# Thread: What is the third root of i?

1. ## What is the third root of i?

What is the third root of i?

2. Originally Posted by Manizzle
What is the third root of i?
$\displaystyle z^3 = i$

$\displaystyle \Rightarrow r^3 \text{cis} (3 \theta) = \text{cis} \left( \frac{\pi}{2} + 2n \pi\right)$ where n is an integer.

Therefore $\displaystyle r = 1$ and $\displaystyle \theta = \, .....$ (get three values of $\displaystyle \theta$ that give three distinct solutions for z).

This gives the three cube roots of i in polar form. Convert to Cartesian form as required.

3. Originally Posted by Manizzle
What is the third root of i?
.

The general formula that give the n, nth roots of a complex No z is:

....$\displaystyle |z|^\frac{\ 1}{n}[cos\frac{\ t + 2\pi j}{n} +isin\frac{\ t + 2\pi j}{n}]$ = $\displaystyle |z|^\frac{\ 1}{n}$$\displaystyle e^\frac{\ i(t+2\pi j)}{n}...........,where \displaystyle 0\leq j<n And in our case n=3 ,hence \displaystyle 0\leq j<3, t=π/2 ,|z|=1 and for: .........j=0,j=1,j=2.....we get the 3 roots of i............................. 4. Originally Posted by archidi . The general formula that give the n, nth roots of a complex No z is: ....\displaystyle |z|^\frac{\ 1}{n}[cos\frac{\ t + 2\pi j}{n} +isin\frac{\ t + 2\pi j}{n}] = \displaystyle |z|^\frac{\ 1}{n}$$\displaystyle e^\frac{\ i(t+2\pi j)}{n}$...........,where $\displaystyle 0\leq j<n$

And in our case n=3 ,hence $\displaystyle 0\leq j<3$, t=π/2 ,|z|=1 and for:

.........j=0,j=1,j=2.....we get the 3 roots of i.............................
1. j is used in some circumstances to represent $\displaystyle \sqrt{-1}$ so it's not the ideal pronumeral to use here.

2. j could be confused with J which is a common symbol for rational numbers. So again, not the ideal choice of pronumeral.

3. It needs to be made very clear that the j represents the set of integers.

4. The values of these integers is not restricted to 0, 1, ... n. In fact, the required values can be any n consecutive integers, including negative values.