compute $(\sqrt 3 + i)^{14} + ( \sqrt 3 - i)^{14}$

Question

I want to compute $(\sqrt 3 + i)^{14} + ( \sqrt 3 - i)^{14}$.

My attempts: I was thinking about De Moivre's theorem

$$(\cos\theta + i \sin\theta)^n= \cos(n\theta) + i \sin(n\theta)$$

but I don't know how it can be applied here. Any hints/solution will be appreciated.

Thank you.

Write $\sqrt{3}\pm i$ in the form $r(\cos\theta\pm i\sin\theta)$ and then apply de Movire's theorem. — A. Goodier, Sep 04 '18 at 21:09

Ahmad Bazzi · Accepted Answer · 2018-09-04T21:35:33.087

3

You're right but you have have to express it in exponential (polar) form, $$\sqrt{3} + i = 2 e^{i \frac{\pi}{6}}$$ and $$\sqrt{3} - i = 2 e^{-i \frac{\pi}{6}}$$ So that $$(\sqrt{3} + i)^{14} = 2^{14} e^{i \frac{14\pi}{6}}$$ and $$(\sqrt{3} - i)^{14} = 2^{14} e^{-i \frac{14\pi}{6}}$$ Since $(\sqrt{3} + i)^{14}$ and $(\sqrt{3} - i)^{14}$ are complex conjugates, then $(\sqrt{3} + i)^{14} + (\sqrt{3} - i)^{14} = 2\Re\{(\sqrt{3} + i)^{14}\} = 2(2^{14} \cdot\Re\{e^{i \frac{\pi}{6}}\}) $

edited Sep 04 '18 at 21:35

answered Sep 04 '18 at 21:10

Ahmad Bazzi

12,076

thanks @taritgoswami – Ahmad Bazzi Sep 04 '18 at 21:14
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Note that $(2a)^{14} = 2^{14} a^{14} \neq 2 a^{14} $. – Math Lover Sep 04 '18 at 21:33
thanks @MathLover – Ahmad Bazzi Sep 04 '18 at 21:35

score 2 · Answer 2 · answered Sep 04 '18 at 21:10

2

HINT: $\sqrt{3}+i=2(\cos(\pi/6)+i\sin(\pi/6))$, $\sqrt{3}-i=2(\cos(-\pi/6)+i\sin(-\pi/6))$.

answered Sep 04 '18 at 21:10

Przemysław Scherwentke

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compute $(\sqrt 3 + i)^{14} + ( \sqrt 3 - i)^{14}$

2 Answers2