# Math Help - Another difficult complex number question

1. ## [Solved] Another difficult complex number question

The question:

If $z = re^{i\theta}, 0 \le \theta \le \frac{\pi}{2}$, show that $|(1 - i)z^2| = \sqrt{2}r^2$

EDIT: I worked it out. It's actually really easy. >_<

For those interested:

$(1 - i) = \sqrt{2}e^{-i\frac{\pi}{4}}$
$z^2 = r^2e^{i2\theta}$
Multiplying both gives $\sqrt{2}r^2e^{-i\frac{\pi}{4}+{2\theta}}$
Since we want the modulus, we get $\sqrt{2}r^2$

2. You should change the title to [Solved]Another difficult complex number question.