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 Bcircle |z|=3⇒z=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