# De Moivre's Theorem

• Nov 17th 2010, 01:08 PM
Quacky
De Moivre's Theorem
I have absolutely no idea where this question should go, so please feel free to move it to the correct subforum.

a)If $\displaystyle z = cos\theta + i sin\theta$, use de Moivre's theorem to show that $\displaystyle \displaystyle z^n + \frac{1}{z^n} = 2cos (n\theta)$

b) Express $\displaystyle (z^2 + \frac{1}{z^2})^3$ In terms of $\displaystyle Cos{6\theta}$ and $\displaystyle Cos{2\theta}$

Part a) I have done correctly, but part b) Is confusing me. My working so far:
$\displaystyle (z^2 + \frac{1}{z^2})^3$

$\displaystyle = [2Cos(2\theta)]^3$

$\displaystyle =2^3(Cos(2\theta))^3$

$\displaystyle =8(Cos(2\theta))^3$

Then I hit a wall. At first glance, I try to say '$\displaystyle =8Cos6\theta$' but that doesn't fit the question. I see that I have to use the theorem in some way, but I have no idea on the correct approach from here.
• Nov 17th 2010, 02:19 PM
emakarov
Isn't $\displaystyle 8\cos^32\theta$ already expressed in terms of $\displaystyle \cos{2\theta}$?
• Nov 17th 2010, 02:28 PM
Quacky
Yes, but the question means without the exponent involved.
• Nov 17th 2010, 02:33 PM
Plato
Quote:

Originally Posted by Quacky
Yes, but the question means without the exponent involved.

Why do you think that?
• Nov 17th 2010, 02:36 PM
Quacky
Quote:

Originally Posted by Plato
Why do you think that?

Experience with the textbook and past questions, mainly. I doubt the question would include the part about $\displaystyle Cos6\theta$ otherwise?
• Nov 17th 2010, 03:36 PM
TheCoffeeMachine
$\displaystyle \displaystyle \cos^3{t} = \cos{t}\cos^2{t} = \cos{t}\left(\frac{1+\cos{2t}}{2}\right) = \frac{\cos{t}+\cos{t}\cos{2t}}{2} = \frac{\cos{t}+\frac{\cos(2t-t)+\cos(2t+t)}{2}}{2}$

$\displaystyle \displaystyle = \frac{\cos{t}+\frac{\cos{t}+\cos{3t}}{2}}{2} = \frac{\frac{2\cos{t}+\cos{t}+\cos{3t}}{2}}{2} = \frac{2\cos{t}+\cos{t}+\cos{3t}}{4} = \frac{3\cos{t}+\cos{3t}}{4}$.

$\displaystyle \therefore ~ \displaystyle \cos^3{t} = \frac{3\cos{t}+\cos{3t}}{4}$.

Put $\displaystyle t = 2\theta$ and multiply by $\displaystyle 8$, you get what you want. (Wink)
• Nov 18th 2010, 10:00 AM
Quacky
Thanks for the help. That's exactly what I needed.
• Nov 18th 2010, 01:06 PM
Quote:

Originally Posted by Quacky
I have absolutely no idea where this question should go, so please feel free to move it to the correct subforum.

a)If $\displaystyle z = cos\theta + i sin\theta$, use de Moivre's theorem to show that $\displaystyle \displaystyle z^n + \frac{1}{z^n} = 2cos (n\theta)$

b) Express $\displaystyle (z^2 + \frac{1}{z^2})^3$ In terms of $\displaystyle Cos{6\theta}$ and $\displaystyle Cos{2\theta}$

Part a) I have done correctly, but part b) Is confusing me. My working so far:

$\displaystyle (z^2 + \frac{1}{z^2})^3$

$\displaystyle = [2Cos(2\theta)]^3$

$\displaystyle =2^3(Cos(2\theta))^3$

$\displaystyle =8(Cos(2\theta))^3$

Then I hit a wall. At first glance, I try to say '$\displaystyle =8Cos6\theta$' but that doesn't fit the question. I see that I have to use the theorem in some way, but I have no idea on the correct approach from here.

(b)

Using the Binomial Expansion,

$\displaystyle \displaystyle\left(z^2+\frac{1}{z^2}\right)^3=\bin om{3}{0}\left(z^2\right)^3+\binom{3}{1}\left(z^2\r ight)^2\frac{1}{z^2}+\binom{3}{2}z^2\left(\frac{1} {z^2}\right)^2+\binom{3}{3}\left(\frac{1}{z^2}\rig ht)^3$

$\displaystyle =z^6+3z^2+3\left(\frac{1}{z^2}\right)+\frac{1}{z^6 }$

$\displaystyle =\left(z^6+\frac{1}{z^6}\right)+3\left(z^2+\frac{1 }{z^2}\right)$

$\displaystyle =2Cos6\theta+3\left(2Cos2\theta}\right)$

$\displaystyle =2Cos6\theta+6Cos2\theta$