The point of intersection of the curves $\arg (\mathrm{z}-3\mathrm{i})=\frac{3\pi }{4}$ and $\arg (2z+1-2i)=\frac{\pi }{4}$ is

Let $z=x+iy,$ then $\arg (z-3i)=\arg (x+iy-3i)=\frac{3\pi }{4}$

$\Rightarrow x<0,y-3>0\left(\because \frac{3\pi }{4}\right.\text{ is in II quadrant) }$

and $\frac{y-3}{x}=\tan \frac{3\pi }{4}=-1$

$\Rightarrow y=-x+3\mathrm{\forall }x<0\text{ and }y>3$                …(1)

and $\arg (2z+1-2i)=\arg \left[\right(2x+1)+i(2y-2\left)\right]=\frac{\pi }{4}$

$\Rightarrow 2x+1>0,2y-2>0\left(\because \frac{\pi }{4}\text{ is in I quadrant }\right)$

and $\frac{2y-2}{2x+1}=\tan \frac{\pi }{4}=1\Rightarrow 2y-2=2x+1$

$\Rightarrow y=x+\frac{3}{2}\mathrm{\forall }x>-\frac{1}{2},y>1$                 …(2)

From equations (1) and (2), we get graph 

It is clear from the graph that two lines do not intersect.

$\therefore$        No point of intersection. 

Caution : It is most likely that the students after getting two straight lines, solve them to get the point of

intersection $\left(\frac{3}{4},\frac{9}{4}\right).$ Clearly the principal values of arguments must be considered.