Question

find the indicated power using De Moivre's Theorem

(1 + i)^20

Answer 1

(1 + i )^20 = { ( 1 + i )^2 } ^10

{ 1 + i² + 2i } ^10

{ 1 - 1 + 2i } ^10

{ 2i } ^ 10 = = ( 2i x 2i x 2i x 2i x 2i ) ^2 = ( 32i^5 ) ^2

= 32i^5 x 32i^5 = 1024i^10 = 1024 x -1 = -1024

Answer 2

thank you

Answer 3

No problem :) Any other questions?

Answer 4

This way, you did not use De Moivre Theorem

Answer 5

I just have one more question. I need help to find the indicated roots, and graph the roots in the complex plane.

the fifth roots of 32

Answer 6

The answer to that is 2

Answer 7

And what about the complex roots?

Answer 8

\[
(1+i)^{20}=\left(\sqrt{2} \left(\cos \left(\frac{\pi }{4}\right)+i \sin
\left(\frac{\pi }{4}\right)\right)\right)^{20}=2^{10} (\cos (5 \pi )+i
\sin (5 \pi ))=-2^{10}=-1024
\]

Answer 9

Mmm yah I was thinking the same thing elia +_+\[\Large\bf\sf \left(1+\mathcal i\right)^{20}\quad=\quad \left(\sqrt{2}\left[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\mathcal i\right]\right)^{20}\]
\[\Large\bf\sf 2^{10}\left(\cos\frac{\pi}{4}+\mathcal i \sin\frac{\pi}{4}\right)^{20}\quad=\quad 2^{10}\left(\cos\frac{20\pi}{4}+\mathcal i\sin\frac{20\pi}{4}\right)\]

Answer 10

@SithsAndGiggles
Hmm I'm confused...
Did you combine both questions together or something? +_+
It looks like you took the 5th root of 1+i.

Answer 11

haha yeah i did... Sorry about that!

Answer 12

XD

Answer 13

I'll just show myself out.

Answer 14

For a *general* complex number \(a+ib=r(\cos\theta+i\sin\theta)\), the \(n\)th roots are
\[r^{1/n}\left(\cos\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)+i\sin\cos\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)\right)\]
where \(k=0,1,2,...,n-1\).

Answer 15

i sin cos( ... ) ?

Oh man you're on a roll today sith XD hehe jk