# Homework Help: De Moivre's Theorum

1. Apr 25, 2007

### catteyes

De Moivre's Theorum - (just needs checking)

1. The problem statement, all variables and given/known data

(1+i)^20

2. Relevant equations

De Moivre's theorum: [r(cos theta + isin theta)]^n= r^n(cos ntheta + isin ntheta):uhh: (i think)

3. The attempt at a solution
x= 1
y=1
r= 1
theta= 45 degrees
[1(cos 45 + i sin 45)]^20 = 1^20[cos(20*45) + i sin (20*45)]
therefore:
1(cos 900 + i sin 900)
(-1 + i 0)

Last edited: Apr 25, 2007
2. Apr 25, 2007

### Vagrant

Yes looks like it.

3. Apr 25, 2007

### da_willem

are you sure?

4. Apr 25, 2007

### George Jones

Staff Emeritus
Check out da_willem's comment.

De Moivre's theorem is the standard way to do this. Another way is

$$\left( 1 + i \right)^{20} = \left( \left( \left( 1 + i \right)^2 \right)^2 \right)^5.$$

Work from the inside to the outside.

5. Apr 25, 2007

### HallsofIvy

If z= 1+ i, then $r= |z|= \sqrt{(1+i)(1-i)}= \sqrt{2}$, NOT 1. Your twentieth power is missing a factor of $r^{20}= (\sqrt{2})^20= 2^{10}= 1024$.

6. Apr 26, 2007

### catteyes

or is it the square root of 2?

7. Apr 26, 2007

### Dick

How would you figure out the answer to that question?