Graphing Relations on the Complex Plane

Question

I am able to grasp most complex relations and their respective depiction on the complex plane however I am unable to get my head around relations such as these:

$Arg(z + 1 + 2i) - Arg(z-1-3i) = \pi$

$Arg(z-2) = Arg(z+3i)$

Any help is appreciated.

Answer 1

Let's start with the second one. Pick a point $z$ in the complex plane. Then $z_1=z-2$ is the point two units to the left of $z$ and $z_2=z+3i$ is point three units above $z$. The statement that $\arg{z_1}=\arg{z_2}$ is that $z_1$ and $z_2$ lie on the same ray from the origin.

Draw the points in the complex plane; $z,z_1,z_2$ are the vertices of a right triangle. The requirement is that the hypotenuse pass through the origin, and that $z_1, z_2$ are on the same side of the origin. Once you've translated the triangle so that $z$ is a solution, notice that you can translate the diagram along the line parallel to the hypotenuse. Therefore, the relation locates $z$ along a line, except for a forbidden interval of length $\sqrt{2^2+3^2}$ where $z_1, z_2$ would lie on opposite sides of the origin.

You can analyze the first one in a similar manner, but now the points $z_3=z+1+2i$ and $z_4=z-i-3i$ must lie on the same line through the origin but on opposite sides. (We don't have a right triangle this time, but that doesn't matter.) Now the equation restricts $z$ to a line segment.

Answer 2

$\arg z$ is the oriented angle between the vector $z$ and the real axis. If $\omega = \arg(z - z_1) - \arg(z - z_2)$ is zero, that means that the angle between $z - z_1$ and the axis is equal to the angle between $z - z_2$ and the axis, therefore the vectors are parallel. The solution set is the line $z_1 z_2$ minus the segment $[z_1, z_2]$.

$\omega = \pi$ means that the vectors are antiparallel. Then you have to evaluate $\omega$ for an arbitrary point on the segment $(z_1, z_2)$. If the result is $+\pi$ rather than $-\pi$, the solution set is $(z_1, z_2)$. Otherwise the solution set is empty.