2022-01-17

If the equation ${\left(z+1\right)}^{7}+{z}^{7}=0$ has roots ${z}_{1},{z}_{2},\cdots {z}_{7}$
then what is the value of $\sum _{r=1}^{7}\mathbb{R}\left({Z}_{r}\right)?$

nick1337

Expert

Step 1
I would choose a different approach, saving some labour:
The polynomial:
$p\left(z\right)=\left(z+1{\right)}^{7}+{z}^{7}=2{z}^{7}+7{z}^{6}+\cdots =2\left({z}^{7}+\frac{7}{2{z}^{6}}+\cdots \right)$
And may also be written as:
$p\left(z\right)=2q\left(z\right)=2\prod _{k=1}^{7}\left(z-{z}_{k}\right)$
The sum of the roots of the monomial $q\left(z\right)$ is equal to minus the coefficient of ${z}^{6}$, as can be seen by expanding $q\left(z\right)$
The sum of the roots of $q\left(z\right)$ equals the sum of roots of $p\left(z\right)$:
$Re\left(\sum _{k=1}^{7}{z}_{k}\right)=\left(\sum _{k=1}^{7}{z}_{k}\right)=-\frac{7}{2}$

star233

Expert

Step 1
You can observe that z=0 is not a solution, so the equation can be rewritten as
$\left(1+\frac{1}{z}{\right)}^{7}=-1$
Hence we get
$1+\frac{1}{{z}_{k}}={w}_{k}$
for $1\le k\le 7$, where ${w}_{k}=\mathrm{exp}\left(\left(\pi +2k\pi \right)\frac{i}{7}\right)$
Write ${\alpha }_{k}=\frac{\pi +2k\pi }{14}$, so we have
${z}_{k}=\frac{1}{{e}^{2{\beta }_{k}i}-1}=\frac{{e}^{-{\beta }_{k}i}}{2i\mathrm{sin}{\beta }_{k}}$
and therefore
${z}_{k}=-\frac{i}{2\mathrm{sin}{\beta }_{k}}\left(\mathrm{cos}{\beta }_{k}-i\mathrm{sin}{\beta }_{k}\right)$
and we obtain $Re{z}_{k}=\frac{1}{2}$

Do you have a similar question?