Question

How do you find the power `[2(cos(pi/4)+isin(pi/4)]^5` and express the result in rectangular form?

Answer 1

`[2(cos((5pi)/4)+isin((5pi)/4))]^5=-16sqrt2-i16sqrt2`

Explanation:

According to DeMoivre's theorem if `z=r(costheta+isintheta)`

then `z^n=r^n(cosntheta+isinntheta)`

Hence, if `z=2(cos(pi/4)+isin(pi/4))`

`z^5=2^5(cos((5pi)/4)+isin((5pi)/4))`

= `32(cos(pi+pi/4)+isin(pi+pi/4))`

= `32(-cos(pi/4)-isin(pi/4))`

= `32(-1/sqrt2-i1/sqrt2)`

= `-32/sqrt2-i32/sqrt2`

= `-16sqrt2-i16sqrt2`