Question

Let A ={ z E C : Im (z+2/z-2)≥ 1}.

(a) Sketch the set A in the complex plane.

(b) Is z = −2i a boundary point of A? Provide reasons for your answer.

(c) Is this set open, closed, both or neither? Provide reasons for your answer.

Answer 1

At first, we set: $z=x+iy$ with real $x,y.$ We receive: $\frac{z+2}{z-2}=\frac{x+iy+2}{x+iy-2}=\frac{x+iy+2}{x+iy-2}=\frac{(x+iy+2)(x-2-iy)}{(x-2)^2+y^2}=\frac{(x-2)(x+2)+y^2+i(y(x-2)-y(x+2))}{(x-2)^2+y^2}$.

From the latter considerations we receive that: $Im\left(\frac{z+2}{z-2}\right)=\frac{-4y}{(x-2)^2+y^2}$. In order to plot the graph, we need to plot the curve: $\frac{-4y}{(x-2)^2+y^2}=1$. The graph is below:

a). All points that are inside of the circle drawn above belong to set $A$. It can be obtained from the inequality: $\frac{-4y}{(x-2)^2+y^2}\geq1$.

b). Substitute the point into equation of the circle: $\frac{4\cdot2}{2^2+2^2}=1$. Thus, it is a boundary point. It can be also check in a formal way: any neighborhood of the point $z=-2i$ contains points that belong to $A$ and points that do not belong to $A$.

c). The set is closed because it contains all boundary points. We can consider the set $\mathbb{C}/{A}$, which is determined by the inequality$Im\left(\frac{z+2}{z-2}\right)<1$. For any point that belongs to $\mathbb{C}/{A}$ there is also a small neighborhood that belongs to $\mathbb{C}/{A}$. Namely, points: $z+\varepsilon$ also belong to $\mathbb{C}/{A}$ for small values of complex $\varepsilon$($|\varepsilon|$ is small). Therefore, the set $\mathbb{C}/{A}$ is opened and, respectively, the set $A$ is closed.