question_answer

The  point  of intersection  of the  curve arg \[(2z-8\,i)\] and arg \[(3z+1-3\,i)=\frac{\pi }{4}\] is

A)  \[\frac{4}{3}(1+2\,i)\]              

B)  \[\frac{4}{3}(1-2\,i)\]

C)  \[\frac{-4}{3}(1+2\,i)\]             

D)  \[\frac{1}{2}(4+3\,i)\]

Correct Answer: A

Solution :

\[\arg \,(2z-8i)\,=\frac{3\pi }{4}\,\Rightarrow \,\arg (z-4i)\,+\arg (2)\,=\frac{3\pi }{4}\] \[\Rightarrow \,\,\arg (z-4i)\,=\frac{3\pi }{4}\] Similarly, \[\arg (3z+1-3i)=\frac{\pi }{4}\] \[\Rightarrow \,\,\arg \left( z+\frac{1}{3}-i \right)=\frac{\pi }{4}\] \[\therefore \] equation of the two rays are \[(y-0)\,=-1(x-4)\,\Rightarrow \,x+y=4\]and \[(y-1)\,=1\left( x+\frac{1}{3} \right)\,\Rightarrow \,x-y=\frac{-4}{3}\,\] and\[y=\frac{8}{3}\] \[\therefore \] Point of intersection \[=\frac{4}{3}(1+2i)\]