How to visualize complex domains

Question

I was hoping if someone can help me visualize complex domains. I know how simplex ones like $|z|<1$ or $\text{Re}z < 1$ look like but for the more complicated ones such as $$\text{Im } z < 2|z|$$ or $$|z-1|<|z+i|$$ I am completely lost. Can anyone give me some hints or tips to help me better understand what is going on?

Answer 1

Write $z=x+\mathrm{i}y$. Then $$ \text{Im } z < 2|z|\Rightarrow y<2\sqrt{x^2+y^2}\ , $$ which (in the plane $(x,y)$) has the solution: $y<0$ $\forall x$ or $$ y>0\ \wedge y^2<4(x^2+y^2)\Rightarrow 4x^2+3y^2>0\ , $$ which is verified by any point $(x,y)$ in the upper half-plane. So, in summary every complex number $z\neq 0$ is such that $\text{Im } z < 2|z|$ (visually, the region in the $(x,y)$ plane corresponding to $\text{Im } z < 2|z|$ is the whole plane itself (excluding the origin)!)

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For the second case, we have $$ |z-1|<|z+i|\Rightarrow |x+\mathrm{i}y-1|<|x+\mathrm{i}y+\mathrm{i}| $$ which corresponds to $$ \sqrt{(x-1)^2+y^2}<\sqrt{x^2+(y+1)^2}\ . $$ Squaring $$ (x-1)^2+y^2<x^2+(y+1)^2\Rightarrow y>-x\ , $$ so the complex numbers $z$ satisfying the relation $|z-1|<|z+i|$ are those lying (in the plane $(x,y)$) to the right of the straight line through the origin and with slope $-1$.