If $\arg \left(z\right)<0,\text{ then }\arg (-z)-\arg \left(z\right)=$

(A) $\arg z\text{ is - ive, say }-\theta \text{, then }z=r{e}^{-i\theta }$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-z=r{e}^{\pi i}{e}^{-i\theta }=r{e}^{i(\pi -\theta )}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\arg (-z)=\pi -\theta =\pi +\arg z\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\arg (-z)-\arg z=\pi \end{array}$

(B) Let $z=r{e}^{i\theta }\text{. Given arg }z+\arg \omega =\pi$

$\therefore \arg \omega =\pi -\theta \therefore \omega =r_1{e}^{i(\pi -\theta )}$

Now $\overline{z}+i\overline{\omega }=0\Rightarrow r{e}^{-i\theta }+i{\mathrm{\Gamma }}_1{e}^{-r(\pi -\theta )}=0$

or $r(\cos \theta -i\sin \theta )+i{r}_1[\cos (\pi -\theta )$

$-i\sin [\pi -\theta )]=0$

or $r(\cos \theta -i\sin \theta )+r_1(-i\cos \theta +\sin \theta )=0$

$\therefore$ Equating real and imaginary parts, 

$r\cos \theta +r_1\sin \theta =0$

$-r\sin \theta -r_1\cos \theta =0$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{r}{r_1}=-\tan \theta =\frac{r_1}{r}\text{ or }{r}^2=r_1^2\therefore r=r_1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan \theta =-1\text{ or }\theta =\arg z=3\pi /4\Rightarrow \left(\mathrm{c}\right)\end{array}$