Prove: $\overline{(\frac{z_1}{z_2})}=\frac{\overline{z_1}}{\overline{z_2}}$ 
Prove: $$\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}$$ 

$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{z_1\cdot z_2^{-1}}=\overline{z_1}\cdot \overline{z_2^{-1}}=\overline{z_1}\cdot \overline{z_2}^{-1}=\frac{\overline{z_1}}{\overline{z_2}}$$
Just to be sure is it valid?
 A: Note: The step $\overline{z_2^{-1}}=\overline{z_2}^{-1}$ need a proof, if you follow my solution you can see why.
Solution: You know that $|z|^2=z\cdot \bar{z}$
Given $|z_2|\neq 0$, $\dfrac{z_1}{z_2}=\dfrac{z_1\bar{z_2}}{|z_2|^2}$
Then $\overline{\Big(\dfrac{z_1}{z_2}\Big)}=\dfrac{\bar{z_1}{z_2}}{|z_2|^2}=\dfrac{\bar{z_1}}{\bar{z_2}}$
A: other approach
put $$z_1=r_1e^{it_1} $$
and $$z_2=r_2e^{it_2} $$
then
$$z=\frac {z_1}{z_2}=\frac {r_1}{r_2}e^{i (t_1-t_2)} .$$
$$\overline {z}=\frac {r_1}{r_2}e^{-i (t_1-t_2)} $$
$$=\frac {r_1e^{-it_1}}{r_2e^{-it_2}} $$
$$=\frac {\overline {z_1}}{\overline {z_2}} $$
A: I think your proof is not  full. Why $\overline{z^{-1}}=\overline{z}^{-1}$? Why $\overline{z_1z_2^{-1}}=\overline{z_1}\overline{z^{-1}}$?
I think it's better to make the following.
Let $z_1=a+bi$ and $z_2=c+di$, where $\{a,b,c,d\}\subset\mathbb R$ and $c^2+d^2\neq0$. 
Thus,
$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{\left(\frac{a+bi}{c+di}\right)}=\overline{\left(\frac{(a+bi)(c-di)}{c^2+d^2}\right)}=$$
$$=\overline{\left(\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i\right)}=\frac{ac+bd}{c^2+d^2}-\frac{bc-ad}{c^2+d^2}i=$$
$$=\frac{(a-bi)(c+di)}{c^2+d^2}=\frac{a-bi}{c-di}=\frac{\overline{z_1}}{\overline{z_2}}.$$
A: Let's do this one in two parts.

If $z$ and $w$ are complex numbers, then $\overline{zw}=\overline{z}\,\overline{w}$.

Suppose $z=a+bi$ and $w=c+di$ with $a,b,c,d$ real numbers. Then
$$
\overline{zw} = \overline{(a+bi)(c+di)} = \overline{ac-bd + i(ad+bc)}
= ac-bd-i(ad+bc)
= (a-bi)(c-di)
= \overline{z}\,\overline{w}.
$$

If $z$ is a nonzero complex number, then $1/\overline{z}=\overline{(1/z)}$.

Suppose $z=a+bi$ with $a$ and $b$ real numbers such that $(a,b)\ne(0,0)$. Then
$$
\overline{\left(\frac{1}{z}\right)}
= \overline{\left(\frac{1}{a+bi}\right)}
= \overline{\left(\frac{a-bi}{a^2+b^2}\right)}
= \overline{\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i}
= \frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}i
= \frac{a+bi}{a^2+b^2}
= \frac{1}{a-bi}
= \frac{1}{\overline{z}}.
$$

Now
$$
\overline{\left(\frac{z}{w}\right)}
= \overline{\left(z\frac{1}{w}\right)} = \overline{z}\overline{\left(\frac{1}{w}\right)}
= \overline{z} \frac{1}{\overline{w}}
= \frac{\overline{z}}{\overline{w}}.
$$
This is essentially what you did.
