Simplify-1-3-i-1-3-i-10-

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

Simplify {(\frac{1 + \sqrt{3} i}{1 - \sqrt{3} i})}^{10}

Answered by JDamian last updated on 07/Sep/23

{(\frac{z}{z^{*}})}^{n} = {(\frac{∣ z ∣ \cdot e^{i φ}}{∣ z ∣ \cdot e^{- i φ}})}^{n} = e^{i 2 n φ}

Commented by MM42 last updated on 08/Sep/23

\underset{―}{\underset{⏟}{⋚}}

Commented by pete last updated on 07/Sep/23

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

some further explanation

Commented by MM42 last updated on 08/Sep/23

y o u s h o u l d s t u d y m o r e a b o u t c o m p l e x n u m b e r

Commented by pete last updated on 07/Sep/23

I have heard you, sir .

Sir can you recommend a book for me ?

Answered by Frix last updated on 08/Sep/23

\frac{1 + \sqrt{3} i}{1 - \sqrt{3} i} = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = z

x^{3} = 1 \Rightarrow x = 1 \lor x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} 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} i

1 \pm \sqrt{3} i =∣ 1 \pm \sqrt{3} i ∣ e^{i \tan^{- 1} \pm \sqrt{3}} = {2 e}^{\pm i \frac{π}{3}}

\frac{1 + \sqrt{3} i}{1 - \sqrt{3} i} = \frac{{2 e}^{i \frac{π}{3}}}{{2 e}^{- i \frac{π}{3}}} = e^{i \frac{2 π}{3}}

{(e^{i \frac{2 π}{3}})}^{10} = e^{i \frac{20 π}{3}} = e^{i (6 π + \frac{2 π}{3})} = e^{i \frac{2 π}{3}}

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