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

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

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

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

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

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

3. Let $\displaystyle z_1=a_1+ib_1$

and $\displaystyle z_2=a_2+ib_2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by flower3
$\displaystyle { Z\scriptstyle1} = -5$
$\displaystyle { Z\scriptstyle2} = -1-i$
in the example you provided,
$\displaystyle \theta_1=\pi$
$\displaystyle \theta_2=\frac{5\pi}{4}$
$\displaystyle arg(Z_1Z_2)=\pi +\frac{5\pi}{4}-2\pi=\frac{\pi}{4}$

6. (3)
then we have,

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

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

it varies from 0 to $\displaystyle 2\pi$

Are you familiar with Argand Plane?

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

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