Question

Let $\arg \left(z\right)$ represent the principal argument of the complex number $Z$.
Then, $|z|=3$ and $\arg (z-1)-\arg (z+1)=\frac{\pi }{4}$ intersect

• A. at infinitely many points
• B. nowhere
• C. exactly at one point
• D. exactly at two points

Answer 1

The correct option is B nowhere

$|z|=3\arg (z-1)-\arg (z+1)=\frac{\pi }{4}\angle AKL=\angle ACB=\frac{\pi }{4}\Rightarrow LK=AL=\alpha =1\Rightarrow K=(0,1)$
Radius $=\sqrt{2}$

$PL=PK+KL=\sqrt{2}+1\Rightarrow P=(0,1+\sqrt{2})$

Therefore, the number of intersection is $0.$