Question Number 85261 by Umar last updated on 20/Mar/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}+\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$ \\ $$$$\mathrm{help}\:\mathrm{please} \\ $$
Commented by mathmax by abdo last updated on 20/Mar/20
$${A}=\left(\mathrm{1}+\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:{but}\:\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$${A}\:=\left(\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:=^{\mathrm{4}} \sqrt{\mathrm{8}}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$
Answered by MJS last updated on 20/Mar/20
$$\mathrm{1}+\sqrt{−\mathrm{3}}=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} =\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{i} \\ $$
Commented by Umar last updated on 20/Mar/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Rio Michael last updated on 20/Mar/20
$$\left(\mathrm{1}\:+\:\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}\:} =\:\left(\mathrm{1}\:+\:\sqrt{\mathrm{3}}\mathrm{i}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)\:+\:\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\mathrm{By}\:\mathrm{demoives}\:\mathrm{theorem},\:\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}\right)\:+\:\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}\right)\right)\:=\:\sqrt[{\mathrm{4}}]{\mathrm{8}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\: \\ $$$$\mathrm{these}\:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}\:\mathrm{and}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{gotton}. \\ $$$$\:\:\mathrm{OR} \\ $$$$\mathrm{you}\:\mathrm{could}\:\mathrm{expand} \\ $$$$\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:\mathrm{1}\:+\:{nx}\:+\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} \:+\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} \:+\:… \\ $$$$\Rightarrow\:\left(\mathrm{1}\:+\:\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\:\mathrm{1}\:+\:\frac{\mathrm{3}\sqrt{−\mathrm{3}}}{\mathrm{4}}\:+\:\frac{\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}!}\left(\:\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} +\:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:+\:\frac{\mathrm{3}\sqrt{−\mathrm{3}}}{\mathrm{4}}\:+\:\frac{\mathrm{9}}{\mathrm{32}}\:+\:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Umar last updated on 20/Mar/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$