Answer to Question #261860 in Calculus for Pankaj

Question

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.

Answer 1

Solution:

We know that

$i^{1/3}=\cos(4k+1)\dfrac{\pi}6+i\sin(4k+1)\dfrac{\pi}6; k=0,1,2$

So, for k=0

$i^{1/3}=\cos\dfrac{\pi}6+i\sin \dfrac{\pi}6 \\=\dfrac{\sqrt3}2+\dfrac i2$

For k=1

$i^{1/3}=\cos\dfrac{5\pi}6+i\sin \dfrac{5\pi}6 \\=-\cos\dfrac{\pi}6+i\sin \dfrac{\pi}6 \\=-\dfrac{\sqrt3}2+\dfrac i2$

For k=2

$i^{1/3}=\cos\dfrac{9\pi}6+i\sin \dfrac{9\pi}6 \\=\cos\dfrac{3\pi}2+i\sin \dfrac{3\pi}2 \\=-\cos\dfrac{\pi}2+i(-\sin \dfrac{\pi}2) \\=-i$

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.