# 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 \(x\)

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

Already have an account? Log in here.