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I have this question which I'm stuck on, here's the question and what I did.

Find the smallest positive integer m such that $\left(\sqrt{3}+i\right)^m=\left(\sqrt{3}-i\right)^m$.

I expanded out each side using De Moivre's, $\cos\left(30m\right)+i\sin\left(30m\right)=\cos\left(-30m\right)+i\sin\left(-30m\right)$.

I tried to compare when $\cos\left(30m\right)=\cos\left(-30m\right)$ and $\sin\left(30m\right)=\sin\left(-30m\right)$ but none of the quadrants work, the answer is $m=6$, which corresponds to Quadrant 3 and 4 (tan and cos).

I'm in year 10 just learning complex, so if there are harder methods to this, don't show me.

Akiva Weinberger
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user145591
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    If you want something to be in display mode, write $$...$$ rather than $...$. – Akiva Weinberger Nov 09 '14 at 00:25
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    Note that $\cos (-x) = \cos x$, and $\sin (-x) = -\sin x$, whatever $x$ is. So the real parts are always equal, and for the imaginary parts, you want $m$ such that $\sin (m\cdot 30^\circ) = -\sin (m\cdot 30^\circ)$. Now, $a = -a$ holds if and only if $a = 0$. So $\sin (m\cdot 30^\circ) = 0$. – Daniel Fischer Nov 09 '14 at 00:25
  • Note that when $m=6$, we have $30m^\circ=180^\circ$. – Akiva Weinberger Nov 09 '14 at 00:27
  • Why can't 30m=0, m = 0? – user145591 Nov 09 '14 at 00:40
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    $m=0$ would work, except that the question asks you to find the smallest positive integer $m$, and $0$ is not a positive integer. – MJD Nov 09 '14 at 02:32

1 Answers1

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$$\left(\sqrt{3}+i\right)^m=\left(\sqrt{3}-i\right)^m$$ $$2^m\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^m=2^m\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^m$$

$$e^{i\theta}=\cos\theta+i\sin\theta$$

Using above formula we can write $$2^m(e^{i\pi/6})^m=2^m(e^{-i\pi/6})^m$$

$$e^{mi\pi/6}=e^{-mi\pi/6}$$

$$e^{2mi\pi/6}=e^{mi\pi/3}=1$$ $$e^{mi\pi/3}=e^{2ik\pi}$$

$$\frac{mi\pi}{3}=2ik\pi$$ $$\frac{m}{3}=2k$$ $$m=6k$$

where $k\in\mathbb Z$

Aditya Hase
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