Question

Compute the indicated power by using de Moivre's theorem.


`\sqrt[4]{1-i√(3) }`
please hurry

Answer 1

I assume you know that there are n total n-th roots for any non-zero complex number.

(If you're looking for "the" principal root, you'll have to specify which branch you're calling the principal branch of the n-th root function.)

Start by writing
`1 - i\sqrt3` in exponential form. We have modulus/magnitude

`|1 - i\sqrt3| = √(1^2 + (-\sqrt3)^2) = \sqrt4 = 2`

and
`1-i\sqrt3` lies in the fourth quadrant of the complex plane, so its argument is

`\arg(1-i\sqrt3) = \tan^(-1)\left(-\sqrt3\right) = -\frac\pi3`

Then the exponential form is

`1 - i \sqrt3 = 2 \exp\left(-i\frac\pi3\right)`

(where
`\exp(x) = e^x`, if you're not familiar with the notation)

By de Moivre's theorem, we have the fourth roots

`\sqrt[4]{1 - i\sqrt3} = \sqrt[4]{2} \exp\left(i \frac{-\frac\pi3+2k\pi}4\right)`

where
`k\in\{0,1,2,3\}`, so that we have a choice of

`\sqrt[4]{1 - i\sqrt3} = \begin{cases} \sqrt[4]{2} \exp\left(-i\frac\pi{12}\right) \\\\ \sqrt[4]{2} \exp\left(i(5\pi)/(12)\right) \\\\ \sqrt[4]{2} \exp\left(i(11\pi)/(12)\right) \\\\ \sqrt[4]{2} \exp\left(i(17\pi)/(12)\right) = \sqrt[4]{2} \exp\left(-i(7\pi)/(12)\right) \end{cases}`

(I rewrote the exponent to the last root just to be consistent about having each argument between -π and π radians)

If you want these in rectangular form (a + bi), sorry, that's where I draw the line; it can be done with simple trigonometry and algebra, but it's rather tedious, and the exponential forms are far more compact.