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Question

lf |z|=3, then the points representing the complex numbers 1+4z lie on a

A
line
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B
circle
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C
parabola
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D
hyperbola
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Solution

The correct option is B circle
|z|=3z=3(cosθ+isinθ)(1)
x+iy=1+4z(2)
Now from (1) and (2),
x+iy=1+12(cosθ+isinθ)
Comparing both side,
x=1+12cosθ
and y=12sinθ
x+1=12cosθ and y=12sinθ
Now take square of (x+1) and y, then add
(x+1)2+y2=144 is a circle

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