Answer to Question #337302 in Complex Analysis for Genius

Question #337302

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.


1
Expert's answer
2022-05-05T10:54:28-0400

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.


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