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Question Number 85261 by Umar last updated on 20/Mar/20

Find the value of (1+(√(−3)))^(3/4) help please

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}+\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$ \\ $$$$\mathrm{help}\:\mathrm{please} \\ $$

Commented by mathmax by abdo last updated on 20/Mar/20

A=(1+(√(−3)))^(3/4) =(1+i(√3))^(3/4) but 1+i(√3)=2((1/2)+i((√3)/2))=2e^((iπ)/3) ⇒ A =(2 e^((iπ)/3) )^(3/4) =2^(3/4) e^((iπ)/4) =^4 (√8)( ((√2)/2) +i((√2)/2))

$${A}=\left(\mathrm{1}+\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:{but}\:\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$${A}\:=\left(\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:=^{\mathrm{4}} \sqrt{\mathrm{8}}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$

Answered by MJS last updated on 20/Mar/20

1+(√(−3))=1+(√3)i=2e^(i(π/3)) (2e^(i(π/3)) )^(3/4) =2^(3/4) e^(i(π/4)) =2^(1/4) +2^(1/4) i

$$\mathrm{1}+\sqrt{−\mathrm{3}}=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} =\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{i} \\ $$

Commented by Umar last updated on 20/Mar/20

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Rio Michael last updated on 20/Mar/20

(1 + (√(−3)))^((3/4) ) = (1 + (√3)i)^(3/4) = 2(cos((π/3)) + isin((π/3)))^(3/4) By demoives theorem, 2^(3/4) (cos((π/3)×(3/4)) + isin((π/3)×(3/4))) = (8)^(1/4) (((√2)/2) + ((√2)/2)i) these can be simplified and the answer gotton. OR you could expand (1 + x)^n = 1 + nx + ((n(n−1))/(2!))x^2 + ((n(n−1)(n−2))/(3!))x^3 + ... ⇒ (1 + (√(−3)))^(3/4) = 1 + ((3(√(−3)))/4) + (((3/4)(((−1)/4)))/(2!))( (√(−3)))^2 + ... = 1 + ((3(√(−3)))/4) + (9/(32)) + ...

$$\left(\mathrm{1}\:+\:\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}\:} =\:\left(\mathrm{1}\:+\:\sqrt{\mathrm{3}}\mathrm{i}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)\:+\:\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\mathrm{By}\:\mathrm{demoives}\:\mathrm{theorem},\:\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}\right)\:+\:\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}\right)\right)\:=\:\sqrt[{\mathrm{4}}]{\mathrm{8}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\: \\ $$$$\mathrm{these}\:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}\:\mathrm{and}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{gotton}. \\ $$$$\:\:\mathrm{OR} \\ $$$$\mathrm{you}\:\mathrm{could}\:\mathrm{expand} \\ $$$$\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:\mathrm{1}\:+\:{nx}\:+\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} \:+\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} \:+\:... \\ $$$$\Rightarrow\:\left(\mathrm{1}\:+\:\sqrt{−\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:=\:\mathrm{1}\:+\:\frac{\mathrm{3}\sqrt{−\mathrm{3}}}{\mathrm{4}}\:+\:\frac{\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}!}\left(\:\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} +\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:+\:\frac{\mathrm{3}\sqrt{−\mathrm{3}}}{\mathrm{4}}\:+\:\frac{\mathrm{9}}{\mathrm{32}}\:+\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by Umar last updated on 20/Mar/20

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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