# If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number.

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 $$0$$.

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

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 $$ad=bc$$.

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

• How did you get from $- \sin \dfrac{(11m - n) \pi}{6}$ to $\sin \dfrac{(m + n) \pi}{6}$? Why are they equal?
– user592938
Jan 8, 2020 at 0:32
• @user1502 $-\sin \frac{(11m-n)\pi}6 = \sin\frac{(12m-(m+n))\pi}6 = - \sin\frac{(m+n)\pi}6$. Jan 8, 2020 at 0:45

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 $$n+m \equiv 0 \pmod{6}$$.

$$\sqrt {3} - i$$ 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 $$n+m$$

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 $$\frac{(n+m) \pi}6\equiv 0\mod \pi\iff (n+m)\pi \equiv 0\mod 6\pi\iff n+m\equiv 0\mod 6.$$

• should it be $e^{\color{red}-mi\pi/6}$ in the denominator? Jan 6, 2020 at 0:14
• @J.W.Tanner: Yes, of course! Thanks for pointing it! Jan 6, 2020 at 0:33

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 $$n+m \equiv 0 \bmod 6$$.