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I have the equation

$z^{8}=\bar{z}$

I had to solve it, and to find the sum and product of the solutions. I did all that and found that the solutions are (in degrees): cis of: 0, 40, 80, 120, ..., 320.

The sum was 0 and the product 1 (I used arithmetic and geometric series).

Now I need to tell what would happen if the equation was:

$z^{n}=\bar{z}$ where n is natural.

I understand that it depends if n is odd or even, but I am not sure how it affect the sum and product. Does it affect the solutions themselves ?

Thank you

user3275222
  • 613

2 Answers2

4

You have $|z|^8=|z|$, so either $z=0$ or $|z|=1$.

In the latter case, the equation becomes $z^9=1$, because $\bar{z}=z^{-1}$.

The general case is exactly the same. The final problem is to find the sum of the $(n+1)$-th roots of $1$, which is fairly easy.

egreg
  • 238,574
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So, as noted, $z = 0$ is always a solution. For $z \ne 0$, if $z^n = \bar{z}$, $$ z^{n+1} = |z|^2 = 1. $$ So the solutions are the $n+1$th roots of unity. The sum of these is always zero, which can be seen from $$ e^{\frac{2\pi i}{n+1}}\sum_{m=0}^n e^{\frac{2\pi i m}{n+1}} = \sum_{m=1}^n e^{\frac{2\pi i m}{n+1}} + e^{2\pi i \frac{n+1}{n+1}} = \sum_{m=1}^n e^{\frac{2\pi i m}{n+1}} + 1 = \sum_{m=0}^n e^{\frac{2\pi i m}{n+1}} $$ and $e^{\frac{2\pi i}{n+1}} \ne 1$. Their product, on the other hand, is $$ \prod_{m=0}^ne^{\frac{2\pi i m}{n+1}} = \exp\left(\frac{2\pi i}{n+1}\sum_{m=0}^nm\right) = \exp\left(\frac{2\pi i}{n+1}\sum_{m=0}^nm\right) = e^{i\pi n} = (-1)^n $$ So we have

  • Sum of solutions: 0
  • Product of nonzero solutions: $(-1)^n$
  • Product of all solutions: 0
eyeballfrog
  • 22,485