(ij) * cos[i * log((a + ib)/(a - ib))] = (a ^ 2 - b ^ 2)/(a ^ 2 + b ^ 2)

asked by guest
on Nov 18, 2024 at 3:27 am



You asked:

Investigate the equation: ijcos⁡(ilog⁡(a+iba−ib))=a2−b2a2+b2i j \cos\left( i \log\left( \frac{a + i b}{a - i b} \right) \right) = \frac{{a}^{2} - {b}^{2}}{{a}^{2} + {b}^{2}}.

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