Cos x= 3i what is the value of x?
A relation we can use is,
e^(a + bi) = e^a * (cos(b) + i.sin(b))
Set b = π/2, then
e^(a + π/2.i) = e^a * (cos(π/2) + i.sin(π/2))
e^(a + π/2.i) = e^a * (0 + i.1) = e^a.i
e^(a + π/2.i) = e^a.i
We want cos(x) = 3i, therefore set e^a = 3, i.e. a = ln(3)
Then
e^(ln(3) + π/2.i) = 3i = cos(x)
giving,
x = arccos(e^(ln(3) + π/2.i))
If we want a general answer, then make b = π/2 + 2nπ, n = 1,2,3,... giving the general solution as,
x = arccos(e^(ln(3) + (π/2 + 2nπ).i)), n = 1,2,3,...