Complex Angles

Solution 1: Converting to polar coordinates, we have $|w|^2 = \left(\sqrt{3}\right)^2 + (-1)^2 = 4 \Rightarrow |w| = 2$ $w = 2\left(\frac{\sqrt{3}}{2} -\frac{i}{2}\right) = 2\left(\cos(30^\circ) -i\sin(30^\circ)\right)$ and $|z|^2 = 2^2 + 2^2 = 8 \Rightarrow |z| = 2\sqrt{2}$ $z = 2\sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 2\sqrt{2}\left(\cos(45^\circ) + i\sin(45^\circ)\right)$ .

Multiplying these two, $\begin{aligned} wz &= \left[2\left(\cos(30^\circ) -i\sin(30^\circ)\right)\right] \left[2\sqrt{2}\left(\cos(45^\circ) + i\sin(45^\circ)\right)\right] \\ &= 4\sqrt{2}\left[\cos(30^\circ)\cos(45^\circ) + \sin(30^\circ)\sin(45^\circ) - i\left(\sin(30^\circ)\cos(45^\circ) - \sin(45^\circ)\cos(30^\circ)\right) \right] \\ &= 4\sqrt{2}\left(\cos(15^\circ) - i\sin(-15^\circ)\right) \\ &= 4\sqrt{2}\left(\cos(15^\circ) + i\sin(15^\circ)\right) \\ \end{aligned}$

Hence $r = 4\sqrt{2}$ and $a = 15^\circ$ .

Solution 2: We have $z = 2\sqrt{2}\left(\cos 45^\circ + i\sin 45^\circ \right)$ and $w = 2\left(\cos(-30^\circ) +i\sin(-30^\circ)\right)$ . Applying Euler's formula, we have $wz = \left(2e^{i(-30^\circ)}\right)\left(2\sqrt{2}e^{i(45^\circ)}\right) = 4\sqrt{2}\left(e^{i(15^\circ)}\right) = 4\sqrt{2}\left(\cos(15^\circ) + i\sin(15^\circ)\right)$ . Hence $r = 4\sqrt{2}$ and $a = 15 ^\circ$ .

Note: Students who initially multiplied $z$ and $w$ directly would be unable to proceed without knowing the value of $\sin 15^\circ$ and $\cos 15 ^ \circ$ .