Solving sin z=2: Equating Real and Imaginary Parts

[neginf]

Homework Statement

Solve sin z=2 by
(a) equating the real and imaginary parts
(b) using the formula for arcsin z.

Homework Equations

(a)
sin z = sin x * cosh y + i * cos x * sinh y
arccosh z = log[z + sqrt(z^2 - 1)]

(b)
arcsin z = -i * log [i * z + sqrt(1 - z^2)]

The Attempt at a Solution

(a)
sin z = sin x * cosh y + i * cos x * sinh y
= 2 + 0 * i

so sin x * cosh y = 2 and cos x * sinh y = 0
for the imaginary part to be 0, x = pi/2 + 2 * n * pi or y = 0
for the real part to be 2, x = pi/2 + 2 * pi * n and y = arccosh 2 , y = 0 won't work
y = arccosh 2
= log[2 + sqrt(2^2 -1)]
=log[2 + sqrt(3)]
=ln |2 + sqrt(3)| + i * arg(2 + sqrt(3))
=ln |2 + sqrt(3)|
so z = x + i * y
= (pi/2 + 2 * pi * n) + i * ln(2 + sqrt(3))

(b)
z = arcsin 2
=-i * log (i * 2 + sqrt(1 - 2^2))
= -i * log(i * (2 + sqrt(3)))
= -i * (ln|i * (2 + sqrt(3))| + i * arg(i * (2 + sqrt(3))))
= -i * (ln|2 + sqrt(3)|) + i * (pi/2 + 2 * pi * n)
= (pi/2 + 2 * pi * n) - i * ln(2 + sqrt(3))

What did I do wrong to get opposite signs of the imaginary parts in (a) and (b) ?

 

[vela]

Remember cosh is an even function, and there are two solutions to cosh y=2.

 

[neginf]

Thank you Vela.

In (b) I still get just a minus in front of the imaginary part instead of plus or minus.

What am I missing ?

 

[vela]

Do you know how to obtain those formulas for arccosh and arcsin?

It also might help if you realize that 1/(2+√3) = 2-√3.

 

[neginf]

Thank you again Vela.