How to get sin(z) = 2 with negative imaginary part

[neginf]

Homework Statement

Solve sin(z)=2.

Homework Equations

sin z=(e^i*z - e^-i*z)/2*i
sin z=sin x * cosh y + i*cos x * sinh y (haven't tried this way yet)

The Attempt at a Solution

Starting with the first relevant equation, I got z=pi*(1/2 + 2*n) + i*ln(2+sqrt(3)).
The book says that another solution is z=pi*(1/2 + 2*n) - i*ln(2+sqrt(3)).
How do you get that ?

 

[bennyska]

because cosh is an even function.

 

[I like Serena]

Assuming you solved your equation with the quadratic formula, you should have 2 solutions.

If you work it out, you'll see that they match the 2 solutions that you gave.

 

[HallsofIvy]

The solution to the quadratic equation $y^2- 4iy- 1= 0$
is
$$y= \frac{4i\pm\sqrt{-16+ 4}}{2}= 2i\pm i\sqrt{3}$$
Did you use both "+" and "-"?