Precalculus, Chapter 6, 6.5, Section 6.5, Problem 68

(3-2i)^8
Take note that De Moivre's Theorem is used to compute the powers and roots of a complex number. The formula is:
[ r(cos theta +isintheta)]^n = r^n(cos(ntheta) + isin(ntheta))
Notice that its formula is in trigonometric form. So to compute (3-2i)^8 , it is necessary to convert the complex number z= 3-2i to trigonometric form z=r(cos theta+isin theta ).
To convert z=x+yi to z=r(costheta +isintheta) , apply the formula
r=sqrt(x^2+y^2) and theta = tan^(-1)y/x
So,
r=sqrt(3^2+(-2)^2)=sqrt(9+4)=sqrt13

theta = tan^(-1) (-2)/3=-33.69007^o
Since x is positive and y is negative, theta is located at the fourth quadrant. So the equivalent positive angle of theta is:
theta =360^o +(-33.69007^o)=326.30993^o
Hence, the trigonometric form of the complex number
z=3-2i
is
z=sqrt13(cos326.30993^o + isin326.30993^o)
Now that it is in trigonometric form, proceed to apply the formula of De Moivre's Theorem to compute z^8 .
z^8=(3-2i)^8
=[sqrt13(cos326.30993^o +isin326.30993^o)]^8
=(sqrt13)^8(cos(8xx326.30993^o) +isin(8xx326.30993^o))
=28561(cos(8xx326.30993^o) +isin(8xx326.30993^o))
= -239+28560i

Therefore, (3-2i)^8=-239+28560i .