Question

Find the root for (-1+i)^1/3​

Answer 1

E. None of the above.

Explanation:

The n-th root of a complex number can be calculated by the De Moivre's Theorem:

`z^(1/n) = r^(1/n)\cdot \left[\cos \left((\theta + 2\cdot k\cdot \pi)/(n) \right)+i\,\sin \left((\theta + 2\cdot k\cdot \pi)/(n)\right)\right]`, for
`k \in \mathbb{N}_(O)` (1)

Where:

`n` - Grade of the root.

`r` - Norm of the complex number.

`\theta` - Direction of the complex number, measured in radians.

`k` - Index of the solution.

If we know that
`z^(1/3) = (-1+i)^(1/3)`, then the root of the number is:

Norm

`r = \sqrt{(-1)^(2)+1^(2)}`

`r = √(2)`

Direction

`\theta = \tan^(-1) \left((1)/(-1)\right)`

`\theta = (3\pi)/(4)`

If
`k = 0`, then the root of this number is:

`z^(1/3) = 2^(1/6)\cdot \left[\cos \left(((3\pi)/(4) )/(3) \right)+i\,\sin \left(((3\pi)/(4) )/(3) \right)\right]`

`z^(1/3) = 2^(1/6)\cdot \left[\cos (\pi)/(4)+i\,\sin (\pi)/(4) \right]`

`z^(1/3) = 2^(1/6)\cdot \left[\cos 45^(\circ)+i\,\sin 45^(\circ)\right]`

Hence, the right answer is E.