Question

Let $z_k(k=0,1,2,\dots ,6)$ be the roots of the equation $(z+1{)}^7+z^7=0$, then $\sum _{k=0}^6\text{Re}\left(z_k\right)$ is equal to

• A. $0$
• B. $\frac{3}{2}$
• C. $-\frac{7}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

we have ${\left(z_k+1\right)}^7+z_k^7=0$
$\Rightarrow {\left(z_k+1\right)}^7=-z_k^7$
$\Rightarrow {\left|z_k+1\right|}^7={\left|z_k\right|}^7$
$\Rightarrow \left|z_k+1\right|=\left|z_k\right|$
$\Rightarrow {\left|x_k+i{y}_k+1\right|}^2={\left|x_k+i{y}_k\right|}^2$
$\Rightarrow {\left(x_k+1\right)}^2+y_k^2=x_k^2+y_k^2$
$\Rightarrow 2x_k+1=0$
or $x_k=-\frac{1}{2}$
Thus, $\sum _{k=0}^6\text{Re}\left(z_k\right)=\sum _{k=0}^6{x}_k=-\frac{7}{2}$.