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

Comparing the complex number&nbsp;3+i&nbsp;with standard form of complex number&nbsp;(x+iy)Where “x” is real part and “y” is imaginary part of the complex number.So,&nbsp;x=3, y=1The polar form of the complex number&nbsp;3+i&nbsp;is:r(cosθ+isinθ)To calculate "r":r=+&nbsp;&nbsp;2y2=32+12=3+1=2To calculate&nbsp;θθ=tan-1yx=tan-113=tan-1tanπ6θ=π6tan-1tanθ=03+i3=2cosπ6+isinπ63=8cosπ6+isinπ63Apply Demoivre's theorem:&nbsp;rcosθ+isinθn=rncosnθ+isinnθ=8cos3π6+isin3π6=8cosπ2+isinπ2=80+i·1=8i3+i3=0+8iReal part&nbsp;=0Imaginarty part&nbsp;=8

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Answered question

2020-11-05

Solve by using De Moivre's Theorem (find real and imaginary part of complex number):

{(\sqrt{3} + i)}^{3}

Answer & Explanation

Faiza Fuller

Skilled2020-11-06Added 108 answers

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

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