Find sum of roots of equation $z^6 = z^{18} = -1$
Roots of $z^6=-1$
Roots of $z^{18}=-1$
From the circle I can see, that:
$\{z : z^6=-1\} \cap \{z : z^{18}=-1\} = \{z : z^6=-1\}$
So, I can notice, that sum of roots $z^6=-1$ is equal to zero. However, I solved that with help of WolframAlpha - drawing a circle with 18 points (or calculating all roots) is not very encouraging. Is there the more reliable method to solve such problem?
Edit: correct circles are:
"Purity Of Essence, Mandrake, Purity Of Essence . . . "; General Jack Ripper, in Stanley Kubrick's Dr. Strangelove . . .
It is in the interest of preserving Mathematical Purity Of Essence that I feel obliged to point out that neither of the two root-plots provided by our OP stil is correct, for they each depict both $1$ and $-1$ as being roots of $z^6 = -1$ and $z^{18} = -1$ whereas in fact neither $1$ nor $-1$ are roots of either equation: $1^6 = (-1)^6 = 1^{18} = (-1)^{18} = 1 \ne -1$. I don't know what's up with Wolfram Alpha if it is saying that these are the root-plots of $z^6 = -1 = z^{18}$; but for all positive integers $n$ it is easy to type $z^n - 1$ instead of $z^n + 1$, so perhaps these plots are the output of an input error. I would here recite that ancient mantra of computer programmers, GIGO, GIGO, GIGO: Garbage In, Garbage Out, except that these cool plots are redeemable by a mere rotation of either $\pi / 6$ or $\pi / 18$, that is, multiplication of all their complex-valued points by $e^{i\pi / 6}$ or $e^{i \pi /18}$, according to whether the plot corresponds to the equation $z^6 = 1$ or $z^{18} = 1$, respectively. Indeed, if $z^6 = 1$ then $(e^{i\pi / 6}z)^6 = e^{i\pi}z^6 = (-1)1 = -1$, with a similar result holding in the case $z^{18} = 1$. So rather than being garbage, these plots are merely like broken toys which can easily be fixed with some magic tape or a dab of glue.
Having said these things, there is an easy way to determine the sum of the roots of any equation of the form $z^n + a = 0$, for any integer $n \ge 1$ and any $a \in \Bbb C$. If $n = 1$, then $z = -a$, so the sum is $-a$ in this case. For $n \ge 2$, let $\alpha_1, \alpha_2, . . . \alpha_n$ be the roots of $z^n + a = 0$. Then we can write $z^n + a$ as the product
$z^n + a = \prod_1^n (z - \alpha_i), \tag{1}$
and when the right-hand side is multiplied out we obtain
$z^n + a = \prod_1^n (z - \alpha_i) = z^n - (\sum_1^n \alpha_i)z^{n - 1} + \sum_{i, j = 1, i \ne j}^n \alpha_i \alpha_j + . . . + (-1)^n \prod_1^n \alpha_i; \tag{2}$
comparing coefficients we see that $\sum_1^n \alpha_i = 0$, $(-1)^n \prod_1^n \alpha_i = a$, and a host of other things besides. Much simpler than working with trigonometric polynomials, as I think you will probably agree! Reliable, too!
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
The sum of the roots is the sum of the cosines and sines: $\sin(\pi/9) = -\sin(\pi/9) = -\sin(17\pi/9)$ and so forth. This gives the imaginary part of the sum as $0$; similarly, $\cos(\pi/9) = -\cos(8\pi/9)$ and so forth; so the real part is also $0$. Just think in terms of components.
