Question

Complex Numbers
Number of values of z (real or complex) simultaneously satisfying the system of equations
1+z+z^2 .... z^17 = 0 and 1+z+z^2 .... z^13 = 0

Answer 1

@rational

Answer 2

@ParthKohli

Answer 3

I'm glad I was thought worthy of this. *bows*\[\dfrac{z^{18} - 1}{z -1}=0\]\[\dfrac{z^{14} -1}{z-1}=0\]As \(z=1\) is not a solution, the problem boils down to:\[z^{18}=1\]\[z^{14} =1 \]

Answer 4

So the answer should be gcd(18,14) = 2?

Answer 5

lemme show my work..
first 14 terms or 1st exp=0
then z^14(1+z+z^2+z^3) = 0
that gives 4 solution z=0 and three from the cubic
z=0 is rejected
but the answer is 1

Answer 6

Oh, hold on.

Answer 7

We started with the assumption that z = 1 is not a solution (obviously).

z^14 = 1 and z^18 = 1 have solutions 1 and -1 by simple inspection.

But 1 is eliminated.

Answer 8

So only -1 satisfies this, yes?

Answer 9

yeah..but whats problem in my solution?

Answer 10

Hmm, tricky solution.

Answer 11

and we are talking about complex number so we can get 14 root of x^14= 1
and 18 of x^18 = 1
using DMT.. so will there be any common root
is there is then more roots

Answer 12

\(\large \color{black}{\begin{align} 1+z+z^2+\cdots+z^{17}=0\hspace{.33em}\\~\\
1+z+z^2+\cdots+z^{13}=0\hspace{.33em}\\~\\
\implies z^{14}+z^{15}+z^{16}+z^{17}=0\hspace{.33em}\\~\\
\implies z^{14}\left(1+z+z^{2}+z^{3}\right)=0\hspace{.33em}\\~\\
\implies 1+z+z^{2}+z^{3}=0\hspace{.33em}\\~\\
\implies 1+z+z^2(1+z)=0\hspace{.33em}\\~\\
\implies (1+z^2)(1+z)=0\hspace{.33em}\\~\\

z=-1,-i,i\hspace{.33em}\\~\\
\end{align}}\)

Answer 13

\[z^2(z + 1) + 1(z+1) = (z^2+1)(z+1)\]The reason why \(\pm i \) is rejected is because the sum of any four consecutive powers is zero. But I'd have to come up with a better explanation.