Question

find the root for (-1+i)^1/3​

Answer 1

Answer:

E. None of the above.

Step-by-step explanation:

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

$z^{1/n} = r^{1/n}\cdot \left[\cos \left(\frac{\theta + 2\cdot k\cdot \pi}{n} \right)+i\,\sin \left(\frac{\theta + 2\cdot k\cdot \pi}{n}\right)\right]$, 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

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

$r = \sqrt{2}$

Direction

$\theta = \tan^{-1} \left(\frac{1}{-1}\right)$

$\theta = \frac{3\pi}{4}$

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

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

$z^{1/3} = 2^{1/6}\cdot \left[\cos \frac{\pi}{4}+i\,\sin \frac{\pi}{4} \right]$

$z^{1/3} = 2^{1/6}\cdot \left[\cos 45^{\circ}+i\,\sin 45^{\circ}\right]$

Hence, the right answer is E.