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.
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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.