# Thread: How to solve sin(z)=2 (Complex Analysis)

1. ## How to solve sin(z)=2 (Complex Analysis)

The problem wants me to solve sin(z)=2 by equating the real and imaginary parts.

Using $\displaystyle sin(z)=sin(x+iy)$, the definition of sin(z), and some algebra, I've gotten as far as:
$\displaystyle e^{xi}e^{-y}-e^ye^{-ix}=4i$.
This is where I'm stuck. I know I'll have to set the imaginary part to 4 and the real part to 0, but how do I separate them from this point...?

2. Originally Posted by paupsers
The problem wants me to solve sin(z)=2 by equating the real and imaginary parts.

Using $\displaystyle sin(z)=sin(x+iy)$, the definition of sin(z), and some algebra, I've gotten as far as:
$\displaystyle e^{xi}e^{-y}-e^ye^{-ix}=4i$.
This is where I'm stuck. I know I'll have to set the imaginary part to 4 and the real part to 0, but how do I separate them from this point...?
$\displaystyle \sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$
So let $\displaystyle \sin(x)\cosh(y)=0$ and $\displaystyle \cos(x)\sinh(y)=4$.