DeMoivre's Theorem and nth Roots of Complex Numbers?

Question

Calculate the given product and express your answer in the form a + bi.

1. (√3/2 + 1/2i)^10

Find the nth roots of the given number in polar form.

2. 81(cos pie/12 + i sin pie/12); n = 4


3. 1 + i ; n = 2

Please show all work and formulas used.

Thank you

Answer 1

DeMoivre's theoem says that, for a complex number in polar form z=rcis(w) [using cis(w) as shorthand for cos(w) + isin(w)] and any n,
z^n = r^n*cis(nw).

1. √3/2 + 1/2i in polar form is cis(pi/6), so by DeMoivre's theorem
cis(pi/6)^10 = cis(10pi/6)
= cis(5pi/3)
= 1/2 -√3/2i.

2. Applying DeMoivre's theorem with n=1/4 gives
[81cis(pi/12)]^(1/4) = 81^(1/4)cis(pi/(12*4))
= 3cis(pi/48)
= 2.99 + 0.196i.

3. Applying DeMoivre's theorem to 1+i =√2cis(pi/4) with n=1/2 gives
(√2cis(pi/4))^(1/2) = 2^(1/4)cis(pi/8)
= 1.099 + 0.455i.

Answer 2

1. sqrt(3)/2 + (1/2)i = cos(pi/6) + i sin(pi/6). By DeM,
[(cos(pi/6) + i sin(pi/6)]^10 = cos[10*(pi/6)] + i sin[10*(pi/6)] =
cos(5pi/3) + i sin(5pi/3) = 1/2 + i(-sqrt(3)/2) .