If $\mathrm{n}\in \mathrm{N}>1$, then the sum of real part of roots of ${\mathrm{z}}^{\mathrm{n}}=(\mathrm{z}+1{)}^{\mathrm{n}}$ is equal to

The equation ${\mathrm{z}}^{\mathrm{n}}=(\mathrm{z}+1{)}^{\mathrm{n}}$ will have exactly n - 1 roots. We have,

${\left(\frac{\mathrm{z}+1}{\mathrm{z}}\right)}^{\mathrm{n}}=1\text{ or }\left|\frac{\mathrm{z}+1}{\mathrm{z}}\right|=1$

or $|\mathrm{z}+1|=\left|\mathrm{z}\right|$

Therefore, z lies on the right bisector of the segment connecting the points (0, 0) and (-1, 0). Thus, Re(z) = -1/2. Hence, roots are collinear and will have their real parts equal to -1/2.

Hence, sum of the real parts of roots is (-1/2)(n - 1).