Blog: `2(sqrt(3) + i)^10` Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

Given $\displaystyle{2}{{\left(\sqrt{{3}}+{i}\right)}}^{{10}}$

Let $\displaystyle{z}={{\left(\sqrt{{3}}+{i}\right)}}^{{10}}$

$\displaystyle{r}=\sqrt{{{{\left(\sqrt{{3}}\right)}}^{{2}}+{{\left({1}\right)}}^{{2}}}}=\sqrt{{{3}+{1}}}=\sqrt{{4}}={2}$

$\displaystyle\theta={\arctan{{\left(\frac{{1}}{\sqrt{{3}}}\right)}}}=\frac{\pi}{{6}}$

DeMoivre's Theorem

$\displaystyle{{z}}^{{n}}={{\left[{r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}\right]}}^{{n}}={{r}}^{{n}}{\left({\cos{{n}}}\theta+{i}{\sin{{n}}}\theta\right)}$

$\displaystyle{{z}}^{{10}}={{\left[{2}{\left({\cos{{\left(\frac{\pi}{{6}}\right)}}}+{i}{\sin{{\left(\frac{\pi}{{6}}\right)}}}\right)}\right]}}^{{10}}$

$\displaystyle{{z}}^{{10}}={{2}}^{{10}}{\left({\cos{{10}}}{\left(\frac{\pi}{{6}}\right)}+{i}{\sin{{10}}}{\left(\frac{\pi}{{6}}\right)}\right)}$

$\displaystyle{{z}}^{{10}}={1024}{\left[{\cos{{\left(\frac{{{5}\pi}}{{3}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{3}}\right)}}}\right]}}}\right.}$

$\displaystyle{{z}}^{{10}}={1024}{\left[\frac{{1}}{{2}}+{\left(-\frac{\sqrt{{3}}}{{2}}\right)}{i}\right]}$

$\displaystyle{{z}}^{{10}}={512}-{512}\sqrt{{3}}{i}$

$\displaystyle{2}{z}={2}{\left({512}-{512}\sqrt{{3}}{i}\right)}={1024}-{1024}\sqrt{{3}}{i}$

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Thursday, June 4, 2015

$\displaystyle{2}{{\left(\sqrt{{{3}}}+{i}\right)}}^{{10}}$ Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

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