Question

How many roots ${(1-\sqrt{3}i)}^{1/4}$ has

Answer 1

For any complex number

$z=R.e^{i(2k\pi +\theta )}$

$z^{\frac{1}{n}}=R^{\frac{1}{n}}.e^{i\frac{2k\pi +\theta }{n}}$

$=r.e^{i\frac{2k\pi }{n}}.e^{i\frac{\theta }{n}}$

Where $k=1,2,\mathrm{3..}n$

Thus

$z^{\frac{1}{n}}=\left(r{e}^{i\frac{\theta }{n}}\right).e^{i\frac{2k\pi }{n}}$

$z^{\frac{1}{n}}=z_1.e^{i\frac{2k\pi }{n}}$ where $k=1,2,\mathrm{3..}n...\left(i\right)$

Hence

$z^{\frac{1}{n}}$ has exactly $n$ roots.

$\therefore (1-\sqrt{3}i{)}^{1/4}$ has $4$ roots.