Given two non-zero complex numbers *z*_{1} and *z*_{2} such that , show that the arguments of *z*_{1} and *z*_{2} differ by π/2.

Recall that for any complex number *z*, , thus

Now, recall that for any complex number *z*, , and since , we see that

, and so

,

and we proved last week that *z*_{1} and *z*_{2} have arguments differing by π/2 (and thus give perpendicular vectors) if and only if .

Consider the geometric interpretation. Given a parallelogram with sides given by the vectors corresponding to *z*_{1} and *z*_{2}, then *z*_{1}+*z*_{2} and *z*_{1}–*z*_{2} correspond to the diagonals; thus, the above is equivalent to stating that if the diagonals of a parallelogram are of equal length, then the parallelogram is a rectangle.

### Like this:

Like Loading...

*Related*

Tags: Complex Numbers, Geometry, Math, Monday Math

This entry was posted on December 20, 2010 at 1:58 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

March 19, 2012 at 10:21 am |

Hi!

I really like your website, I think it’s really useful, but unfortunately, I cannot view any of the formulae, there’s a boxed cross.

If you have any idea what the problem is, can you please let me know?

Thanks,

panayiota