Precalculus, Chapter 6, 6.5, Section 6.5, Problem 67

(-1+i)^6
De Moivre's Theorem is used to compute the powers and roots of a complex number. The formula is:
[r(costheta + isintheta)]^n=r^n(cos(nxxtheta) + isin(n xx theta))
To be able to apply it, convert the complex number z=-1+i to trigonometric form.Take note that that the trigonometric form of
z=x+yi
is
z=r(costheta + isintheta)
where
r =sqrt(x^2+y^2) and theta=tan^(-1) y/x
Applying these two formulas, the values of r and theta of z=-1+i are:
r=sqrt((-1)^2+1^2)=sqrt2
theta = tan^(-1) (1/(-1))=tan^(-1) (-1) = -45^o
Since x is negative and y is positive, theta is located at the second quadrant. So the equivalent positive angle of theta is:
theta =180^o +(-45^o)=135^o
Then, plug-in the values of r and theta to the trigonometric form
z=r(costheta + isintheta)
z=sqrt2(cos135^o +isin135^o)
Now that z=-1+i is in trigonometric form, proceed to compute z^6 .
z^6=(-1+i)^6
=[sqrt2(cos135^o + isin135^o)]^6
= (sqrt2)^6(cos(6xx135^o)+isin(6xx135^o)
=8(cos810^o + isin810^o)
=8(0 + 1i)
=8i

Therefore, (-1+i)^6=8i .