Proof: $\arg(z\cdot w)= arg(z) + arg(w)$ and $\arg\left( \frac{z}{w} \right)=\arg(z) − \arg(w) $ Note: I've seen similar questions but I'd like to ask you for review the way of proving shown below.



$\arg(z\cdot w)= arg(z) + arg(w)$


Let's say that $arg(z) = \alpha$ and $arg(w) = \beta$.
Then: $z = |z|(cos\alpha + isin\alpha)$
$z\cdot w = |z|\cdot |w|[(cos \alpha\cdot \cos\beta −\sin \alpha \cdot \sin β)+i(\cos \alpha \cdot sin \beta +\cos \beta\cdot
\sin \alpha)]$
$z \cdot w = |z|\cdot|w|[\cos(\alpha + \beta) + i\sin(\alpha + \beta)]$


Point 4. based on: 

$\cos(\alpha + \beta) = \cos\alpha\cdot\cos\beta -\sin\alpha\sin\beta$ 

and

$\sin(\alpha + \beta) = \sin\alpha\cdot\cos\beta + \sin\beta\cos\alpha$



$\arg\left( \frac{z}{w} \right)=\arg(z) − \arg(w) $

$z = w \cdot \frac{z}{w}$

Using the previous equation: $\arg(z\cdot w)= arg(z) + arg(w)$

$\arg(z) = \arg(w) + \arg\left( \frac{z}{w} \right)$

$\arg\left( \frac{z}{w} \right) = \arg(z) - \arg(w)$
Your work [above] is based on trigonometry, but trig is based on complex arithmetic. Sometimes it helps to cut out the trig and use Euler's identity 
$$exp(a+bi)=cos(a)+I*sin(b)$$ which of course is obvious in the series definitions of $exp, cos, sin$. Let $z=exp(a1+i*b1)$ and $w=exp(a2+i*b2)$ now by third-grade algebra $$z*w=exp^{a1+i*b1}*exp^{a2+i*b2}=exp^{(a1+i*b1)+(a2+i*b2)}=exp^{(a1+a2)+i*(b1+b2)}.$$ Since $b1, b2, (b1+b2)$ are your arg's you immediately get your first equation. The others are obvious using the same approach.
Your work [above] is based on trigonometry, but trig is based on complex arithmetic. Sometimes it helps to cut out the trig and use Euler's identity 
$$exp(a+bi)=cos(a)+I*sin(b)$$ which of course is obvious in the series definitions of $exp, cos, sin$. Let $z=exp(a1+i*b1)$ and $w=exp(a2+i*b2)$ now by third-grade algebra $$z*w=exp^{a1+i*b1}*exp^{a2+i*b2}=exp^{(a1+i*b1)+(a2+i*b2)}=exp^{(a1+a2)+i*(b1+b2)}.$$ Since $b1, b2, (b1+b2)$ are your arg's you immediately get your first equation. The others are obvious using the same approach.