If the roots of $(z - 1)^{n} = i (z + 1)^{n}$ are plotted on the argand diagram, they are:

On a parabola

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Concyclic

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Collinear

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The vertices of a triangle

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Solution

The correct option is D Collinear
$(z - 1)^{n} = i (z + 1)^{n}$
$\Rightarrow (\frac{z - 1}{z + 1}) = (i)^{1 / n}$
$\Rightarrow \frac{z - 1}{z + 1} = e^{(2 k π + \frac{π}{2}) \frac{1}{n}} = c o s (2 k + 1) \frac{π}{2 n} + i s i n (2 k + 1) \frac{π}{2 n}$
$= c o s α + i s i n α$ where $α = (2 k + 1) \frac{π}{2 n}$
Applying componendo & dividendo
$\Rightarrow \frac{2 z}{- 2} = \frac{c o s α + i s i n α + 1}{c o s α + i s i n α - 1} = \frac{2 c o s^{2} \frac{α}{2} + i 2 s i n \frac{α}{2} c o s \frac{α}{2}}{- 2 s i n^{2} \frac{α}{2} + i 2 s i n \frac{α}{2} c o s \frac{α}{2}}$
$- z = (\frac{1}{i}) (\frac{c o s \frac{α}{2}}{s i n \frac{α}{2}})$
$\Rightarrow z = i c o t \frac{α}{2}$
which is a purely imaginary number.
$\Rightarrow$ all roots lie on imaginary axis
$\Rightarrow$ all roots are collinear.

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