Proving if $z$ is an n'th root, $\bar z$ is also an n'th root 
Let $n>0$ be an even number, and let $z$ be an $n$'th root of a real number. Is $\bar z$ also an $n$'th root of this number?

My answer is yes. The way I solved this was to consider a complex number $z = a+bi$ on the (polar) form $z = re^{i\theta}$.
Solving $z^n = re^{i\theta}$, I get $z = \sqrt[n] r e^{{(i\theta+2\pi m)}/n}$, $0 \leq m \leq n-1$.
Since $n>0$ is an even number, we will always have an even number of solutions. If we then draw the solutions in the complex plane, we will always have that for each solution that $z$ that has an angle $\phi=\theta$, we will always find $\bar z$, also a solution with $\phi=-\theta$ on the complex plane.
However, my answer is more a geometric interpreration/realization than a rigorious algebraic proof.
I wanted to try to prove this in a more general algebraic form: My attempt was to assume $z=a+bi$ is an $n$'th root of a real number. We now want to show that $z=a-bi$ is also an $n$'th root of the same real number.
I tried setting up $(a+bi)^2 = a^2-b^2+2iab$ and $(a-bi)^2 = a^2-b^2-2iab$, noticing that the real part of these numbers are always the same, the imaginary part is different in sign.
Now it's easy to realize that $(2iab)^n=(-2iab)^n$ for all $n>0$ given that $n$ is even.
But I have not proved that if we have a complex number $z = a+bi$ (being an $n$'th root of a real number, there also exists a $\bar z = a-bi$ such that $\bar z$ is also an $n$'th root of the same real number.
If this were the case, shouldn't $(a+bi)^n=(a-bi)^n$ for all even $n>0$? This is obviously not the case, I can find $a$ and $b$ not satisfying this equation.
I guess I'm confused about what I really have to do to prove this.
 A: It's really easy to prove it with the comments, but I don't think you're allowed to use these stronger results to resolve your problem. 
So I think you should stay with your initial idea of using $z = re^{i\theta}$.
So, suppose:


*

*$z = re^{i\theta}$

*$\exists n \in \mathbb Z, z^n = r^ne^{ni\theta} = x$
Since $x^n \in \mathbb R$, you can conclude that $n\theta \equiv 0 \pmod \pi$
You also have: 
$\bar{z} = re^{-i\theta}$
So:
$\bar{z}^n = r^ne^{-ni\theta}$
From there, you can use the parity of n to prove $\frac{z^n}{\bar{z}^n}=\frac{x}{\bar{z}^n}=1$
Edit: As from [Hagen von Eitzen] comment, you don't even need n to be even
A: Lemma:  $(\overline z)^{2k} = \overline {(z^{2k})}$
Pf: If $z = a+bi$ then $(a+bi)^{2k} = \sum\limits_{j=0}^{2k}{2k \choose j}a^{j}b^{2k-j}i^{2k-j}$.
Now if $j$ is even, then $2k-j$ is even, and $i^{2k-j}$ is real, and ${2k \choose j}a^{j}b^{2k-j}i^{2k-j}$ is real.
If $j$ odd, then $2k-j$ is odd, and $i^{2k-j}$ is purely imaginary and  ${2k \choose j}a^{j}b^{2k-j}i^{2k-j}$ is purely imaginary.
So $(a+b)^{2k} = \sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) + i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})$
And so $\overline {(z^{2k})} =  \sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) - i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})$
Now if we expand out $(\overline z)^{2k}=(a-bi)^{2k}=\sum\limits_{j=0}^{2k}(-1)^j{2k \choose j}a^{j}b^{2k-j}i^{2k-j}$
Now $(-1)^j$ is $1$ if even and $-1$ if odd and $i^{2k-j}$ is real if even and imaginary if odd so:
$(\overline z)^{2k}=(a-bi)^{2k}=\sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) - i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})=$
$\overline{(z^{2k})}$
....
And that's it really.  If $z$ is an $n$th root then $z^n = m\in \mathbb R$ so $z^n = m = m + 0i = m - 0i = \overline m = \overline {z^n} = (\overline z)^n$.  So $\overline z$ is an $n$th root of $m$.
.....
o
Or, if it's not to geometric:
If $z = re^{i\theta}$ then $\overline z = re^{-i\theta}$ and and $z^n = r^ne^{in\theta}$ and $\overline z^n = r^ne^{-in\theta}$
If $z^n \in \mathbb R$ then $n\theta = k\pi $ for some integer $k$.  And so $-n\theta = -k\pi$.  Now $-k\pi \equiv k\pi \pmod {2\pi}$ so $(\overline z)^n = z^n$.
