Prove $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ for two complex numbers 
If $z_1 = r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2+i\sin\theta_2)$ prove that $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ and that $\arg\left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2$.

I've done most of this, but I got stuck at the end:
$$\begin{align}
\frac{z_1}{z_2}&=\frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2(\cos\theta_2+i\sin\theta_2)}\\
&=\frac{r_1}{r_2}\cdot\frac{(\cos\theta_1+i\sin\theta_1)(\cos\theta_2-i\sin\theta_2)}{1}\\
&=\frac{r_1}{r_2}\cdot[(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)]\\
&=\frac{r_1}{r_2}\cdot[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]\\
\\
\left|\frac{z_1}{z_2}\right|&=\sqrt{\frac{r_1^2}{r_2^2}\cdot[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]^2}
\end{align}$$
But in order for this to work, $[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)]^2$ must equal $1$ (so that $\left|\frac{z_1}{z_2}\right|=\frac{r_1}{r_2}$), but it doesn't. Where am I going wrong?
 A: The step you take to the last equation isn't right; recall that the absolute value of a complex number is defined as
$$|a+bi|=\sqrt{a^2+b^2}$$
but you've just written it as
$$|z|=\sqrt{z^2}$$
which only works for real $z$. If you use the proper (first) identity, noting that the second to lasts equation is easy to separate into imaginary and real parts, you will get to
$$\left|\frac{z_1}{z_2}\right|=\sqrt{\frac{r_1^2}{r_2^2}\cdot[\cos(\theta_1-\theta_2)^2+\sin(\theta_1-\theta_2)^2]}$$
which will work out as you expect.
A: If $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$ it's just a matter of computations with the addition formulas to show that
$$
z_1z_2=r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))
$$
so
$$
|z_1z_2|=r_1r_2=|z_1|\,|z_2|,\qquad
\arg(z_1z_2)=\theta_1+\theta_2=\arg z_1+\arg z_2
$$
For the quotient, set $w_1=z_1/z_2$ and $w_2=z_2$; then
$$
|w_1w_2|=|w_1|\,|w_2|
$$
or
$$
|z_1|=\left|\frac{z_1}{z_2}\right|\,|z_2|
$$
For the arguments it's the same.
In order to show that $|z_1z_2|=|z_1|\,|z_2|$ you can also prove first that
$$
\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}
$$
which is just computations (use $z_1=a_1+ib_1$, $z_2=a_2+ib_2$). Then
$$
|z_1z_2|^2=z_1z_2\cdot\overline{z_1z_2}=
z_1z_2\overline{z_1}\overline{z_2}=
z_1\overline{z_1}z_2\overline{z_2}=
|z_1|^2|z_2|^2
$$
and the conclusion follows by taking square roots.
A: Well $\frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}$. And $|e^{i(\theta_1-\theta_2)}|=1$ So $|\frac{z_1}{z_2}|=\frac{r_1}{r_2}=\frac{|z_1|}{|z_2|}$ . all is used is Pythagorean's to get $|\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}|=1$
A: Using the identity $|z|^2=z\overline{z}$, we have
$$
\begin{aligned}
\left|\frac{z_{1}}{z_{2}}\right| ^{2} &=\left(\frac{z_{1} }{z_{2}} \right)\overline{\left(\frac{z_{1}}{z_{2}}\right)}\\& =\left(\frac{z_{1}}{z_{2}}\right)\left(\frac{\bar{z}_{1}}{\bar{z}_{2}}\right)\\
&=\frac{z_{1} \bar{z}_{1}}{z_{2} \bar{z}_{2}} \\
&=\frac{\left|z_{1}\right|^{2}}{\left|z_{2}\right|^{2}} \\
&=\left(\frac{|z_{1} \mid}{\left|z_{2}\right|}\right)^{2}
\end{aligned}
$$
We can conclude that $$
\left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|}
$$
