We know that $|z|=\sqrt{a^{2}+b^{2}}$. Show that $|z_{1} \cdot z_{2}|= |z_{1}| \cdot |z_{2}|$ 
For $z=a+bi \in \mathbb{C}$ you define the modulus as
  $|z|=\sqrt{a^{2}+b^{2}}$. Show that $|z_{1} \cdot z_{2}|= |z_{1}|
\cdot |z_{2}|$ for all $z_{1},z_{2} \in \mathbb{C}$

It's another task from an old exam and I have no idea how this could be solved.
What I have tried:
Assume we have $|z_{1}z_{2}|^{2}$. This would equal $$(z_{1}z_{2})(\overline{z_{1}z_{2}})=(z_{1}z_{2})(\bar{z_{1}} \bar{z_{2}})=z_{1}\bar{z_{1}}z_{2}\bar{z_{2}}=|z_{1}|^{2}|z_{2}|^{2}$$
To sum up, we now know that $$|z_{1}z_{2}|^{2}=|z_{1}|^{2}|z_{2}|^{2}$$
And if we take the square root on both sides:
$$|z_{1}z_{2}|=|z_{1}||z_{2}|$$
Not sure if we can just take the square root but it should work because the modulus provides us only positive values? Actually it shouldn't even matter because we are in $\mathbb{C}$.
I'm not sure if this is correct? Please show me another solution if I'm wrong.
 A: Your proof is okay, if you already know that
$$\overline{z_1\cdot z_2} = \overline{z_1}\cdot\overline{z_2},\quad\quad \forall z_1,z_2\in\mathbb C $$
that 
$$|z|\in[0,\infty),\quad\quad \forall z\in\mathbb C $$
that $\sqrt{\cdot}:[0,\infty)\to[0,\infty)$ is invertible,
and that
$$|z|^2 = z\cdot \overline{z},  \quad\quad \forall z\in\mathbb C.$$
A: You can also do it the hard way, although the method with $|z|^2=z\bar{z}$ is far better.
If $z_1=a+bi$ and $z_2=c+di$, then
$$
z_1z_2=(ac-bd)+(ad+bc)i
$$
so
\begin{align}
|z_1z_2|&=\sqrt{(ac-bd)^2+(ad+bc)^2} \\
&=\sqrt{a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2}\\
&=\sqrt{a^2c^2+b^2c^2+a^2d^2+b^2d^2\mathstrut}\\
&=\sqrt{(a^2+b^2)c^2+(a^2+b^2)d^2}\\
&=\sqrt{(a^2+b^2)(c^2+d^2)}\\
&=\sqrt{a^2+b^2\mathstrut}\,\sqrt{c^2+d^2\mathstrut}\\[3px]
&=|z_1|\,|z_2|
\end{align}
A: You are absolutely correct!
Alternatively, you could treat the $z_1$ and $z_2$ in polar form as follows:
We know that any complex number can be rewritten in polar form as real, positive number situated on the real axis multiplied (rotated) by $e^{i\theta}=cos(\theta) + i\sin(\theta)$. Then,
$$|z_1||z_2| = |r_1e^{i\theta_1}||r_2e^{i\theta_2}|$$
$$|z_1||z_2| = |r_1||r_2||e^{i\theta_1}e^{i\theta_2}|$$
$$|z_1||z_2| = |r_1r_2e^{i(\theta_1 + \theta_2)}| = |z_1z_2|$$
This is merely a reformulation in terms of polar coordinates, but it gives extra insight into how the modulus's (radii) multiply.
A: Taking root is okay since $|z|\in\Bbb R_+$ and $\sqrt{\,\cdot\,}\colon\Bbb R_+\to\Bbb R_+$ is bijective. What you did is standard way to show it: first you show that conjugation is multiplicative, then norm $|z|^2=z\bar z$ is as well and it works even in more general setting of quadratic field/ring extensions.
