Using DeMoivre's Theorem to put (1 - i)^12 in standard form

[Failenn]

the problem is to put solve a complex number using DeMoivere's theorm and put in standard form.

I know that standard form is a + bi.

(1- i)^12.

first I solved for r and got r is = to sqrt of 2, then i plugged in the values into the equation

[sqrt 2 (cos pi/4 + i sin pi/4)^12
(sqrt 2)^12 (cos 12pi/4 + i sin 12pi/4)
64 (cos 3pi + i sin 3pi)

this is where i got stuck because 3pi I think is 180 degrees.

 

[DrMike]

Re: Using DeMoivre's Theorem

sqrt 2 (cos pi/4 + i sin pi/4)^12

Don't you mean sqrt 2 (cos pi/4 - i sin pi/4)^12 ? You've put the complex number in the wrong quadrant...

64 (cos 3pi + i sin 3pi)

this is where i got stuck because 3pi I think is 180 degrees.

Well, it's actually 540 degrees, but the cosine and sine are the same.