Answer to Question #315512 in Algebra for عبدالرحمن

Question

Evaluate the cube root of z=27cis(240^ \circ) . Then raise them to the cube. Show the steps of your reasoning.

2. Evaluate [ \sqrt[5]{3} ( \frac{ \sqrt{3}}{2}+ \frac{i}{2} ) ]^{10} .

3. Find \frac{z_1}{z_2} in polar form:

z_1=21cis(135^ \circ) z_2=3cis(75^ \circ)

Answer 1

$Q1. Evaluate the cube root of z=27cis(240^ \circ). Then raise them to the cube. Show the steps of your reasoning$

Using DeMoivre's Theorem, one of the cube roots is 27^1/3cis(240°/3) = 3cis80°

The 3 cube roots of the given number are equally spaced around the circle centered at (0,0) with radius 3.

360° / 3 = 120°

Another cube root is 3cis(80°+120°) = 3cis200°

The third cube root is 3cis(80°+2(120°) = 3cis320° [Answer]

$Q2. Evaluate [ \sqrt[5]{3} ( \frac{ \sqrt{3}}{2}+ \frac{i}{2} ) ]^{10} .$

convert to polar form to get $\sqrt[5]{3}$ cis30°

So, [ $\sqrt[5]{3}$ cis30°]¹⁰ = ($\sqrt[5]{3}$ )¹⁰cis[10(30°)] = 237046875cis300° [Answer]

$Q3. Find \frac{z_1}{z_2} in polar form:$

$z_1=21cis(135^ \circ)\\ z_2=3cis(75^ \circ)$

Divide 21 by 3 and subtract 75° from 135°

So, z₁ / z₂ = 7cis60° [Answer]