| z − a k | ≤ R k where z = x + i y (complex number) and a k and R k are real numbers for k = 1 , … , 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?

Question

|  z  −      a    k    |  ≤      R    k  where   z  =  x  +  i  y (complex number) and       a    k   and       R    k   are real numbers for   k  =  1  ,  …  ,  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?

Angelo Murray · Accepted Answer

The union of a bunch of sets is the complement of the intersection of the complements of the sets. So   |  z  −      a    k    |  &amp;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.

Dwllane4 · Accepted Answer

Think about the union of   x  &amp;gt;  10 and   x  &amp;lt;  3, say. Can you see that it&#039;s the complement of the intersection of   x  ≤  10 and   x  ≥  3?

| z − a k </msub> | ≤ R

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New Questions in Algebra I