# DeMoivre's Theorem

• December 4th 2009, 01:49 PM
thekrown
DeMoivre's Theorem
4. b) Use De Moivre's theorem to express (-square root 2 / 2 + i square root 2 / 2 )^50 in the form a + bi

I found out on the unit circle the point is at 135 degrees. Once applying the exponent rules the point becomes 270 degrees (0,-1).

At this point I am stuck trying to convert to a + bi form.

I believe it is radius 2^25 so we get: 2^25 (0 - 1i) for an answer of 2^25i. Yes?
• December 4th 2009, 02:42 PM
Chris L T521
Quote:

Originally Posted by thekrown
4. b) Use De Moivre's theorem to express (-square root 2 / 2 + i square root 2 / 2 )^50 in the form a + bi

I found out on the unit circle the point is at 135 degrees. Once applying the exponent rules the point becomes 270 degrees (0,-1).

At this point I am stuck trying to convert to a + bi form.

I believe it is radius 2^25 so we get: 2^25 (0 - 1i) for an answer of 2^25i. Yes?

Rewrite the expression as $\left(\cos\!\left(\tfrac{3\pi}{4}\right)+i\sin\!\l eft(\tfrac{3\pi}{4}\right)\right)^{50}$.

Applying DeMoivre's Theorem, we get

$\cos\!\left(\tfrac{75}{2}\pi\right)+i\sin\!\left(\ tfrac{75}{2}\pi\right)$.

Due to the periodic behavior of sine and cosine, this is equivalent to $\cos\!\left(\tfrac{3\pi}{2}\right)+i\sin\!\left(\t frac{3\pi}{2}\right)=-i$

Does this make sense?
• December 4th 2009, 02:53 PM
thekrown

I don't see where cos 75 pi / 2 and i sin 75 pi / 2 come from.
• December 4th 2009, 02:58 PM
Chris L T521
Quote:

Originally Posted by thekrown

No, its 1 because $\left|\cos\!\left(\tfrac{3\pi}{4}\right)+i\sin\!\l eft(\tfrac{3\pi}{4}\right)\right|=\sqrt{\cos^2\!\l eft(\tfrac{3\pi}{4}\right)+\sin^2\!\left(\tfrac{3\ pi}{4}\right)}=\sqrt{1}=1$.

Quote:

I don't see where cos 75 pi / 2 and i sin 75 pi / 2 come from.
By DeMoivre's Theorem,

\begin{aligned}\left(\cos\!\left(\tfrac{3\pi}{4}\r ight)+i\sin\!\left(\tfrac{3\pi}{4}\right)\right)^{ 50}&=\cos\!\left(\tfrac{3\pi}{4}\cdot 50\right)+i\sin\!\left(\tfrac{3\pi}{4}\cdot 50\right)\\ &=\cos\!\left(\tfrac{150\pi}{4}\right)+i\sin\!\lef t(\tfrac{150\pi}{4}\right)\\ &=\cos\!\left(\tfrac{75\pi}{2}\right)+i\sin\!\left (\tfrac{75\pi}{2}\right)\end{aligned}
• December 4th 2009, 04:33 PM
thekrown
Okay I understand where the 3 pi / 2 comes from, it is 270 degrees. I had the wrong radius, I think I was using the initial radius maybe?

I believe I made the mistake of not using the formula to find the radius. I'm having trouble with this but I'll give it another shot.

If I have trouble I will let you know. Thanks.
• December 4th 2009, 04:39 PM
thekrown
On the unit circle the point is at 0,-1 for 270 degrees or cos 3 pi / 2 + sin 3 pi / 2, yes?

If so can I use the formula r = square root of a^2 + b^2 which is the same as square root 0^2 + (-1)^2 = 1?