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I have a system of inequalities $$|z-a_k|\le R_k$$ where $z=x+iy$ (complex number) and $a_k$ and $R_k$ are real numbers for $k=1, \dots, n$. Basically the inequality above shows circle with center $a_k$ and radius $R_k$. The question here is, if I write $n$ inequalities as a system of inequalities and then solve this system, the solution will be the intersection of $n$ inequalities. But I want to find the union of $n$ inequalities. Is there any way to do that?

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1 Answer 1

The union of a bunch of sets is the complement of the intersection of the complements of the sets. So $|z-a_k|\gt R_k$ gives the complement of the disk, the system of such inequalities gives the intersection of the complements, then you want the complement of that.

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@ Gerry Myerson: Is there any reference for proving this idea? And also can we say the same thing even for system of equalities? –  kotoll Aug 6 '12 at 20:50
    
What is there to prove? Think about the union of $x\gt10$ and $x\lt3$, say. Can you see that it's the complement of the intersection of $x\le10$ and $x\ge3$? –  Gerry Myerson Aug 7 '12 at 0:42

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