The scheme with \( u^n=I\exp{(i\omega\tilde\Delta t\, n)} \) inserted gives $$ -I\exp{(i\tilde\omega t)}\frac{4}{\Delta t^2}\sin^2(\frac{\tilde\omega\Delta t}{2}) + \omega^2 I\exp{(i\tilde\omega t)} = 0 $$ which after dividing by \( I\exp{(i\tilde\omega t)} \) results in $$ \frac{4}{\Delta t^2}\sin^2(\frac{\tilde\omega\Delta t}{2}) = \omega^2 $$ Solve for \( \tilde\omega \): $$ \tilde\omega = \pm \frac{2}{\Delta t}\sin^{-1}\left(\frac{\omega\Delta t}{2}\right) $$