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The locus of z satisfying the inequity $\bigg|\large\frac{z+2i}{2z+i}\bigg| < 1,$ where $z=x+iy,$ is :

\[\begin {array} {1 1} (1)\;x^2 +y^2 < 1 & \quad (2)\;x^2-y^2 < 1 \\ (3)\;x^2+y^2 > 1 & \quad (4)\;2x^2+3y^2 < 1 \end {array}\]
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1 Answer

(3) $x^2+y^2 > 1$
answered Nov 7, 2013 by pady_1
edited Jan 2, 2014 by meena.p
 
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