Q:

Complex Numbers....

Suppose a is a complex number such that:
$$a^2+a+\frac{1}{a}+\frac{1}{a^2}+1=0$$
If m is a positive integer, find the value of:
$$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}$$  
My Approach:
After I could not solve it using the usual methods I tried a bit crazier approach. I thought that as the question suggests that the value of the expression does not depend upon the value of m, provided it is positive, hence the graph of the expression on Y-axis and m on X-axis would be parallel to X-axis and thus the slope be zero. So I differentiated it with respect to m and equated it to zero and after factorizing and solving I got $a^m=-1$ or $a^m=1$ or $a^m=\left(\frac{-1}{4}+i\frac{\sqrt15}{4}\right)$ or $a^m=\left(\frac{-1}{4}-i\frac{\sqrt15}{4}\right)$. But if $a^m=1$ then $a=1$ which does not sattisfy the first equaion. Thus the value of $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=\frac{-9}{4}$$
                     OR
 $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=0$$
What other approach would you suggest? What are the flaws in my approach (if any)?

A:

If we multiply
$$a^2 + a + \frac1a + \frac{1}{a^2} + 1 = 0$$
with $a^2$, we obtain (since evidently $a \neq 1$)
$$0 = a^4 + a^3 + a^2 + a + 1 = \frac{a^5-1}{a-1},$$
so $a^5 = 1$,
$$a = e^{(2\pi ik)/5},\quad k \in \{1,2,3,4\}.$$
Then, if $m$ is a multiple of $5$, we have $$(a^2)^m + a^m + \frac1{a^m} + \frac{1}{(a^2)^m} = 1+1+1+1 = 4,$$
and if $m$ is not a multiple of $5$, the four numbers
$$a^{2m},\, a^m,\, a^{-m}\, a^{-2m}$$
are the numbers $a^2,\, a,\, a^{-1},\,a^{-2}$, possibly in a different order, then the sum is $-1$.

