# DeMoivre's theorem

• January 14th 2007, 09:25 PM
mike1
DeMoivre's theorem
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.
• January 15th 2007, 03:52 AM
Soroban
Hello, Mike!

Quote:

Use DeMoivre's theorem to write $\left[3\,\text{cis}\left(\frac{\pi}{2}\right)\right]^2$ in the form $a + bi$

We don't really need DeMoivre's Theorem for this one, but here goes . . .

We have: . $\bigg[3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\bigg]^2\;=\;3^2\bigg[\cos\left(2\!\cdot\!\frac{\pi}{2}\right) + i\sin\left(2\!\cdot\!\frac{\pi}{2}\right)\bigg]$

. . . . . . $= \;9\left[\cos\pi + i\sin\pi\right] \;=\;9[-1 + i(0)] \;=\;\boxed{-9}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could have squared it without DeMoivre's Theorem.

$3\,\text{cis}\frac{\pi}{2}\:=\:3\left(\cos\frac{\p i}{2} + i\sin\frac{\pi}{2}\right) \:=\:3[0 + i(1)] \:=\:3i$

. . Then: . $(3i)^2\:=\:-9$