>>449
(exp(ix)+exp(-ix))/2 = i

z = exp(ix) とおくと
(z + 1/z)/2 = i
z^2 - 2iz + 1 = 0

解の公式より
z = i ± (i^2 - 1)^(1/2) = (1 ± √2)i
ix = log((1±√2)i) = ±log((1+√2)i)
x = ±i*log((1+√2)i)