Complex Number Proof of $\overline{\left(\frac {z_1}{z_2}\right)}=\frac {\overline{z_1}}{\overline{z_2}}$. 
Let $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$ be complex numbers. Let $\overline{z_1}$ and $\overline{z_2}$ be the conjugate of $z_1$ and $z_2$, respectively. Prove: $$\overline{\left(\frac {z_1}{z_2}\right)}=\frac {\overline{z_1}}{\overline{z_2}}$$

My Attempt:
I started with $\overline{\left(\frac {z_1}{z_2}\right)}$ and multiplied it by $\frac {z_1}{z_2}$ to get $\left|\frac {z_1}{z_2}\right|^2=\frac {a_1^2+b_1^2}{a_2^2+b_2^2}$, but now, I don't know where to continue afterwards.
 A: Hint:
You can complete your proof multiplying also the RHS by $\frac{z_1}{z_2}$ then show that  $\left|\frac {z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ ( easy in polar form).
A: $$
\begin{align}
\frac{z_1}{z_2}&=\frac{a_1+ib_1}{a_2+ib_2}\\
\\
&=\frac{a_1+ib_1}{a_2+ib_2}\cdot\frac{a_2-ib_2}{a_2-ib_2}\\
\\
&=\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2+ib_2)(a_2-ib_2)}\\
\\
&=\frac{(a_1+ib_1)(a_2-ib_2)}{(a_2^2+b_2^2)}\\
\\
&=\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}
\end{align}
$$
The 
$\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}$ fraction is a Complex Number with Real denominator ($a_2^2+b_2^2$) and conjugation of this fraction only changes the Numarator $(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)$.
So
$$
\begin{align}
\overline{\Big(\frac{z_1}{z_2}\Big)}&=\overline{\Big(\frac{a_1+ib_1}{a_2+ib_2}\Big)}\\
\\
&=\overline{
\Big(\frac{(a_1a_2+b_1b_2)+i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}\Big)}
\\
\\
&=\frac{(a_1a_2+b_1b_2)-i(-a_1b_2+a_2b_1)}{(a_2^2+b_2^2)}
\\
\\
&=\frac{(a_1a_2+b_1b_2)+i(a_1b_2-a_2b_1)}{(a_2^2+b_2^2)}
\\
\\
&=\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2^2+b_2^2)}
\\
\\
&=\frac{(a_1-ib_1)(a_2+ib_2)}{(a_2-ib_2)(a_2+ib_2)}
\\
\\
&=\frac{a_1-ib_1}{a_2-ib_2}
\cdot\frac{a_2+ib_2}{a_2+ib_2}
\\
\\
&=\frac{a_1-ib_1}{a_2-ib_2}
\\
\\
&=\frac{\overline{z_1}}{\overline{z_2}}
\end{align}
$$
A: If $w_1=re^{i\theta}$ and $w_2=\rho e^{i\phi}$ then
$$\overline{w_1}\ \overline{w_2}=re^{-i\theta} \rho e^{-i\phi}=r\ \rho e^{-i(\theta+\phi)}=\overline{w_1\ w_2}$$
so
$$\boxed{\overline{w_1}\ \overline{w_2}=\overline{w_1\ w_2}}$$
now let $w_1=\dfrac{z_1}{z_2}$ and $z_2$. Also straightforward shows
$$\overline{\left(\frac {w_1}{w_2}\right)}=\dfrac{r}{\rho} e^{-i(\theta-\phi)}=\dfrac{re^{-i\theta}}{\rho e^{-i\phi}}=\dfrac{\overline{w_1}}{\overline{w_2}}$$
A: we get $$\frac{a_1+b_1i}{a_2+b_2i}=\frac{a_1a_2+b_1b_2+(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ and the complement is
$$\overline{\frac{a_1+b_1i}{a_2+b_2i}}=\frac{a_1a_2+b_1b_2-(a_2b_1-a_1b_2)i}{a_2^2+b_2^2}$$ 
and $$\frac{a_1-b_1i}{a_2-b_2i}=\frac{(a_1-b_1i)(a_2+b_2i)}{(a_2-b_2i)(a_2+b_2i)}=...$$
can you proceed?
