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Given,$(4+i)(z+\overline{z})-(3+i)(z-\overline{z})+26i=0$ then value of $\mid z\mid^2$ is

(A) 13

(B) 17

(C) 11

(D) 26

1 Answer

Let $z=x+iy\Rightarrow \overline{z}=x-iy$
$(4+i)(2x)-(3+i)(2iy)+26i=0$
$\Rightarrow 4x+y=0,x-3y+13=0$
$\Rightarrow x=-1,y=4$
$\Rightarrow \mid z\mid^2=x^2+y^2=17$
Hence (B) is the correct answer.
answered Apr 9, 2014 by sreemathi.v
 

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