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

hexacordoK 2020-11-05 Answered
Solve by using De Moivre's Theorem (find real and imaginary part of complex number):
(3+i)3
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Expert Answer

Faiza Fuller
Answered 2020-11-06 Author has 108 answers

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\(=8\)

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