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Find $(1+\sqrt{3} i)^8$. Express the answer in rectangular form.

Solution

Verified$\begin{gathered} {\left( {1 + \sqrt 3 i} \right)^8} \Rightarrow n = 8 \\ \textcolor{#4257b2}{{\text{Write }}1 + \sqrt 3 i{\text{ in polar form}}} \\ 1 + \sqrt 3 i \Rightarrow x = 1,y = \sqrt 3 \\ \textcolor{#4257b2}{{\text{Apply }}r = \sqrt {{x^2} + {y^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right),{\text{ so}}} \\ r = \sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} = 2 \\ \theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{1}} \right) = \frac{\pi }{3} \\ \textcolor{#4257b2}{ {\text{Polar form }}z = r\left( {\cos \theta + i\sin \theta } \right)} \\ z = 2\left[ {\cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right)} \right] \\ \textcolor{#4257b2}{ {\text{Apply De Moivre}}'{\text{s theorem}}} \\ {z^n} = {r^n}\left( {\cos n\theta + i\sin n\theta } \right) \\ \textcolor{#4257b2}{ {\text{Let }}n = 8,r = 2,\theta = \frac{\pi }{3},{\text{ so}}} \\ {\left( {1 + \sqrt 3 i} \right)^8} = {\left( 2 \right)^8}\left[ {\cos 8\left( {\frac{\pi }{3}} \right) + i\sin 8\left( {\frac{\pi }{3}} \right)} \right] \\ \textcolor{#4257b2}{{\text{Simplify}}} \\ {\left( {1 + \sqrt 3 i} \right)^8} = 256\left[ {\cos \left( {\frac{{8\pi }}{3}} \right) + i\sin \left( {\frac{{8\pi }}{3}} \right)} \right] \\ {\left( {1 + \sqrt 3 i} \right)^8} = 256\left( { - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) \\ {\left( {1 + \sqrt 3 i} \right)^8} = - 128 + 128\sqrt 3 i \\ \end{gathered}$

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