Answer in Calculus for Pankaj #261860

all the cube root of iota (i) in a complex number (C) are z1= cosπ/2 + i sinπ/2 , z2= cosπ/6 +i sinπ/6 and z3= cos5π/6 + i sin5π/6 .

Verify if the statement is true or false. Give reason for your answer in the form of a short proof or a counterexample.

Solution:

We know that

“i^{1/3}=cos(4k+1)dfrac{pi}6+isin(4k+1)dfrac{pi}6; k=0,1,2”

So, for k=0

“i^{1/3}=cosdfrac{pi}6+isin dfrac{pi}6n\=dfrac{sqrt3}2+dfrac i2”

For k=1

“i^{1/3}=cosdfrac{5pi}6+isin dfrac{5pi}6n\=-cosdfrac{pi}6+isin dfrac{pi}6n\=-dfrac{sqrt3}2+dfrac i2”

For k=2

“i^{1/3}=cosdfrac{9pi}6+isin dfrac{9pi}6n\=cosdfrac{3pi}2+isin dfrac{3pi}2n\=-cosdfrac{pi}2+i(-sin dfrac{pi}2) n\=-i”

Given ones are: z1= cosπ/2 + i sinπ/2 , z2= cosπ/6 +i sinπ/6 and z3= cos5π/6 + i sin5π/6 .

z1 is incorrect, rest are correct.

So, it is false.