If (root(3)+i)^10 = a+bi then a and b are

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\left(\sqrt{3}+\mathrm{i}\right)}^{10}=\mathrm{a}+\mathrm{ib} \\ \mathrm{Now} \\ \mathrm{z}=\sqrt{3}+\mathrm{i} \\ \left|\mathrm{z}\right|=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=2 \\ \mathrm{\alpha }={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6} \\ \mathrm{Hence} \sqrt{3}+\mathrm{i}=2\left(\cos \frac{\mathrm{\pi }}{6}+\mathrm{isin}\frac{\mathrm{\pi }}{6}\right) \\ \mathrm{Hence} {\left(\sqrt{3}+\mathrm{i}\right)}^{10}=\mathrm{a}+\mathrm{ib} \mathrm{becomes} \\ \Rightarrow {\left[2\left(\cos \frac{\mathrm{\pi }}{6}+\mathrm{isin}\frac{\mathrm{\pi }}{6}\right)\right]}^{10}=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}{\left(\cos \frac{\mathrm{\pi }}{6}+\mathrm{isin}\frac{\mathrm{\pi }}{6}\right)}^{10}=\mathrm{a}+\mathrm{ib} \\ \mathrm{Using} \mathrm{De}-\mathrm{Movier}\text{'}\mathrm{s} \mathrm{theorem} \\ \Rightarrow 2^{10}\left(\cos \frac{10\mathrm{\pi }}{6}+\mathrm{isin}\frac{10\mathrm{\pi }}{6}\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}\left(\cos \frac{5\mathrm{\pi }}{3}+\mathrm{isin}\frac{5\mathrm{\pi }}{3}\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}\left(\cos \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)+\mathrm{isin}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}\left(\cos \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)+\mathrm{isin}\left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}\left(\cos \frac{\mathrm{\pi }}{3}-\mathrm{isin}\frac{\mathrm{\pi }}{3}\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^{10}\left(\frac{1}{2}-\mathrm{i}\frac{\sqrt{3}}{2}\right)=\mathrm{a}+\mathrm{ib} \\ \Rightarrow 2^9-\mathrm{i}{2}^9\sqrt{3}=\mathrm{a}+\mathrm{ib} \\ \mathrm{Comparing} \\ \mathrm{a}=2^9 \\ \mathrm{b}=2^9\sqrt{3} \end{gathered}$$

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