Number of points of intersection of $\arg(z-2-7i)=\cot^{-1}2$ and $\arg\!\left(\dfrac{z-5i}{z+2-i}\right)=\pm\dfrac{\pi}{2}$
Step-by-Step Solution
Key Concept: First curve: ray from $(2,7)$ at angle $\arctan(1/2)$ (northeast). Second curve: circle on diameter joining $5i$ and $-2+i$, center $(-1,3)$, radius $\sqrt5$. Distance from $(-1,3)$ to $(2,7)$ is $5>\sqrt5$, and the ray goes further away.
0 intersection points.
Correct Answer: (A) 0