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Find the values of $n$ such that $z^n=(1+\sqrt3i)^n$ is a real number.

My reasoning: The power will be real iff $\sin\arg z=0$. Since $\sin 0,\sin\pm\pi,\sin\pm2\pi,\dots=0$, $3\mid n$. Is it correct?

$$z=re^{ia}=2e^{i\pi/3}$$ $$z^n=2^n\left(\cos\frac{n\pi}3+i\sin\frac{n\pi}3\right)$$ $$\to\{n\mid n=3k, k\in\Bbb Z\}$$

Parcly Taxel
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BananaBuisness
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1 Answers1

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An alternate view for $a_{n} = (1+ \sqrt{3} i)^{n}$ is: \begin{align} a_{0} &= 1 \\ a_{1} &= 1 + \sqrt{3} i \\ a_{2} &= -2 + 2 \sqrt{3} i \\ a_{3} &= -8. \end{align} It is determined that $n \in {0, 3, \cdots}$. Consider $n \to 3n$ then \begin{align} a_{3n} &= (1 + \sqrt{3} i)^{3n} = [(1 + \sqrt{3} i)^{3}]^{n} = (-8)^n = (-1)^{n} \, 2^{3n}. \end{align}

This yields the same result as that given by the proposer's solution.

Leucippus
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  • So my approach is wrong? – BananaBuisness Oct 19 '16 at 05:27
  • @AsafFisher Your approach is correct. If, in your solution, you let $n \to 3n$ the result will be the same as the "alternate" view I presented. The method I presented is a longer path that verifies when $a_{n}$ is a real value. – Leucippus Oct 19 '16 at 17:02