Question #315512

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)

1
Expert's answer
2022-03-24T19:06:45-0400

Q1.Evaluatethecuberootofz=27cis(240).Thenraisethemtothecube.ShowthestepsofyourreasoningQ1. 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 271/3cis(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[35(32+i2)]10.Q2. Evaluate [ \sqrt[5]{3} ( \frac{ \sqrt{3}}{2}+ \frac{i}{2} ) ]^{10} .


convert to polar form to get 35\sqrt[5]{3} cis30°


So, [ 35\sqrt[5]{3} cis30°]10 = (35\sqrt[5]{3} )10cis[10(30°)] = 237046875cis300° [Answer]



Q3.Findz1z2inpolarform:Q3. Find \frac{z_1}{z_2} in polar form:


z1=21cis(135)z2=3cis(75)z_1=21cis(135^ \circ)\\ z_2=3cis(75^ \circ)


Divide 21 by 3 and subtract 75° from 135°


So, z1 / z2 = 7cis60° [Answer]

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