# Complex number plane multiplication

• Mar 28th 2010, 03:53 PM
meggnog
Complex number plane multiplication
We know that multiplying the plane of complex numbers by a constant complex number u multiplies all distances by the absolute value of u.

Assuming that u is not equal to 1, and hence that uz is not equal to z when z is not equal to 0, deduce that multiplication by u is a rotation.
• Mar 28th 2010, 06:02 PM
Prove It
Quote:

Originally Posted by meggnog
We know that multiplying the plane of complex numbers by a constant complex number u multiplies all distances by the absolute value of u.

Assuming that u is not equal to 1, and hence that uz is not equal to z when z is not equal to 0, deduce that multiplication by u is a rotation.

Let $\displaystyle u = \cos{\theta_1} + i\sin{\theta_1}$

and $\displaystyle z = \cos{\theta_2} + i\sin{\theta_2}$.

Therefore $\displaystyle uz = (\cos{\theta_1} + i\sin{\theta_1})(\cos{\theta_2} + i\sin{\theta_2})$

$\displaystyle = \cos{(\theta_1 + \theta_2)} + i\sin{(\theta_1 + \theta_2)}$.

Since the angle has changed, that means you have a rotation.