# 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 $$x$$
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