$\displaystyle sin(z)=\mathbf{i}$

$\displaystyle \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}=\mathbf{i}\Rightarrow e^{2\mathbf{i}z}+2e^{\mathbf{i}z}-1=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}=\frac{2\pm\sqrt{8}}{2}=-1\pm\sqrt{2}$

$\displaystyle \displaystyle \mathbf{i}z=ln(-1+\sqrt{2})\Rightarrow z=-\mathbf{i}(ln(-1+\sqrt{2})+\mathbf{i}arg(z))\Rightarrow -\mathbf{i}(ln(-1+\sqrt{2})+\mathbf{i}(0+2\pi k), \ k\in\mathbb{Z}$

$\displaystyle \displaystyle z=2\pi k -\mathbf{i}ln(-1+\sqrt{2})$

The books answers:

$\displaystyle \displaystyle z=(2k+1)\pi-\mathbf{i}ln(1+\sqrt{2})$

Why the discrepancy?