Note that $\sinh$ is a strictly increasing positive unbounded function on $(0,\infty)$ so for say $n \ge 3$, the equation $\sinh 1+ \sinh 2+...\sinh (n-1) =\sinh x$ has a unique positive solution $x_n> n-1$. 

Using that $\sin {iy}=i\sinh y$ for real $y$, the above means that $i$ is a root of the equation $\sin z+\sin 2z +..\sin {(n-1)z}-\sin x_nz=0$

For $n=2$ we can adapt this and use that $2\sinh 1=\sinh x_2$ for a unique $x_2>1$ and get that $i$ is a root of $2\sin z-\sin x_2z$