# De Moivre's Theorem

• Apr 14th 2009, 03:56 PM
Struggle_Through
De Moivre's Theorem
I have the problem: (1+i)/\5 (One plus i, raised to the 5th). I know r= square root of 2 and arctan= pi/4. Then we have square root of 2 raised to the 5th. This is where I run into a problem. How exactly does this work out. For example i know how to get the cube 8 by 8/\(1/3). I just don't know how to raise an nth power to something that is squared.
Thanks.
• Apr 14th 2009, 04:30 PM
nzmathman
$\displaystyle (\sqrt{2})^5 = (2^{1/2})^5 = 2^{5/2}$

Or you can write this as $\displaystyle \sqrt{2^5} = \sqrt{32}$

And simplifying you can get $\displaystyle = 4\sqrt{2}$

But as far as I know it doesn't really matter which way you write it...
• Apr 15th 2009, 05:48 AM
Hello Struggle_Through
Quote:

Originally Posted by Struggle_Through
I have the problem: (1+i)/\5 (One plus i, raised to the 5th). I know r= square root of 2 and arctan= pi/4. Then we have square root of 2 raised to the 5th. This is where I run into a problem. How exactly does this work out. For example i know how to get the cube 8 by 8/\(1/3). I just don't know how to raise an nth power to something that is squared.
Thanks.

You don't actually say what it is you have to do with $\displaystyle (1+i)^5$, but I hope you may find this helpful:

$\displaystyle (1+i)^2 = 1 +2i + i^2 = 2i$

$\displaystyle \Rightarrow (1+i)^4 = (2i)^2 = -4$

$\displaystyle \Rightarrow (1+i)^5 = -4(1+i)$

$\displaystyle (1+i)^5=\sqrt{2^5} ( \frac {\sqrt{2}} {2}+i \frac {\sqrt{2}} {2} )^5=$
$\displaystyle =\sqrt{2^4\cdot2}(cos\frac{\pi}{4}+isin\frac{\pi}{ 4})^5=4\sqrt{2}(cos\frac{5\pi}{4}+isin\frac{5\pi}{ 4})$
$\displaystyle =4\sqrt{2}(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=-4-4i=-4(1+i)$
$\displaystyle (cos(x)+isin(x))^n=cos(nx)+isin(nx)$