The question:

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

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

For those interested:

$\displaystyle (1 - i) = \sqrt{2}e^{-i\frac{\pi}{4}}$

$\displaystyle z^2 = r^2e^{i2\theta}$

Multiplying both gives $\displaystyle \sqrt{2}r^2e^{-i\frac{\pi}{4}+{2\theta}}$

Since we want the modulus, we get $\displaystyle \sqrt{2}r^2$