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Question Number 197089 by pete last updated on 07/Sep/23

$$\mathrm{Simplify}{\left(\frac{1+\sqrt{3}\mathrm{i}}{1-\sqrt{3}\mathrm{i}}\right)}^{10}$$

Answered by JDamian last updated on 07/Sep/23

$${\left(\frac{z}{z^{\ast }}\right)}^n={\left(\frac{\cancel{\mid z\mid }\cdot e^{i\phi }}{\cancel{\mid z\mid }\cdot e^{-i\phi }}\right)}^n=e^{i2n\phi }$$

Commented by MM42 last updated on 08/Sep/23

$$\underset{―}{\underbrace{⋚}}$$

Commented by pete last updated on 07/Sep/23

$$\mathrm{thank}\mathrm{you}\mathrm{sir},\mathrm{but}\mathrm{i}\mathrm{will}\mathrm{be}\mathrm{glad}\mathrm{if}\mathrm{you}\mathrm{can}\mathrm{offer}$$$$\mathrm{some}\mathrm{further}\mathrm{explanation}$$

Commented by MM42 last updated on 08/Sep/23

$$youshouldstudymoreaboutcomplexnumber$$

Commented by pete last updated on 07/Sep/23

$$\mathrm{I}\mathrm{have}\mathrm{heard}\mathrm{you},\mathrm{sir}.$$$$\mathrm{Sir}\mathrm{can}\mathrm{you}\mathrm{recommend}\mathrm{a}\mathrm{book}\mathrm{for}\mathrm{me}?$$

Answered by Frix last updated on 08/Sep/23

$$\frac{1+\sqrt{3}\mathrm{i}}{1-\sqrt{3}\mathrm{i}}=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}=z$$$$x^3=1\Rightarrow x=1\vee x=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}\Rightarrow z^3=1$$$$z^{10}=z^3\times z^3\times z=1\times 1\times z=z=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}$$$$$$$$1\pm \sqrt{3}\mathrm{i}=\mid 1\pm \sqrt{3}\mathrm{i}\mid {\mathrm{e}}^{\mathrm{i}{\tan }^{-1}\pm \sqrt{3}}={2\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{3}}$$$$\frac{1+\sqrt{3}\mathrm{i}}{1-\sqrt{3}\mathrm{i}}=\frac{{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{3}}}{{2\mathrm{e}}^{-\mathrm{i}\frac{\pi }{3}}}={\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}$$$${\left({\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}\right)}^{10}={\mathrm{e}}^{\mathrm{i}\frac{20\pi }{3}}={\mathrm{e}}^{\mathrm{i}(6\pi +\frac{2\pi }{3})}={\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}$$

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