# Polar form

## Homework Statement

solve below using de Moivre's theorem, in polar form
z^3 = i

## Homework Equations

r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

## The Attempt at a Solution

r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle

danago
Gold Member

## Homework Statement

solve below using de Moivre's theorem, in polar form
z^3 = i

## Homework Equations

r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

## The Attempt at a Solution

r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle

It is not true that i = 1, which is essentially what you have done in solving the equation. 1 is simply the modulus of the complex number i.

Try writing the complex number i in polar form i.e. in the form $$R cis \theta$$, and then taking the cube root of both sides of the equation.