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

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

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

$$A={(1+\sqrt{-3})}^{\frac{3}{4}}={(1+i\sqrt{3})}^{\frac{3}{4}}but1+i\sqrt{3}=2(\frac{1}{2}+i\frac{\sqrt{3}}{2})=2e^{\frac{i\pi }{3}}\Rightarrow$$$$A={\left(2e^{\frac{i\pi }{3}}\right)}^{\frac{3}{4}}=2^{\frac{3}{4}}{e}^{\frac{i\pi }{4}}{=}^4\sqrt{8}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$$

Answered by MJS last updated on 20/Mar/20

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

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

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

Commented by Umar last updated on 20/Mar/20

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

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