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Question

Consider the equation az+b¯¯¯z+c=0, where a,b,cϵZ. If |a||b|, then z represents

A
Circle
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B
Straight line
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C
One point
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D
Ellipse
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Solution

The correct option is A Straight line
a¯¯¯z+b¯¯¯z+c=0 (1)
or ¯¯¯a¯¯¯z+¯¯bz+¯¯c=0 (2)
Eliminating ¯¯¯z from (1) and (2), we geet
z=c¯¯¯ab¯¯c|b|2|a|2
If |a||b|, then z represents one point on the Argand plane. If |a|=|b| and ¯¯¯acb¯¯c, then no such z exists. Adding (1) and (2),
(¯¯¯a+b)¯¯¯z+(a+¯¯b)z+(c+¯¯c)=0
This is of the form A¯¯¯z+¯¯¯¯Az+B=0, where B=c+¯¯c=0 is real.
Hence, locus of z is a straight line.

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