Posted by **Jake** on Sunday, January 14, 2007 at 3:35pm.

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

cosacoshb-isinasinhb=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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