# Solve sin(z)=2

1. Jul 1, 2011

### neginf

1. The problem statement, all variables and given/known data

Solve sin(z)=2.

2. Relevant 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)

3. 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 ?

2. Jul 1, 2011

### bennyska

Re: sin(z)=2

because cosh is an even function.

3. Jul 1, 2011

### I like Serena

Re: sin(z)=2

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.

4. Jul 1, 2011

### HallsofIvy

Re: sin(z)=2

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 "-"?