On simplification, \( \frac{1}{1+z^{a-b}+z^{a-c}} + \frac{1}{1+z^{b-c}+z^{b-a}} + \frac{1}{1+z^{c-a}+z^{c-b}} \) would reduce to?
Answer (1)
To simplify the expression 1 + z a − b + z a − c 1 + 1 + z b − c + z b − a 1 + 1 + z c − a + z c − b 1 , we need to analyze it carefully.
This expression includes three fractions. Each fraction has the same structure, with the denominators shifting based on the powers of z .
The key idea to simplify this is to try substituting specific values for z such as z = 1 and see if it simplifies to a recognizable form.
Substitute z = 1 : Each denominator simplifies since z n = 1 n = 1 for any integer n .
The expression becomes:
1 + 1 + 1 1 + 1 + 1 + 1 1 + 1 + 1 + 1 1
Which simplifies further to:
3 1 + 3 1 + 3 1 = 1
Interpretation and Conclusion: Since substituting z = 1 reduces the expression to 1, and given the symmetry of the expressions in the denominators, we can infer that for symmetry-based expressions like this, where the terms are rotated cyclically, substituting z = e 2 πi /3 might also yield similar results, reinforcing the simplification to 1.
Thus, upon simplification, the entire expression reduces to 1 given appropriate substitutions and symmetry considerations.
