Question

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

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

Answer 1

The correct option is C $-\frac{7}{2}$

Given: $z_k(k=0,1,2,............6)zk(k=0,1,2,............6)$be the roots of equation ${(z+1)}^7+z^7=0$

To find: ${\sum }_{k=0}^{6}Re\left(z_k\right)$

Soln:

If $z_k$ is a root of the equation,

${(z+1)}^7+z^7=0$ then it satisfies the equation.

Taking modules both sides

$\therefore {|z_k+1|}^7={|-z_k|}^7$

$\left|{(z_k+1)}^7\right|=\left|{-z_k}^7\right|$

$|z_k+1|=\left|z_k\right|$

that is

$|z_k+1|=|z_k-0|$

So that $z_k$ is equidistant from the points$(0,0)$ and$(-1,0)$ is the complex plan.

Therefore, it lies on the perpendicular bisector of the line segment joining $(0,0)$ and$(-1,0)$

that is$z_k$ lies on the line$x=\frac{-1}{2}$

To find the solution of the $z_k$ note that.${(z+1)}^7+z^7=0$ May be written as

${\left[\frac{-(z+1)}{z}\right]}^7=1$

Therefore we have, $\frac{-(z+1)}{z}=e^{i\left(\frac{2k\pi }{7}\right)}$

for$k=0,1,2,......6$

Now, let $\lambda =\frac{2\pi }{7}$and solve for $z$ to get

$z_k=\frac{-1}{1+e^{i\lambda k}}=\frac{-1}{1+\cos \lambda k+i\sin \lambda k}$

$z_k=-\frac{1+\cos \lambda k-i\sin \lambda k}{{(1+\cos \lambda k)}^2+i{\sin }^2\lambda k}$

$z_k=-\frac{1+\cos \lambda k-i\sin \lambda k}{2(1+\cos \lambda k)}=\frac{-1}{2}+\frac{i\sin \lambda k}{2(1+\cos \lambda k)}$

$\therefore Re\left(z_k\right)=\frac{-1}{2}$

${\sum }_{k=0}^{k=6}Re\left(z_k\right)={\sum }_{k=0}^{k=6}=\frac{-1}{2}{\sum }_{k=0}^{k=6}Re\left(z_k\right)=\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2}$

$=-\frac{7}{2}$

Hence, the answer is $\frac{-7}{2}$