Question

Compute the indicated power by using de Moivre's theorem.
$\sqrt[4]{1-i\sqrt{3} }$
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

$|1 - i\sqrt3| = \sqrt{1^2 + (-\sqrt3)^2} = \sqrt4 = 2$

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

$\arg(1-i\sqrt3) = \tan^{-1}\left(-\sqrt3\right) = -\dfrac\pi3$

Then the exponential form is

$1 - i \sqrt3 = 2 \exp\left(-i\dfrac\pi3\right)$

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

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

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

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

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

(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.