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Which point in Argand plane is equidistant from roots of equation $(z+1)^4=16z^4$?

$\begin{array}{1 1}(A)\;\big(\large\frac{-1}{3}\normalsize, 0\big)&(B)\;\big(\large\frac{1}{3}\normalsize, 0\big)\\(C)\;\big(\large\frac{1}{\sqrt 5}\normalsize, 0\big)&(D)\;(0,0)\end{array} $

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$\Rightarrow \big[(z+1)^2\big]^2=(47^2)^2$
$\Rightarrow (z+1)^2=4z^2$ and $(z+1)^2=-4z^2$
$\Rightarrow 3z^2-2z-1=0$ and $5z^2+2z+1=0$
$\Rightarrow z=-1,3$ and $z=-1\pm 2i$
Clearly $z=-1$ does not satisfy.
$\Rightarrow$ Roots of z are $-1\pm 2i,3$
Thus equidistant oint is $(\large\frac{1}{3}$$,0)$
Hence (B) is the correct answer.
answered Apr 10, 2014 by sreemathi.v

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