Question Number 197089 by pete last updated on 07/Sep/23
$$\mathrm{Simplify}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}}\right)^{\mathrm{10}} \\ $$
Answered by JDamian last updated on 07/Sep/23
$$\left(\frac{{z}}{{z}^{\ast} }\right)^{{n}} =\left(\frac{\cancel{\mid{z}\mid}\centerdot{e}^{{i}\varphi} }{\cancel{\mid{z}\mid}\centerdot{e}^{−{i}\varphi} }\right)^{{n}} ={e}^{{i}\mathrm{2}{n}\varphi} \\ $$
Commented by MM42 last updated on 08/Sep/23
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
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
$${you}\:{should}\:{study}\:{more}\:{about}\:{complex}\:{number} \\ $$
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
$$\frac{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}={z} \\ $$$${x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{z}^{\mathrm{3}} =\mathrm{1} \\ $$$${z}^{\mathrm{10}} ={z}^{\mathrm{3}} ×{z}^{\mathrm{3}} ×{z}=\mathrm{1}×\mathrm{1}×{z}={z}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$ \\ $$$$\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i}=\mid\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i}\mid\mathrm{e}^{\mathrm{i}\:\mathrm{tan}^{−\mathrm{1}} \:\pm\sqrt{\mathrm{3}}} =\mathrm{2e}^{\pm\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}}=\frac{\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} }{\mathrm{2e}^{−\mathrm{i}\frac{\pi}{\mathrm{3}}} }=\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{10}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{20}\pi}{\mathrm{3}}} =\mathrm{e}^{\mathrm{i}\left(\mathrm{6}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$