# Solve by using De Moivre's Theorem (find real and imaginary part of complex number): (sqrt3+i)^3

Solve by using De Moivre's Theorem (find real and imaginary part of complex number):
${\left(\sqrt{3}+i\right)}^{3}$
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Faiza Fuller

Comparing the complex number $$\displaystyle{\left(\sqrt{{3}}+{i}\right)}$$ with standard form of complex number $$(x+iy)$$
Where “x” is real part and “y” is imaginary part of the complex number.
So, $$\displaystyle{x}=\sqrt{{3}}, {y}={1}$$
The polar form of the complex number $$\displaystyle{\left(\sqrt{{3}}+{i}\right)}{i}{s}:$$
$$r(\cos \theta + i \sin \theta)$$
To calculate "r":
$$\displaystyle{r}=\sqrt{{^{2}+{y}^{{2}}}}=\sqrt{{{\left(\sqrt{{3}}\right)}^{{2}}+{\left({1}\right)}^{{2}}}}=\sqrt{{{3}+{1}}}={2}$$
To calculate $$\displaystyle\theta$$
$$\displaystyle\theta={{\tan}^{{-{{1}}}}{\left(\frac{{y}}{{x}}\right)}}={{\tan}^{{-{{1}}}}{\left(\frac{{1}}{\sqrt{{3}}}\right)}}={{\tan}^{{-{{1}}}}{\left({\tan{{\left(\frac{\pi}{{6}}\right)}}}\right)}}$$
$$\displaystyle\theta=\frac{\pi}{{6}}$$
$$\displaystyle{{\tan}^{{-{{1}}}}{\left({\tan{\theta}}\right)}}={0}$$
$$\displaystyle{\left(\sqrt{{3}}+{i}\right)}^{{3}}={\left[{2}{\left({\cos{{\left(\frac{\pi}{{6}}\right)}}}+{i}{\sin{{\left(\frac{\pi}{{6}}\right)}}}\right)}\right]}^{{3}}={8}{\left({\cos{{\left(\frac{\pi}{{6}}\right)}}}+{i}{{\sin{{\left(\frac{\pi}{{6}}\right)}}}^{{3}}}\right.}$$
Apply  Demoivre's  theorem: $$\displaystyle{\left[{r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}\right]}^{{n}}={r}^{{n}}{\left({\cos{{n}}}\theta+{i}{\sin{{n}}}\theta\right)}$$
$$\displaystyle={8}{\left({\cos{{\left(\frac{{{3}\pi}}{{6}}\right)}}}+{i}{\sin{{\left(\frac{{{3}\pi}}{{6}}\right)}}}\right)}={8}{\left({\cos{{\left(\frac{\pi}{{2}}\right)}}}+{i}{\sin{{\left(\frac{\pi}{{2}}\right)}}}={8}{\left({0}+{i}\cdot{1}\right)}={8}{i}\right.}$$
$$\displaystyle{\left(\sqrt{{3}}+{i}\right)}^{{3}}={0}+{8}{i}$$
Real  part$$=0$$
Imaginarty  part$$=8$$