# 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.

• The biggest mistake in your MathJax was to have too many small, separate segments. The more math you put together in one pair of dollar signs, the better. Apr 24, 2021 at 11:01
• @arthur ok thanks. Apr 24, 2021 at 11:07
• see my comments following Kavi Rama Murthy's answer. The convoluted inference based on the premise is that $a_7 = 0$ and $a_{14} = 0.$ Apr 24, 2021 at 13:55
• I just left an answer in response to Kavi Rama Murthy's editing of his answer. Apr 25, 2021 at 3:15

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 $$0 = a_{14}$$.

• thanks a lot for clarification. Apr 25, 2021 at 6:24
• And I also agree that the variable k was overloaded, but I am sure it did not change gist of my query. Apr 25, 2021 at 6:25

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 $$k=7$$.

• I am confused by your answer. I independently came to the exact same conclusion as the OP. That is, when you compute the 20-th degree polynomial, the coefficients of each term are zero, except for the coefficients that pertain to the terms $x^0, x^7$ and $x^{14}$. For those terms, I agree with the OP that the coefficients are each $(7)$. So you have that $7a_0 + 7a_7x^7 + 7a_{14}x^{14} = ka_0$. I don't see how that you deduce from this that $k = 7$, since that inference would require that $(a_7x^7 + a_{14}x^{14}) = 0,$ for any value of $(x).$ Apr 24, 2021 at 13:49
• One interpretation of the problem that would resolve the issue is that since you are given that the final sum equals $ka_0$, for any value of $(x)$, you are forced to conclude that $a_7 = 0$ and $a_{14} = 0$. A somewhat convoluted inference, but still feasible. Is this what you are thinking? Apr 24, 2021 at 13:53
• I just left an answer in response. In short, I agree that the variable $k$ is overloaded in the original presentation of the question, but still think that the convoluted inference is necessary. Apr 25, 2021 at 3:14