# Union of system of inequalities

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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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.
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