Number of points having integral coordinates

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SUMMARY

The discussion focuses on determining the number of integral coordinate points within the intersection of three sets defined by complex numbers: A, B, and C. The sets are defined as A = {z:|z+1|≤2+Re(z)}, B = {z:|z-1|≥1}, and C={z: |z-1|/|z+1|≥1}. The intersection A ∩ B ∩ C is bounded by the parabola y²=2(x+3/2) and the Y-axis, yielding a total of 6 integral coordinate points: (-1,0), (0,0), (-1,1), (-1,-1), (0,-1), and (0,1).

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utkarshakash
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Homework Statement


Let A,B,C be three sets of complex numbers as defined below

A = {z:|z+1|\leq2+Re(z)}, B = {z:|z-1|\geq1} and
C=\left\{z: \frac{|z-1|}{|z+1|}\geq 1 \right\}

The number of point(s) having integral coordinates in the region A \cap B \cap C is

Homework Equations



The Attempt at a Solution


I worked out and found that A \cap B \cap C is the area bounded by the parabola y^{2}=2(x+\frac{3}{2}) and the Y-axis. So the points having integral coordinates in this region are (-1,0), (0,0), (-1,1) and (-1,-1) which counts up to 4. But the correct answer is 6.
 
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Get out a piece of graph paper and carefully graph your region. Note that boundary points are included in the given regions.
 
jambaugh said:
Get out a piece of graph paper and carefully graph your region. Note that boundary points are included in the given regions.

Ughhh... How can I miss (0,-1) and (0,1)! Thanks.
 

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