# Find all 3 cube roots of i

• Jul 17th 2010, 03:23 PM
FailingTrig
Find all 3 cube roots of i
1. z(1/3) = r^(*1/3) e^(i*pi/2/3)
= cos(pi/6) + i*sin(pi/6)
= squareroot(3/4) + i/2

2. z(1/3) = r^(*1/3) e^(i*(pi/2 + 2*pi)/3)
= cos(5*pi/6) + i*sin(5*pi/6)
= -squareroot(3/4) + i/2

3. z(1/3) = r^(*1/3) e^(i*(pi/2 + 4*pi)/3)
= cos(3*pi/2) + i*sin(3*pi/2)
= 0 - i

Is that correct?

A friend of mine says its supposed to be

k = 0 ==> i^(1/3) = cos(π/6) + i sin(π/6) = (√3 + i)/2
k = 1 ==> i^(1/3) = cos(5π/6) + i sin(5π/6) = (-√3 + i)/2
k = 2 ==> i^(1/3) = cos(3π/2) + i sin(3π/2) = -i.

How do i figure this out =/
• Jul 17th 2010, 05:56 PM
mr fantastic
Quote:

Originally Posted by FailingTrig
1. z(1/3) = r^(*1/3) e^(i*pi/2/3)
= cos(pi/6) + i*sin(pi/6)
= squareroot(3/4) + i/2

2. z(1/3) = r^(*1/3) e^(i*(pi/2 + 2*pi)/3)
= cos(5*pi/6) + i*sin(5*pi/6)
= -squareroot(3/4) + i/2

3. z(1/3) = r^(*1/3) e^(i*(pi/2 + 4*pi)/3)
= cos(3*pi/2) + i*sin(3*pi/2)
= 0 - i

Is that correct?

A friend of mine says its supposed to be

k = 0 ==> i^(1/3) = cos(π/6) + i sin(π/6) = (√3 + i)/2
k = 1 ==> i^(1/3) = cos(5π/6) + i sin(5π/6) = (-√3 + i)/2
k = 2 ==> i^(1/3) = cos(3π/2) + i sin(3π/2) = -i.

How do i figure this out =/

Both sets of answers are correct and equivalent. Surely you can see that all your friend has done is simplified what you got.