MHB Set of points on the complex plane

cbarker1
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Dear Everybody, I am wanting to check the solution to this question:

Sketch the set of points determined by the given conditions:
a.) $\left| z-1+i \right|=1$
b.)$\left| z+i \right|\le3$
c.)$\left| z-4i \right|\ge4$

work:
I know (a.) is a circle with radius 1 and its center at (-1,1) on the complex plane see in figure 1.
I know (b.) is a circle with radius 3 and its center at (0,1) on the complex plane see in figure 2.
I know (c.) is a circle (?) with radius 4 with its center (0,-4) on the complex plane see in figure 3.

Thanks
Cbarker1
 

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  • Problem 5b for Complex Analysis.png
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  • Problem 5c for Complex Analysis.png
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Almost. But "=", "\le", and "\ge" are not the same! In general, |z- a| is the distance from complex number z to complex number a. Yes, the set of points, z, in the complex plane, such that |z- 1+ i|= |z- (1- i)|= 1 is the circle with center at 1- i, (1, -1), and radius 1. Similarly, |z+ i|= 3 is the circle with center at i, (0, 1), and radius 3. But your problem has \le rather than =. In addition to the circle where the distance is equal to 3, this set contains also the interior of the circle, the "disk" with center at i and radius 3.

Likewise, (c) is the circle with center 4i, (0, 4), and all points outside that circle.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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