If `n in N >1` , then the sum of real part of roots of `z^n=(z+1)^n` is equal to

Options

A. `(n)/(2)`
B. `((n-1))/(2)`
C. `-(n)/(2)`
D. `((1-n))/(2)`

Answer

D

Explanation

The equations `z^(n) = (z+ 1)^(n)` will have excactly n - 1 roots. We have
`((z+1)/(z^(n)))= 1 or |(z+1)/(z)|= 1` or `|z+1|= |z|`
Therefore,z lies on the the right bisector of the segment connecting the points (0,0) and (-1,0). Thus `Re(z) = - 1//2`. Hence roots are colliner and will have their real parts of roots is `(-1//2)(n-1)`