Complex number plane multiplication

**meggnog** · Mar 28th 2010, 03:53 PM

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.

**Prove It** · Mar 28th 2010, 06:02 PM

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

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

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

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

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

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