} $$

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