Let $z_1 = 4 \left[ \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right]$ and \(z_2 = 2 \left[

Question

Let $z_1 = 4 \left[ \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right]$ and $z_2 = 2 \left[ \cos \left( -\frac{\pi}{6} \right) + i \sin \left( -\frac{\pi}{6} \right) \right]$. Calculate, in polar form, the complex numbers $\frac{z_1}{z_2}$, $z_2^3$, and $-iz_1^2$.

Accepted Answer

$\dfrac{z_1}{z_2}=2\operatorname{cis}\!\left(\dfrac\pi4\right)$, $z_2^3=8\operatorname{cis}\!\left(-\dfrac\pi2\right)$, $-iz_1^2=16\operatorname{cis}\!\left(-\dfrac\pi3\right)$, where $\operatorname{cis}\theta=\cos\theta+i\sin\theta$.

Examples involving De Moivre's Theorem

Examples involving De Moivre's Theorem

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