Find $\sum_{k=0}^6 \operatorname{Re}(Z_k)$ and $\sum_{k=0}^6\operatorname{Im}(Z_k)$ .If the Equation $(Z+1)^7=Z^7$ has roots$Z_0,Z_2...Z_6$.

Question

Find $\sum_{k=0}^6 \operatorname{Re}(Z_k)$ and $\sum_{k=0}^6 \operatorname{Im}(Z_k)$. If the equation $(Z+1)^7=Z^7$ has roots$Z_0,Z_2...Z_6$.

My approach:

$(\frac{Z+1}{Z})^7=1$

$(\frac{Z+1}{Z})^7=\cos(2k\pi)+i\sin(2k\pi)$

Then by De Moivre's Theorem

$(\frac{Z+1}{Z})=\cos(\frac{2k\pi}{7})+i\cdot\sin(\frac{2k\pi}{7})$

$1+\frac{1}{Z}=\cos(\frac{2k\pi}{7})+i\cdot\sin(\frac{2k\pi}{7})$

$\frac{1}{Z}=-1+\cos(\frac{2k\pi}{7})+i\cdot\sin(\frac{2k\pi}{7})$

$\frac{1}{Z}=-2sin^2(\frac{k\pi}{7})+i\cdot 2sin(\frac{k\pi}{7})\cdot\cos(\frac{k\pi}{7})$

$\frac{1}{Z}=-2i\sin(\frac{k\pi}{7})+i\cdot 2\sin(\frac{k\pi}{7})\cdot\cos(\frac{k\pi}{7})$

$\frac{1}{Z}=-2i\cdot\sin(\frac{k\pi}{7})(\cos(\frac{k\pi}{7})+\cdot i\sin(\frac{k\pi}{7}))$

$Z=\frac{(-1+i\cot(\frac{k\pi}{7}))}{2}=\frac{-1}{2}+\frac{i\cot(\frac{k\pi}{7})}{2}$

Which will have $7$ roots for $k=0,1,2,...6$

If I Sum all roots that must be not defined because $cot0$ is not defined.

But according to me $\sum_{k=0}^6 \operatorname{Re}(Z_k)$ must be $\frac{-7}{2}$ and $\sum_{k=0}^6\operatorname{Im}(Z_k)$ must be not defined.

Am i correct?

My second Approach:

Using binomial Expansion I am getting $0\cdot Z^7+7Z^6+...+1=0$

By Vieta's formula $\frac{7}{0}$ which is not defined. Also because highest degree term is getting cancelled so one root will approach to infinity.

Am I making any mistake. Is any other way to solve this problem?

Answer 1

1. Regarding your second approach: the given equation expands into the degree-$6$ polynomial $$7 Z^6 + 21 Z^5 + 35 Z^4 + 35 Z^3 + 21 Z^2 + 7 Z + 1 = 0,$$ so, Vieta's formula gives the sum of its six—not seven—roots as $-\frac{21}7=-3.$

2. As for your first approach, the mistake was neglecting to exclude $k=0$ at the step immediately preceding “Then by De Moivre's Theorem” (by the way, this isn't accurate: De Moivre's theorem, for technical reasons, applies only for integer powers, not $\frac17$). This is necessary because the preceding line $$\left(\frac{Z+1}{Z}\right)^7=1$$ necessitates that $Z\neq1.$

   Here's my full answer: $$(Z+1)^7=Z^7\\ \left(1+\frac1Z\right)^7=1\\ Z=\left(e^\frac{i2n\pi}7-1\right)^{-1}\quad \left(n\notin7\mathbb Z\right)\\ =\left(\cos\left(\frac{2n\pi}7\right)-1+i\sin\left(\frac{2n\pi}7\right)\right)^{-1}\\ =\frac{\cos\left(\frac{2n\pi}7\right)-1-i\sin\left(\frac{2n\pi}7\right)}{2-2\cos\left(\frac{2n\pi}7\right)}\\ =-\frac12+i\frac{\sin\left(\frac{2n\pi}7\right)}{2\left(\cos\left(\frac{2n\pi}7\right)-1\right)}\\ =-\frac12-i\frac12\cot\left(\frac{2n\pi}7\right).$$ Therefore, $$\sum_{k=1}^6 \operatorname{Re}(Z_k)\\=6\left(-\frac12\right)\\=-3,$$ and, since $\cot$ is an odd function, $$\sum_{k=1}^6 \operatorname{Im}(Z_k)\\=\sum_{n\in\{\pm1,\pm2,\pm3\}}-\frac12\cot\left(\frac{2n\pi}7\right)\\=0.$$