A little bit of product and sum is good for everyone

Algebra Level 4

$\text{Let: }\large Z_x = \sqrt[n]{a + bi} = \sqrt[2n]{a^2 + b^2}~\text{cis}\left(\frac{\arctan \frac{b}{a}}{n} + \frac{2x\pi}{n}\right)$

$\text{Let: }\large Y_y = \sqrt[m]{a + bi} = \sqrt[2m]{a^2 + b^2}~\text{cis}\left(\frac{\arctan \frac{b}{a}}{m} + \frac{2y\pi}{m}\right)$

If the following is true:

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} = \sum_{x = 0}^{n - 1}{Z_x} - \sum_{y = 0}^{m - 1}{Y_y}$

Is $|n - m|$ an odd or even number?

Details and Assumptions

$\large \text{cis}(\theta) = \cos(\theta) + i\sin(\theta)$
The equations at the top are used to calculate all the $n$ and $m$ roots respectively of the complex number $a + bi$
$a + bi \neq 0$
The variables $n$ and $m$ are both non-zero real integers
$x$ is a real integer which obeys the following rule: $0 \leq x < n$
$y$ is a real integer which obeys the following rule: $0 \leq y < m$
$i$ is the imaginary number equal to $\sqrt{-1}$
$|x|$ represents the absolute value of $x$

Neither odd nor even function Odd Not enough information Even

1 solution

Jack Rawlin
Mar 5, 2016

We'll sort out the RHS first.

The sum of all the roots of a complex number will always amount to zero, so the RHS comes to $0 - 0$ or simply $0$ .

$\large \sum_{x = 0}^{n - 1}{Z_x} - \sum_{y = 0}^{m - 1}{Y_y} = 0$

Now the LHS.

The product of all the roots of a complex number will equal the complex number itself or it's negative equivalent. If there are an odd amount of roots then the result is positive, if there's an even amount of roots then the result is negative. In other words:

$\text{When }n \text{ is odd}: \prod_{x = 0}^{n - 1}{Z_x} = (a + bi)$

$\text{When }n \text{ is even}: \prod_{x = 0}^{n - 1}{Z_x} = -(a + bi)$

If both $n$ and $m$ are even or odd their difference will be even. In the case that both $n$ and $m$ are even:

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} =~?$

$\large -(a + bi) + -(a + bi) =~?$

$\large -(a + bi) + -(a + bi) = -2(a + bi)$

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} = -2(a + bi)$

The only way this would be equal to $0$ is if both $a$ and $b$ are $0$ which the question says isn't the case.

So what about if both $n$ and $m$ are odd numbers?

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} =~?$

$\large (a + bi) + (a + bi) =~?$

$\large (a + bi) + (a + bi) = 2(a + bi)$

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} = 2(a + bi)$

As with before, the result is non-zero. Therefore the difference between $n$ and $m$ is not even.

There are two situations which would make the difference odd. When $n$ is odd and $m$ is even, or when $n$ is even and $m$ is odd. Let's try that first one

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} =~?$

$\large (a + bi) + -(a + bi) =~?$

$\large (a + bi) - (a + bi) =~?$

$\large (a + bi) - (a + bi) = 0$

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} = 0$

So the first one gives a result of $0$ . What about the second one, does it give us a different answer?

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} =~?$

$\large -(a + bi) + (a + bi) =~?$

$\large -(a + bi) + (a + bi) = 0$

$\large \prod_{x = 0}^{n - 1}{Z_x} + \prod_{y = 0}^{m - 1}{Y_y} = 0$

Both situations where the difference is odd give an answer which fits the equation. Therefore the difference between $n$ and $m$ is an odd number.

$\large \boxed{\text{The difference between }n\text{ and }m\text{ is an odd number}}$

A little bit of product and sum is good for everyone

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