Question

Find the point of intersection of the curves $arg(z-3i)=3\pi /4$ and $arg(2z+1-2i)=\pi /4$

• A. 1+3i
• B. 3+i
• C. 1-i
• D. given curves do not intersect.

Answer 1

The correct option is D given curves do not intersect.

Give loci are as follows:
$arg(z-3i)=\frac{3\pi }{4}$
which is a ray that starts from 3i and makes an angle an $3\pi /4$ with the positive real axis as shown in Fig. 3.18
$arg(2z+1-2i)=\frac{\pi }{4}$
or $arg\left[2(z+\frac{1}{2}-i)\right]=\frac{\pi }{4}$
or $arg2+arg[z-(-\frac{\pi }{4})]=\frac{\pi }{4}$
or $0+arg[z-(-\frac{1}{2}+i)]\frac{\pi }{4}$
or $arg[z-(-\frac{1}{2}+i)]=\frac{\pi }{4}$
This is a ray that starts from point -1/2 + i and makes an angle $\pi /4$ with the positive real axis as shown in the figure. From the figure, it is obvious that the system of equation has no solution.

Ans: D