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

Use DeMoivre’s Theorem to find the indicated power of the complex number .Write answer in rectangular form.
${\left(\sqrt{3}-i\right)}^{6}$
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The complex number can be written as,
$\left(\sqrt{3}-i\right)=2\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)$
$\left(\sqrt{3}-i\right)=2\left(\mathrm{cos}\left(-\frac{\pi }{6}\right)+i\mathrm{sin}\left(-\frac{\pi }{6}\right)\right)$
therefore, it can be represented as,
$2\left(\mathrm{cos}\left(-\frac{\pi }{6}\right)+i\mathrm{sin}\left(-\frac{\pi }{6}\right)\right)=2{e}^{-i\left(\frac{\pi }{6}\right)}$
therefore,
${\left(\sqrt{3}-i\right)}^{6}={\left[2{e}^{-i\left(\frac{\pi }{6}\right)}\right]}^{6}={2}^{6}{e}^{-i\pi }=64{e}^{-\pi }$
Therefore ,the given complex number can be simplified further as,
${\left(\sqrt{3}-i\right)}^{6}=64\left(\mathrm{cos}\left(-\pi \right)+i\mathrm{sin}\left(-\pi \right)\right)=64\cdot \left(-1\right)+i0\right)=64\cdot \left(-1\right)=-64$
Hence, the given complex number becomes −64.