If the imaginary part of 2z+1iz+1 is −2, then the locus of z is
A
a circle
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B
a straight line
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C
an ellipse
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D
a parabola
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Solution
The correct option is B a straight line Let z=x+iy 2z+1iz+1=2(x+iy)+1i(x+iy)+1 =(2x+1)+2iy(1−y)+ix×(1−y)−ix(1−y)−ix =(2x+1)+2iy(1−y)2+x2×{(1−y)−ix} ∴ Imaginary part =2y(1−y)−x(2x+1)(1−y)2+x2=−2 ⇒2y−2y2−2x2−x=−2x2−2+4y−2y2 ⇒x+2y−2=0 which is a straight line.