Finding (1+i)^20 using De Moivre's theorem

[catteyes]

De Moivre's Theorum - (just needs checking)

Homework Statement

(1+i)^20

Homework Equations

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

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)
the answer: is -1 ??

 

[Vagrant]

Yes looks like it.

 

[da_willem]

r= 1

are you sure?

 

[George Jones]

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.

 

[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 $r^{20}= (\sqrt{2})^20= 2^{10}= 1024$.

 

[catteyes]

are you sure?

or is it the square root of 2?

 

[Dick]

or is it the square root of 2?

How would you figure out the answer to that question?