$w^2v$ in Polar Form

Algebra Level 3

Given $w=3+3i$ and $v=-2 \sqrt{3}+2i,$ $w^2v$ can be expressed as $r(\cos \alpha^{\circ}+i\sin \alpha^{\circ}),$ where $r$ is a real number and $0 \leq \alpha \leq 360.$ What is the value of $r+\alpha?$

222

252

282

312

1 solution

Tom Engelsman
Jul 22, 2018

Let $u = 3 + 3i = 3\sqrt{2} \cdot e^{i \frac{\pi}{4}}$ and $v = -2\sqrt{3} + 2i = 4 \cdot e^{i \frac{5\pi}{6}}$ . The expression $u^{2} v$ computes to:

$u^{2} v = (3\sqrt{2} \cdot e^{i \frac{\pi}{4}})^{2} ( 4 \cdot e^{i \frac{5\pi}{6}}) = (18 \cdot 4)e^{i (\frac{\pi}{2} + \frac{5\pi}{6}}) = 72 \cdot e^{i \frac{4\pi}{3}} = 72(cos(240) + i \cdot sin(240)).$

Hence, $r = 72, \alpha = 240 \Rightarrow r + \alpha = \boxed{312}.$

w 2 v w^2v w 2 v in Polar Form

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$w^2v$ in Polar Form