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If $R_e\bigg(\large\frac{z-8i}{z+6}\bigg)$$=0$, then $z$ lies in the curve

$\begin{array}{1 1} (A) x^2+y^2+6x-8y=0 \\ (B) x^2+y^2+6x+8y=0 \\ (C) x^2+y^2-6x-8y=0 \\ (D) x^2+y^2-6x+8y=0 \end{array} $

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Let $z=x+iy$
$R_e\bigg(\large\frac{z-8i}{z+6}\bigg)=$$0$
$\Rightarrow\:R_e\bigg(\large\frac{x+iy-8i}{x+iy+6}\bigg)$$=0$
$\Rightarrow\:R_e\bigg(\frac{(x+i(y-8))((x+6)-iy}{(x+6)+iy)((x+6)-iy}\bigg)$$=0$
$\Rightarrow\:\large\frac{x^2+6x+y^2-8y}{x^2+12x+y^2}=$$0$
$\Rightarrow\:x^2+y^2+6x-8y=0$
answered Aug 1, 2013 by rvidyagovindarajan_1
 

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