If the roots of $(z-1)^n = i(z+1)^n$ are plotted on the argand plane they are
A. On a parabola
B. Noncyclic
C. Collinear
D. The Vertices of a triangle
I solved as far as getting the nth root of $i =\frac{z-1}{z+1}$. And I thought as the roots would lie on a circle of mod 1 hence the answer must be a circle. But there is no option. Please help. And also suggest what's wrong in my approach.
Take the absolute value on both sides of the equation:$$|(z-1)^n|=|i(z+1)^n|$$ $$|(z-1)^n|=|(z+1)^n|\cdot|i|$$ $$|z-1|^n=|z+1|^n$$ $$|z-1|=|z+1|$$ That last equation implies that $z$ must lie on the imaginary axis.