How to solve the following question from complex number. If $\alpha=e^{\frac{i2\pi}{7}}$ and $f(x)= a_0+\sum_{k=1}^{20} a_kx^k$ then the value of $$f(x)+f(\alpha x)+f(\alpha^2 x)+\cdots+f(\alpha^6 x)$$ is $ka_0$. Then find the value of $k$.
I used a common property of complex numbers which is giving me entire different result. It is as follows.
$$1^k+\alpha^k+(\alpha^2)^k+\cdots+(\alpha^{n-1})^k=0$$ if $k$ in not a multiple of $n$, else $n$; where $\alpha$ is the $n^{th}$ root of unity.
Now from the above fact if can be said that there are some powers of $x$ which are multiple of $7$, hence the required sum will contain coefficients other than $a_0$. Hence I highly doubt the question to be misprinted in my book. Please help whether I am right or wrong.
 A: I think a silly confusion is caused by the use of bad notations in the question. The $k$ in the definition of $f$ is a dummy variable which has nothing to do with the $k$ in the statement that the sum is $ka_0$. Change $k$ to $i$ or some other variable in the definition of $f(x)$.
Note that $(1-\alpha)(1+\alpha+\alpha^{2}+\cdots+\alpha^{6})=1-\alpha^{7}=0$.  So $1+\alpha+\alpha^{2}+\cdots+\alpha^{6}=0$ Now if you compute the sum you wll get $7a_0$ so $k=7$.
A: This answer is an extension of the comments that I left after
Kavi Rama Murthy's answer, and explains why I agree with the OP's analyis
that the problem is puzzling.  I also explain why the problem requires the
convoluted inference that both $a_7$ and $a_{14}$ are equal to zero.
First of all, I definitely agree with almost all of the analysis in
Kavi Rama Murthy's answer:

*

*$1 + \alpha + \alpha^2 + \cdots \alpha^6 = 0.$


*In the original presentation of the problem, the variable $k$ is overloaded.
This is easily resolved by re-expressing $f(x)$ as 
$\displaystyle a_0 + \sum_{r=1}^{20} a_r x^r.$


*For $r \in \Bbb{Z^+}$ such that $r$ is not a multiple of $(7)$ 
$\displaystyle 1 + (\alpha^r) + (\alpha^r)^2 + (\alpha^r)^3 + (\alpha^r)^4 + (\alpha^r)^5 + (\alpha^r)^6 = 0.$ 
This is because Kavi Rama Murthy's analysis against $(\alpha)$ also pertains to
$(\alpha)^r.$ 
Namely that $(1 - [\alpha^r]) \neq 0$, while 
$\displaystyle (1 - [\alpha^r]) \times (1 + [\alpha^r] + [\alpha^r]^2 + \cdots + [\alpha^r]^6)
~=~ 1 - [\alpha^r]^7 = 0.$

However, as explained below, this does not resolve the conflict originally identified by
the original poster.
Let $\displaystyle g(x) = f(x) + f(\alpha x) + f(\alpha^2 x) + \cdots + f(\alpha^6 x)$
$\displaystyle =~ a_0 + \sum_{r=1}^{20} a_r x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^r) x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{2r}) x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{3r}) x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{4r}) x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{5r}) x^r $
$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{6r}) x^r. $
Therefore,
$$g(x) = 7a_0 + \sum_{r=1}^{20} \left[a_r x^r \left(\sum_{s=0}^6 \alpha^{(rs)} \right) \right]. \tag{1}$$
As discussed, as $r$ takes on the values $(1)$ through $(20)$,  
if $r$ is not a multiple of $(7)$, then 
the inner summation for $g(x)$  in equation (1) above, 
$\displaystyle \left(\sum_{s=0}^6 \alpha^{(rs)} \right)$ 
will equal zero.
However, for $r = 7$ or $r = 14$ 
the inner summation for $g(x)$ in equation (1) above, 
$\displaystyle \left(\sum_{s=0}^6 \alpha^{(rs)} \right)$ 
will instead equal $(7).$
Therefore
$$g(x) = 7a_0 + 7a_7(x^7) + 7a_{14}(x^{14}). \tag{2}$$
A premise is given that (presumably) for all values of $x, ~g(x) = ka_0$.
I don't see how this premise can be true, for all values of $x$, unless
both $a_7$ and $a_{14}$ are equal to zero.
Therefore, since the constraint
that for all values of $x, ~g(x) = ka_0$
is a premise, 
one is forced into the convoluted inference that $0 = a_7$ and $0 = a_{14}$.
