show that :
if Re (z1) >0 and Re( z2) >0 then :
$Arg (z1z2) =$
$Arg (z1)+ Arg (z2)$

2. Fact: If $z=r\mathrm e^{\mathrm i\theta}$ where $r>0$ and $-\pi<\theta\leq\pi$ then $r=|z|$ and $\theta=\mathop{arg} z$.

If $\mathop{arg} z_1=\theta$ and $\mathop{arg} z_2=\phi$ then $z_1=|z_1|\mathrm e^{\mathrm i\theta}$ and $z_2=|z_2|\mathrm e^{\mathrm i\phi}$.

Therefore $z_1z_2=|z_1||z_2|\mathrm e^{\mathrm i\theta}\mathrm e^{\mathrm i\phi}=|z_1z_2|\mathrm e^{\mathrm i(\theta+\phi)}$.

If $\mathop{Re}(z_1)>0$ then $-{\textstyle\frac\pi2}<\theta<{\textstyle\frac\pi2}$, and if $\mathop{Re}(z_2)>0$ then $-{\textstyle\frac\pi2}<\phi<{\textstyle\frac\pi2}$. Therefore $-\pi<\theta+\phi<\pi$.

Thus $\mathop{arg}(z_1z_2)=\theta+\phi=\mathop{arg} z_1+\mathop{arg} z_2$.

3. Let $z_1=a_1+ib_1$

and $z_2=a_2+ib_2$

be two complex numbers where $a_1,a_2>0$

$arg(z_1)=tan^{-1}\frac{b_1}{a_1}$

$arg(z_2)=tan^{-1}\frac{b_2}{a_2}$

Now, $z_1z_2=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)$

$arg(z_1z_2)=tan^{-1}\frac{a_1b_2+a_2b_1}{a_1a_2-b_1b_2}$

$arg(z_1z_2)=tan^{-1}\frac{b_1}{a_1}+tan^{-1}\frac{b_2}{a_2}$

I don't know how the property $a_1,a_2>0$ is used.

4. i Noted that $Arg({Z\scriptstyle1} {Z\scriptstyle2}) \leq Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )$
this true or not ??
i take some examples like :
${ Z\scriptstyle1} = -5$
${ Z\scriptstyle2} = -1-i$ ,

5. $z=a+ib=|z|e^{i\theta}$

There are four cases:
(1) $a\geq 0,b\geq 0$,
then we have, $0 \leq \theta \leq \frac{\pi}{2}$

(2) $a\leq 0,b\geq 0$
then we have, $\frac{\pi}{2} \leq \theta \leq {\pi}$

(3) $a\leq 0,b\leq 0$
then we have, $\pi \leq \theta \leq \frac{3\pi}{2}$

(4) $a\geq 0,b\leq 0$
then we have, $\frac{3\pi}{2} \leq \theta \leq 2\pi$

Now,
if sum of arguments( $\theta_1,\theta_2$) is less than $2\pi$,
then
$
Arg({Z\scriptstyle1} {Z\scriptstyle2}) = Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )
$

And,
if sum of arguments( $\theta_1,\theta_2$) is equal to or more than $2\pi$,
then
$
Arg({Z\scriptstyle1} {Z\scriptstyle2}) = Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )-2\pi
$

$
Arg({Z\scriptstyle1} {Z\scriptstyle2}) < Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )
$

Originally Posted by flower3
$
{ Z\scriptstyle1} = -5
$

$
{ Z\scriptstyle2} = -1-i
$
in the example you provided,
$\theta_1=\pi$
$\theta_2=\frac{5\pi}{4}$
$arg(Z_1Z_2)=\pi +\frac{5\pi}{4}-2\pi=\frac{\pi}{4}$

6. (3)
then we have,

(4)
then we have,
these cases are not true because :
$Arg Z=\theta and -\pi \prec \theta \leq \pi$

7. Originally Posted by flower3
these cases are not true because :
$Arg Z=\theta and -\pi \prec \theta \leq \pi$
there is no condtion like this.

it varies from 0 to $2\pi$

Are you familiar with Argand Plane?

Then plot -1-i on it and find $\theta$.

EDIT: you and me are saying the same thing...don't know which is standard.