sin (z) = 2

Solving $\sin z = 2$ we obtain $\begin{aligned} e^{iz} - e^{-iz} & = \; 4i \\ e^{2iz} - 4ie^{iz} - 1 & = \; 0 \\ (e^{iz} - 2i)^2 & = \; -3 \\ e^{iz} & = \; (2 \pm \sqrt{3})i \end{aligned}$ Since $\log z = \ln|z| + i\mathrm{arg}(z)$ , remembering that the argument of a complex number is only defined modulo $2\pi$ , we deduce that $\begin{aligned} iz & = \; \ln(2 \pm \sqrt{3}) + (2n+\tfrac12)\pi i \\ z & = \; (2n + \tfrac12)\pi \pm i\ln(2 + \sqrt{3}) \end{aligned}$ for any integer $n$ . One particular value is $\boxed{\tfrac12\pi - i\ln(2 + \sqrt{3})} = 1.570796... - 1.316957...i$ .