I have this problem and I want to make sure I understand it.
Use DeMoivre's theorem to write [3cis(pi/2)][squared] in the form a + bi without trigonometric functions.
Hello, Mike!
Use DeMoivre's theorem to write $\displaystyle \left[3\,\text{cis}\left(\frac{\pi}{2}\right)\right]^2$ in the form $\displaystyle a + bi$
We don't really need DeMoivre's Theorem for this one, but here goes . . .
We have: .$\displaystyle \bigg[3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\bigg]^2\;=\;3^2\bigg[\cos\left(2\!\cdot\!\frac{\pi}{2}\right) + i\sin\left(2\!\cdot\!\frac{\pi}{2}\right)\bigg]$
. . . . . . $\displaystyle = \;9\left[\cos\pi + i\sin\pi\right] \;=\;9[-1 + i(0)] \;=\;\boxed{-9}$
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We could have squared it without DeMoivre's Theorem.
$\displaystyle 3\,\text{cis}\frac{\pi}{2}\:=\:3\left(\cos\frac{\p i}{2} + i\sin\frac{\pi}{2}\right) \:=\:3[0 + i(1)] \:=\:3i $
. . Then: .$\displaystyle (3i)^2\:=\:-9$