# Math Help - De Moivre's Theorem

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

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

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

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

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

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

Then I hit a wall. At first glance, I try to say ' $=8Cos6\theta$' 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.

2. Isn't $8\cos^32\theta$ already expressed in terms of $\cos{2\theta}$?

3. Yes, but the question means without the exponent involved.

4. Originally Posted by Quacky
Yes, but the question means without the exponent involved.
Why do you think that?

5. 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 $Cos6\theta$ otherwise?

6. $\displaystyle \cos^3{t} = \cos{t}\cos^2{t} = \cos{t}\left(\frac{1+\cos{2t}}{2}\right) = \frac{\cos{t}+\cos{t}\cos{2t}}{2} = \frac{\cos{t}+\frac{\cos(2t-t)+\cos(2t+t)}{2}}{2}$

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

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

Put $t = 2\theta$ and multiply by $8$, you get what you want.

7. Thanks for the help. That's exactly what I needed.

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

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

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

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

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

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

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

Then I hit a wall. At first glance, I try to say ' $=8Cos6\theta$' 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,

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

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

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

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

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