Question

If ${\alpha }_i,(i=1,2,3,...,9)$ are the roots of the equation $z^9=(4-z{)}^9$, then the value of $\sum _{i=1}^9\text{Re}\left({\alpha }_i\right)$ is
($z$ being a complex number)

Answer 1

$z^9+(z-4{)}^9=0$
$\Rightarrow 2z^9+{}^9{C}_1{z}^8(-4)+...=0$
$\Rightarrow 2z^9-36z^8+...=0$
As the polynomial has real coefficients, then the roots occurs in complex conjugate pair.

$\sum _{i=1}^9\text{Re}\left({\alpha }_i\right)=\sum _{i=1}^9{\alpha }_i=-\frac{-36}{2}=18$