Use DeMoivre’s Theorem to find the indicated power of the complex number .Write answer in rectangular form. (sqrt3-i)^6

Question
Complex numbers
Use DeMoivre’s Theorem to find the indicated power of the complex number .Write answer in rectangular form.
$$\displaystyle{\left(\sqrt{{3}}-{i}\right)}^{{6}}$$

2020-12-18
The complex number can be written as,
$$\displaystyle{\left(\sqrt{{3}}-{i}\right)}={2}{\left(\frac{\sqrt{{3}}}{{2}}-{i}\frac{{1}}{{2}}\right)}$$
$$\displaystyle{\left(\sqrt{{3}}-{i}\right)}={2}{\left({\cos{{\left(-\frac{\pi}{{6}}\right)}}}+{i}{\sin{{\left(-\frac{\pi}{{6}}\right)}}}\right)}$$
therefore, it can be represented as,
$$\displaystyle{2}{\left({\cos{{\left(-\frac{\pi}{{6}}\right)}}}+{i}{\sin{{\left(-\frac{\pi}{{6}}\right)}}}\right)}={2}{e}^{{-{i}{\left(\frac{\pi}{{6}}\right)}}}$$
therefore,
$$\displaystyle{\left(\sqrt{{3}}-{i}\right)}^{{6}}={\left[{2}{e}^{{-{i}{\left(\frac{\pi}{{6}}\right)}}}\right]}^{{6}}={2}^{{6}}{e}^{{-{i}\pi}}={64}{e}^{{-\pi}}$$
Therefore ,the given complex number can be simplified further as,
$$\displaystyle{\left(\sqrt{{3}}-{i}\right)}^{{6}}={64}{\left({\cos{{\left(-\pi\right)}}}+{i}{\sin{{\left(-\pi\right)}}}\right)}={64}\cdot{\left(-{1}\right)}+{i}{0}{)}={64}\cdot{\left(-{1}\right)}=-{64}$$
Hence, the given complex number becomes −64.

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