If z=(3/5)-(4/5)i , what is z^10 and the three cube roots of z

Question

This is a polar coordinate and complex numbers question. (powers and roots of complex numbers)

Answer 1

°z = (3/5) - (4/5)i

Express z in polar form:

l z l = √[(3/5)² + (4/5)²] = 1

Since (3/4, -4/5) is in quadrant 4, θ = 360° - Tan^-1(4/3) = 306.87°

So, z = rcosθ + (rsinθ)i = cos306.87° + isin306.87°

By DeMoivre's Theorem, z¹⁰ = r¹⁰(cos10θ + isin10θ) = -0.988 - 0.151i

The 3 cube roots of z are equally spaced about the circle centered at (0,0) with radius r^1/3 = 1

One of the three is r^1/3(cos(θ/3) + isin(θ/3)) = cos102.29° + isin102.29° = -0.213 + 0.977i

360°/3 = 120°

Second cube root = r^1/3(cos(102.29° + 120°) + isin(102.29+ 120°)) = -0.740 - 0.673i

Third cube root = r^1/3(cos(102.29° + 2(120°)) + isin(102.29° + 2(120°)) = 0.953 - 0.304i