Complex numbers, I've forgotten everything

[lnaa19]

How do I show that (1+i)^19 is equal to 2^9(i-1)?

[paulsterio]

convert to polar use de moivres theorem

[lnaa19]

...how do I do that? ><

[Special At Specialist]

1 + i = sqrt(2) cis(45)
(1 + i)^19 = (sqrt(2))^19 * cis(pi/4 * 15)
(1 + i)^19 = sqrt(2)*2^9 * cis(3pi/4)
Since cis(3pi/4) = -1/sqrt(2) + 1/sqrt(2) *i
(1 + i)^19 = 2^9 * sqrt(2) * (-1/sqrt(2) + 1/sqrt(2) *i)
(1 + i)^19 = 2^9 * -1 + 2^9 * i
(1 + i)^19 = -2^9 + 2^9 *i
(1 + i)^19 = (-2^9)(1 - i)

[lnaa19]

Isn't this more of unit 3&4 than 1&2? I'm a noobie.
Sorry to be a nuisance but thanks for your contibution nonetheless.

[monkeywantsabanana]

(1 + i)^19 = (sqrt(2))^19 * cis(pi/4 * 15)

Is this meant to be:

(1 + i)^19 = sqrt(2)^19 *cis(pi/4 *19) ?

I stumbled across De Moivre's Theorem yesterday when we got our textbooks... correct me if I'm wrong...

[pi]

^^That's right. Must've been a typo

Isn't this more of unit 3&4 than 1&2? I'm a noobie.
Sorry to be a nuisance but thanks for your contibution nonetheless.

Yeh, it is

[lnaa19]

I think I've got it.

(1+i) (1+i)^18
=(1+i) ((1+i)^2)^9
=(1+i) (2i)^9
=(1+i) (2^9)(i)
=2^9(i-1)

Yay !