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Locate the points for which $\; 3 < |z| < 4$

$(a)\;x^{2}+y^{2}=9 \;and\;x^{2}+y^{2} =16\qquad(b)\;x^{2}+y^{2}=4 \;and\;x^{2}+y^{2} =9\qquad(c)\;x^{2}+y^{2}=9 \;and\;x^{2}+y^{2} =3\qquad(d)\;x^{2}+y^{2}=0 \;and\;x^{2}+y^{2} =16$

1 Answer

Answer : $\;x^{2}+y^{2}=9 \;and\;x^{2}+y^{2} =16$
Explanation :
$|z| < 4 \; => \; x^{2}+y^{2} =16\;$ which is the interior of a circle with centre and at origin and radius 4 units ,
And $\;|z| > 3 \; => \; x^{2}+y^{2} =9\;$ which is the interior of a circle with centre and at origin and radius 3 units .
Hence $\; 3 < |z| < 4\;$ is the portion between two circles $\;x^{2}+y^{2} =9\;$ and $\;x^{2} +y^{2} =16\;$
answered Apr 17, 2014 by yamini.v
 

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