Find the point of intersection of the curves `a r g(z-3i)=(3pi)/4a n d arg(2z+1-2i)=pi//4.`

Explanation

We have, `arg(z-3i)=(3pi)/(4)`
which is a ray that starts from 3i and makes an angle `3pi//4` with the positive real axis as shown in the figure.

`arg(2z+1-2i)=(pi)/(4)`
or `arg[2(z+(1)/(2)-i)]=(pi)/(4)`
or `arg2+arg[z-(-(1)/(2)+i)]=(pi)/(4)`
or `0+arg[z-(-(1)/(2)+i)]=(pi)/(4)`
or `arg[z-(-(1)/(2)+i)]=(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, clearly, the system of equations has no solution.