maths 2
posted by Jake .
The original question I had was write arcsin4 in the form a+ib. I manage and understand how to get so far BUT How do I get from
cosacoshbisinasinhb=4
to
2m(pi)+/ iarccosh4
arcsin4 = a + b i >
4 = sin(a + bi)
sin(a + bi) =
sin(a)cos(bi) + cos(a) sin(bi)
cos(bi) = cosh(b)
sin(bi) = i sin(b)
So,
4 = sin(a)cosh(b) + i cos(a) sinh(b)
The imaginary part of the left hand side is zero, so we must have:
cos(a) sinh(b) = 0
We already know that b can't be zero, because we know that arcsin(4) is not real. This means that a must be plus or minus pi/2.
Equatiiong the real parts gives you:
4 = sin(a)cosh(b)
cosh(b) is positive, it thuis follws that sin(a) must be positive and we must thus choose a = pi/2. And we obtain:
b = plus or minus arccosh(4)
It turns out that the correct analytical continuation of the arcsin function defined for real argument yields the minus sign for b here.
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