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A5 For a nonnegative integer
k, let
f(k) be the number of ones in the base 3 representation of
k. Find all complex numbers
z such that

sum_(k=0)^(3^(1010)-1)(-2)^(f(k))(z+k)^(2023)=0

A $5$ For a nonnegative integer $k$ , let $f(k)$ be the number of ones in the base $3$ representation of $k$ . Find all complex numbers $z$ such that $\newline$ $\sum_{k=0}^{3^{1010}-1}(-2)^{f(k)}(z+k)^{2023}=0$

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Complex conjugate theorem

Full solution

Q. A

5

For a nonnegative integer

k

, let

f(k)

be the number of ones in the base

3

representation of

k

. Find all complex numbers

z

such that

\newline

\sum_{k=0}^{3^{1010}-1}(-2)^{f(k)}(z+k)^{2023}=0

Understand function $f(k)$ : To solve this problem, we need to understand the nature of the function $f(k)$ and how it affects the sum. The function $f(k)$ counts the number of ones in the base $3$ representation of $k$ . This means that for each $k$ , the exponent of $-2$ will be determined by how many ones are in the base $3$ representation of $k$ . We need to consider the properties of the sum and the terms involved.
Consider sum of terms: Let's consider the sum in question. We have a sum of terms of the form $(-2)^{f(k)}(z+k)^{2023}$ from $k=0$ to $3^{1010}-1$ . Notice that the sum is over all nonnegative integers $k$ less than $3^{1010}$ . This range includes all possible base $3$ representations with $1010$ digits (including leading zeros).
Cancellation of terms: The sum is equal to zero only if each term in the sum is zero or if the terms cancel each other out. Since $(z+k)^{2023}$ is raised to an odd power, it will not be zero unless $z = -k$ . However, $z$ cannot be equal to $-k$ for all $k$ in the range simultaneously. Therefore, we must rely on the cancellation of terms.
Cycle of base $3$ representations: We need to consider the properties of the exponents of $-2$ , which are determined by $f(k)$ . Since the base $3$ representation of $k$ determines $f(k)$ , we can see that $f(k)$ will cycle through values as $k$ increases. For example, $f(0) = 0$ , $f(1) = 1$ , $f(2) = 0$ , $f(k)$ $0$ (since $3$ in base $3$ is $f(k)$ $3$ ), and so on.
Cancellation over complete cycle: The key insight is that the sum is over a complete cycle of base $3$ representations from $0$ to $3^{1010}-1$ . This means that for every possible pattern of ones in the base $3$ representation, there is an equal number of $k$ 's with that pattern. Therefore, the terms $(-2)^{f(k)}$ will cancel out over the complete cycle since for every $f(k) = n$ , there will be an equal number of $f(k) = n$ terms with a positive and negative sign.
Simplify sum to find $z$ : Given that the terms $(-2)^{f(k)}$ cancel out over the complete cycle, the sum simplifies to the sum of $(z+k)^{2023}$ from $k=0$ to $3^{1010}-1$ . Since we are looking for values of $z$ such that this sum equals zero, we need to consider the possibility that $z$ is a complex number that causes these terms to cancel out.
Complex number consideration: However, since $(z+k)^{2023}$ is a high-degree polynomial in $k$ , and the sum is over a symmetric range of $k$ values, there is no obvious reason why these terms should cancel out for any particular value of $z$ . In fact, for any given $z$ , the terms $(z+k)^{2023}$ will generally not cancel out, as they will have different magnitudes and phases when $z$ is complex.
Terms do not cancel out: Therefore, we conclude that there is no value of $z$ that will satisfy the given equation. The sum cannot be zero for any complex number $z$ because the terms $(z+k)^{2023}$ do not cancel out in the sum from $k=0$ to $3^{1010}-1$ .

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