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

Simplify (((1+(√3)i)/(1−(√3)i)))^(10)

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

Answered by JDamian last updated on 07/Sep/23

((z/z^∗ ))^n =(((∣z∣∙e^(iϕ) )/(∣z∣∙e^(−iϕ) )))^n =e^(i2nϕ)

$$\left(\frac{{z}}{{z}^{\ast} }\right)^{{n}} =\left(\frac{\cancel{\mid{z}\mid}\centerdot{e}^{{i}\varphi} }{\cancel{\mid{z}\mid}\centerdot{e}^{−{i}\varphi} }\right)^{{n}} ={e}^{{i}\mathrm{2}{n}\varphi} \\ $$

Commented by MM42 last updated on 08/Sep/23

⋛

$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

Commented by pete last updated on 07/Sep/23

thank you sir,but i will be glad if you can offer some further explanation

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

Commented by MM42 last updated on 08/Sep/23

you should study more about complex number

$${you}\:{should}\:{study}\:{more}\:{about}\:{complex}\:{number} \\ $$

Commented by pete last updated on 07/Sep/23

I have heard you, sir. Sir can you recommend a book for me?

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

Answered by Frix last updated on 08/Sep/23

((1+(√3)i)/(1−(√3)i))=−(1/2)+((√3)/2)i=z x^3 =1 ⇒ x=1∨x=−(1/2)±((√3)/2)i ⇒ z^3 =1 z^(10) =z^3 ×z^3 ×z=1×1×z=z=−(1/2)+((√3)/2)i 1±(√3)i=∣1±(√3)i∣e^(i tan^(−1) ±(√3)) =2e^(±i(π/3)) ((1+(√3)i)/(1−(√3)i))=((2e^(i(π/3)) )/(2e^(−i(π/3)) ))=e^(i((2π)/3)) (e^(i((2π)/3)) )^(10) =e^(i((20π)/3)) =e^(i(6π+((2π)/3))) =e^(i((2π)/3))

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

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