for which n is $(3+ i\sqrt{3})^{n} $a real number

Question

I understand complex numbers and their exponents, but I'm not sure how to go about this problem or where to start it.

Answer 1

Transform to polar coordinates $r^2=9+3=12$, so $3+i\sqrt{3}=2\sqrt{3}(\frac{\sqrt{3}}{2}+i\frac{1}{2})=2\sqrt{3}e^{i\frac{\pi}{6}}$ To get n to make it real, the exponent has to be an integer multiple of $\pi$ so $\frac{n}{6}$ must be an integer.

Answer 2

Write $3+i\sqrt{3}$ with the $re^{i\theta}$ form.

$$3+i\sqrt{3} = \sqrt{12} e^{i\tfrac{\pi}{6}}$$

$$(3+i\sqrt{3})^{n} = \sqrt{12} e^{i\tfrac{n\pi}{6}}$$

Which is real when $n=6k$ for some integer $k$.

Answer 3

$$(3+ i\sqrt{3})^{n}=({\sqrt{3}})^{n}\Bigl(\sqrt{3}+i\Bigr)^n$$ $$({\sqrt{3}})^{n}\Bigl(\sqrt{3}+i\Bigr)^n=(2\sqrt3)^n\Bigl(\frac{\sqrt3}{2}+\frac{1.i}{2}\Bigr)^n$$ $$(2\sqrt3)^n\Bigl(\frac{\sqrt3}{2}+\frac{1.i}{2}\Bigr)^n=(2\sqrt3)^n\Bigl(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Bigr)^n$$

I'm not sure if you are aware of this identity: $\cos\theta+i\sin\theta=e^{i\theta}$. You can find proofs for this online. $$(2\sqrt3)^n\Bigl(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Bigr)^n=(2\sqrt3)^n\Bigl(e^{i\frac{\pi}{6}}\Bigr)^n$$

Going back to the stated identity:$\cos\theta+i\sin\theta=e^{i\theta}$, we can observe the following: $$\cos\pi+i\sin\pi=e^{i\pi}=-1$$ $$\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=e^{i\frac{\pi}{2}}=i$$

$e^{i\theta}$ is after all, a complex number which is written as $\cos\theta + i\sin\theta$. For it to be completely real, it shouldn't have an imaginary part. So, $\sin\theta=0$. And this is true for all $\theta \in n\pi$, where $n=0,1,2,....$. Thus, $e^0, e^{i\pi},e^{2i\pi},e^{3i\pi}.. $ are all real numbers.

For our question to be real: $$(2\sqrt3)^n\Bigl(e^{i\frac{\pi}{6}}\Bigr)^n=(2\sqrt3)^ne^{ik\pi}$$

Thus for all $n=6k$, the given complex number is real