As

${\displaystyle \sin z=\frac{1}{2i}(e^{\mathrm{iz}}-e^{-\mathrm{iz}})=2}$

then, substituting $w=e^{\mathrm{iz}}$, we have

${\displaystyle w-\frac{1}{w}=4i.}$

So $w^2-4\mathrm{iw}-1=0$ and the solutions of this quadratic equation are:

${\displaystyle w=e^{\mathrm{iz}}=\frac{4i\pm \sqrt{-}12}{2}=i(2\pm \sqrt{3}).}$

Taking logarithms gives

${\displaystyle \mathrm{iz}={\log }_{e}i(2\pm \sqrt{3})={\log }_{e}i+{\log }_e(2\pm \sqrt{3})=i\frac{\pi }{2}+{\log }_e(2\pm \sqrt{3}).}$

Dividing by $i$ gives the solutions $z=\frac{\pi }{2}-i{\log }_e(2\pm \sqrt{3})$ but since $\sin z$ is periodic with period $2\pi$ the set of all solutions is given by

${\displaystyle z=\frac{\pi }{2}-i\log (2\pm \sqrt{3})+2n\pi .}$

[Editor's note:

The same method works to give solutions for $\sin z=a$ where $a$ is any complex number.

The step in the solution above ${\log }_{e}i=\frac{\pi }{2}$ follows from the definition of the complex logarithm function using $|i|=1$ and $\arg i=\frac{\pi }{2}$. The logarithms of $z$ are the numbers $\lambda$ such that $e^{\lambda }=z$, that is:

${\displaystyle \lambda =\log |z|+i(\arg z+2\pi n)}$

for integer $n$ . It is easy to check:

${\displaystyle e^{\log |z|+i(\arg z+2\pi n)}=e^{\log |z|}\times e^{i\arg z}\times e^{2\pi \mathrm{ni}}=|z|e^{i\arg z}=z}$

giving the complex number $z$ in modulus-argument form. Every complex number has infinitely many complex logarithms each differing by an integer multiple of $2\pi i$ because $e^{2\pi \mathrm{ni}}=1$.]