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I am given the following number $z$:

$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$

with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:

$$\sqrt{3} + i = 2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) = 2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$

$$\sqrt{3} - 1 = 2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) = 2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) = 2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$

So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:

$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n} {\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$

$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}} {\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$

But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.

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    $\begingroup$ Multiply both the numerator and denominator by the conjugate of the denominator. Then you will be able to isolate real and imaginary parts. $\endgroup$ Jan 5, 2020 at 23:56

6 Answers 6

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Hint: Imaginary part of $\frac {a+ib} {c+id}$ equals imaginary part of $\frac {(a+ib) (c-id)} {|c+id|^{2}}$ which is $\frac {bc-ad} {c^{2}+d^{2}}$ and this is $0$ iff $ad=bc$.

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You may continue as follows,

$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}} {\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$ $$=2^{n - m} \cdot \dfrac{\left(\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}\right) \left(\cos \dfrac{11m \pi}{6} - i \sin \dfrac{11m \pi}{6}\right)} {\cos^2 \dfrac{11 m \pi}{6} + \sin^2 \dfrac{11 m \pi}{6}}$$

Then, set the the imaginary part of the numerator to zero,

$$I=\sin \dfrac{n \pi}{6}\cos \dfrac{11m \pi}{6} - \cos \dfrac{n \pi}{6}\sin \dfrac{11m \pi}{6} = -\sin\dfrac{(11m -n)\pi }{6}=\sin\dfrac{(m +n)\pi }{6}=0 $$

which leads to $\dfrac{(m+n)\pi }{6}= k\pi$. Thus, the relationship between $m$ and $n$ is

$$m+n=6k$$

with $k=0,1,2,...$

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  • $\begingroup$ How did you get from $- \sin \dfrac{(11m - n) \pi}{6}$ to $\sin \dfrac{(m + n) \pi}{6}$? Why are they equal? $\endgroup$
    – user592938
    Jan 8, 2020 at 0:32
  • $\begingroup$ @user1502 $-\sin \frac{(11m-n)\pi}6 = \sin\frac{(12m-(m+n))\pi}6 = - \sin\frac{(m+n)\pi}6$. $\endgroup$
    – Quanto
    Jan 8, 2020 at 0:45
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We have $\sqrt{3}+i=2e^{i\pi/6}$ and $\sqrt{3}-i=2e^{-i\pi/6}$. So \begin{eqnarray*} z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m} = 2^{n-m} e^{ i \pi (n+m) /6}. \end{eqnarray*} So we require $ n+m \equiv 0 \pmod{6}$.

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$\sqrt {3} - i$ is the conjugate to $\sqrt {3} + i$

$\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{(\sqrt 3 - i)^m(\sqrt 3 + i)^m}$

This will rationalize the denominator $\frac {(\sqrt 3 + i)^n}{(\sqrt 3 - i)^m} = \frac {(\sqrt 3 + i)^{n+m}}{4^{m}}$

We must find where $(\sqrt 3 + i)^{n+m}$ is real

$\sqrt 3 + i = 2(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})\\ (\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})^{n+m}\\ (\sqrt 3 + i)^{n+m} = 2^{n+m}(\cos \frac {(n+m)\pi}{6} + i\sin \frac {(n+m)\pi}{6})$

if $(\sqrt 3 + i)^{n+m}$ is real $\sin \frac {(n+m)\pi}{6} = 0$

$\frac {(n+m)\pi}{6} = k\pi\\ n+m = 6k$

$6$ divides $n+m$

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Hint:

Use the complex exponential notation and congruences: your final fraction is none other than $$z=2^{n-m}\frac{\mathrm e^{\tfrac{ni\pi}6}}{\mathrm e^{\tfrac{-mi\pi}6}}=2^{n-m} \mathrm e^{\tfrac{(n+m)i\pi}6},$$ anf it is a real number if and only if $$\frac{(n+m) \pi}6\equiv 0\mod \pi\iff (n+m)\pi \equiv 0\mod 6\pi\iff n+m\equiv 0\mod 6.$$

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    $\begingroup$ should it be $e^{\color{red}-mi\pi/6}$ in the denominator? $\endgroup$ Jan 6, 2020 at 0:14
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    $\begingroup$ @J.W.Tanner: Yes, of course! Thanks for pointing it! $\endgroup$
    – Bernard
    Jan 6, 2020 at 0:33
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The key insight is to recognize roots of unity in the expression.

We have $$ z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m} = \frac{(2\omega)^n}{(2\omega^5)^m} = 2^{n-5m}\omega^{n-5m} $$ where $\omega^6=-1$. Therefore, we need $n-5m \equiv 0 \bmod 6$, or $n+m \equiv 0 \bmod 6$.

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