If $arg(z - a) = \frac{\pi}{4}$,where $a \in R$,then the locus of $z \in C$ is a

If $z=1+i \sqrt{3}$ then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to

$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$

Match the items of List-$I$ with those of List-$II$:

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

The correct matching is:

Let $z$ be a complex number such that the principal value of argument, $\arg(z) > 0$. Then, $\arg(z) - \arg(-z)$ is

Consider the following statements:
$I$: If $a$ and $b$ are positive real numbers,then $\sqrt{-a} \times \sqrt{-b} = \sqrt{ab}$
$II$: The argument of $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}$ is $120^{\circ}$
Then: