Question

Using DeMoivre's Theorem, find the indicated power of the complex number. Express the result in standard form. (2+2i)^6

Answer 1

Demovers is what:

r^n (cos(nax) + sin(nbx)) ??

im sure i never remember that one right

Answer 2

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx

Answer 3

the theta part is what i was missing :/

Answer 4

given:\[(a+bi)^n\]
\[r=\sqrt{a^2+b^2};\ t=tan^{-1}(\frac{b}{a})\]
\[(a+bi)^n=r^n(cos(nt)+isin(nt))\]

Answer 5

Amistre

Answer 6

@amistre64

Answer 7

Need your help real quick

Answer 8

timmys in the well?

Answer 9

So (2+2i)^6, sqrt(8)=2sqrt(2), and tan^-1=2/2 = pi/4, now what?

Answer 10

Lassie!

Answer 11

now we fill in demovers

Answer 12

so 2sqrt(2)^6(cos(5*pi/4)+isin(5*pi/4))?

Answer 13

\[(a+bi)^n=r^n(cos(nt)+isin(nt))\]
\[(2+2i)^6=(2\sqrt{2})^6(cos(6\frac{pi}4)+isin(6\frac{pi}4))\]

Answer 14

\[(2\sqrt{2})^6(cos(6\frac{pi}4)+isin(6\frac{pi}4))\]
\[(64*8)(cos(\frac{3pi}2)+isin(\frac{3pi}2 ))\]

Answer 15

Sorry I meant 6 in my equation, typoed.

Answer 16

sin(t) = -1 there and cos(t)=0

Answer 17

(64*8)(-i)

Answer 18

480
32
----
512

-512i

Answer 19

Is there a different more easier way to do this?

Answer 20

and what does "sin(t) = -1 there and cos(t)=0" have to do with anything?