Prove that the roots of $(z - 1)^{n} = i (z + 1)^{n}$ when plotted in the Argand plane are collinear.

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Solution

we have $(z - 1)^{n} = i (z + 1)^{n}$
$∴ {| z - 1 |}^{n} = | i | {| z + 1 |}^{n}$
or ${| z - 1 |}^{n} = {| z + 1 |}^{n}$
or $| z - 1 | = | z + 1 |$
or ${| z - 1 |}^{2} = {| z + 1 |}^{2}$
$⟹ (x - 1)^{2} + y^{2} = (x + 1)^{2} + y^{2}$
or $4 x = 0$ or $x = 0$
Therefore, the roots lie on the y-axis.

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