The equation is $$i\sqrt{\omega^2-p^2}=\frac{A}{1+\frac{i\gamma}{\omega}}$$ where $A,p,\gamma\in \mathbb{R}$, and $A,p> 0$ and $\omega$ is complex, the solutions of which we seek.
I want to show that $\omega$ cannot have solutions in the upper half complex plane (except for solutions on the real or imaginary axis).
First, consider the top-right quadrant in $\omega$ plane. It is mapped by the RHS of the equation onto a quarter of a circle centred at origin restricted to the bottom-right quadrant and radius $A$. LHS would map the top-right quadrant to the top-left quadrant. Since the resulting mapped areas of either side of the equation do not coincide no solutions exist in the top-right quadrant.
Repeating the same steps for the top-left quadrant unfortunately does not eliminate it for a region that may contain solutions - if the square root is taken to "turn" the region away from the negative real axis.
Anyone with ideas on how to eliminate the top-left quadrant of the $\omega$ complex plane from possible region of solutions?