2

If z and w are complex numbers such that $|z+w|$ = $|z-w|$, prove that $\arg(z)-\arg(w)= \pm\pi/2$.

Can this be solved algebraically or would a graphic interpretation be better. Both methods would be grateful.

Giovanni
  • 6,321
Stratos1
  • 25

3 Answers3

5

Geometrically, consider the parallelogram with vertices $0,z,w,z+w$. Then the condition is that its two diagonals have the same length, so it must be a rectangle, that is $z$ and $w$ are orthogonal.

Algebraically, from $$(z+w)(\bar{z}+\bar{w})=(z-w)(\bar{z}-\bar{w})$$ one obtains $z\bar{w}+w\bar{z}=0$ and now just switch to polar notation.

Sonner
  • 3,108
2

Since $\arg(0)$ is not defined, let's assume that $z,w\ne0$. Then dividing by $\left|\,w\,\right|$ gives $$ \left|\,\frac zw-1\,\right|=\left|\,\frac zw+1\,\right| $$ This means that $\frac zw$ is equidistant from $-1$ and $1$; that is, $\frac zw$ is pure imaginary. Thus, $$ \arg\left(\frac zw\right)=\pm\frac\pi2 $$ which is the same as $$ \arg(z)-\arg(w)=\pm\frac\pi2 $$

robjohn
  • 345,667
1

$|z+w| = |z-w|$ means that $z$ is equidistant from $\pm w$, and so $z$ lies on the line bisecting the segment $[-w,w]$. This line is perpendicular to the segment, hence the result.

lhf
  • 216,483
  • Your previous answer, which was posted while I was still typing, was very close to mine. As I was typing up something about that, your answer changed. Now I don't need to :-) – robjohn Sep 22 '15 at 16:11