Question

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

• A. $(\frac{3}{4},\frac{9}{4})$
• B. $(\frac{9}{4},\frac{3}{4})$
• C. $(\frac{3}{2},\frac{3}{2})$
• D. Given curves do not intersect

Answer 1

The correct option is D Given curves do not intersect

Given loci are as follows: $\arg (z-3i)=\frac{3\pi }{4}$
which is a ray that starts from $3i$ and makes an angle $\frac{3\pi }{4}$ with positive real axis.

Now,
$\arg (2z+1-2i)=\frac{\pi }{4}\Rightarrow \arg \left[2(z+\frac{1}{2}-i)\right]=\frac{\pi }{4}\Rightarrow \arg 2+\arg [z-(-\frac{1}{2}+i)]=\frac{\pi }{4}\Rightarrow 0+\arg [z-(-\frac{1}{2}+i)]=\frac{\pi }{4}\Rightarrow \arg [z-(-\frac{1}{2}+i)]=\frac{\pi }{4}$

This is a ray that starts from point $(-\frac{1}{2}+i)$ and makes an angle $\frac{\pi }{4}$ with positive real axis as shown in the figure.


From the figure, it is obvious that the system of equations has no solution.