Would anyone know how to calculate the values of the complex variable $s$ such as
$(\frac{s}{\omega_b}+1)^n(\frac{s}{\omega_h}+1)^{1-n}=C_0$
with
$\omega_b \in \mathbb{R}$, $\omega_h \in \mathbb{R}$, $\omega_b <\omega_h$
$n \in \mathbb{R}$ and $0<n<1$
$C_0 \in \mathbb{C}$
Thank you very much for your kind help.
If $C_0=0$ then $s\in\{-\omega_b, -\omega_h\}.$
Else the issue equation can be transformed:
$$(s+\omega_b)^n(s+\omega_h)^{1-n} = C_1, \quad C_1 = C_0\omega_b^n\omega_h^{1-n},\tag1$$
$$\left(\dfrac{s+\omega_b}{C_1}\right)^n\left(\dfrac{s+\omega_h}{C_1}\right)^{1-n} = 1.$$
The equation can be presented in the form of
$$\left(\dfrac1t+a\right)^nt^{n-1} = 1, \quad t = \dfrac {C_1}{s+\omega_h},\quad a = \dfrac{\omega_b-\omega_h}{C_1},\tag2$$
$$t^{1/n}=1+at.\tag3$$
Let
$$a = re^{i\alpha},\quad t =\rho e^{i\varphi}\tag4,$$
then
$$\left(\rho e^{i\varphi}\right)^{1/n} =1+r\rho e^{\varphi+\alpha}.\tag5$$
Taking in account the formulas
$$\tan\dfrac y2 = \dfrac{\sin y}{1+\cos y},\quad u+iv = \sqrt{u^2+v^2}\exp\left(2i\arctan{v\over\sqrt{u^2+v^2}+u}\right),\tag6$$
one can obtain the transcendent system
$$\begin{cases}
\rho^{2/n} = 1+2r\rho\cos(\varphi+\alpha)+r^2\rho^2,\\[4pt]
\tan\dfrac{\varphi+2\pi k}{2n} = \dfrac{r\rho\sin(\varphi+\alpha)}{1+r\rho\cos(\varphi+\alpha)+\rho^{1/n}}\\[4pt]
k\in\mathbb Z.\\[4pt]
\end{cases}\tag7$$
If the exponent $\dfrac1n$ is rational, then the solution set is finite.
Else continued fractions allow to change the exponent to the rational form and to obtain some solutions approximately.
In the general case, the formulas $(7)$ leads to an infinite number of solutions.
HOW TO SOLVE $(7)?$
The system $(7)$ can be presented in the form of $$\begin{cases} 2+2r\rho\cos(\varphi+\alpha) + r^2\rho^2 = 1+ \rho^{2/n}\\[4pt] \tan\dfrac{\varphi+2\pi k}{2n} = \pm\dfrac{\sqrt{(2r\rho - 2r\rho\cos(\varphi+\alpha))(2r\rho + 2r\rho\cos(\varphi+\alpha))}}{2+2r\rho\cos(\varphi+\alpha)+2\rho^{1/n}}\\[4pt] k\in\mathbb Z.\\[4pt] \end{cases}\tag8$$ Some expressions of $(8)$ can be transformed: $$2r\rho - 2r\rho\cos(\varphi+\alpha) = 2r\rho + 2 + r^2\rho^2 - \left(1+\rho^{2/n}\right) = (1+r\rho)^2 - \rho^{2/n},$$ $$2r\rho - 2r\rho\cos(\varphi+\alpha) = \left(1+r\rho-\rho^{1/n}\right)\left(1+r\rho+\rho^{1/n}\right),\tag{9a}$$ $$2r\rho + 2r\rho\cos(\varphi+\alpha) = 2r\rho - 2 - r^2\rho^2 + (1+\rho^{2/n}) = \rho^{2/n} - (1-r\rho)^2,$$ $$2r\rho + 2r\rho\cos(\varphi+\alpha) = \left(-1 + r\rho +\rho^{1/n}\right)\left(1 - r\rho + \rho^{1/n}\right),\tag{9b}$$ $$2 + 2r\rho\cos(\varphi+\alpha) + 2\rho^{1/n} = 2\rho^{1/n} - r^2\rho^2 + \left(1+\rho^{2/n}\right) = \left(1+\rho^{1/n}\right)^2 - r^2\rho^2,$$ $$2 + 2r\rho\cos(\varphi+\alpha) + 2\rho^{1/n} = \left(1-r\rho +\rho^{1/n}\right)\left(1+r\rho +\rho^{1/n}\right),\tag{9c}$$ $$1-\tan^2\dfrac{\varphi+2\pi k}{2n} = 1 - \dfrac{r^2\rho^2-\left(1-\rho^{1/n}\right)^2}{\left(1+\rho^{1/n}\right)^2-r^2\rho^2} = \dfrac{2\left(1 - r^2\rho^2 +\rho^{2/n}\right)}{\left(1+\rho^{1/n}\right)^2-r^2\rho^2},\tag{9d}$$ $$1+\tan^2\dfrac{\varphi+2\pi k}{2n} = 1 + \dfrac{r^2\rho^2-\left(1-\rho^{1/n}\right)^2}{\left(1+\rho^{1/n}\right)^2-r^2\rho^2} = \dfrac{4\rho^{1/n}}{\left(1+\rho^{1/n}\right)^2-r^2\rho^2},\tag{9e}$$ $$\cos\dfrac{\varphi+2\pi k}{n} = \dfrac{1-\tan^2\dfrac{\varphi+2\pi k}{2n}}{1+\tan^2\dfrac{\varphi+2\pi k}{2n}} = \dfrac{1 - r^2\rho^2 +\rho^{2/n}}{2\rho^{1/n}}\tag{10}$$ and the system in the form of $$\begin{cases} \cos(\varphi+\alpha) = \dfrac{-1 - r^2\rho^2 + \rho^{2/n}}{2r\rho}\\[4pt] \cos\dfrac{\varphi + 2\pi k}{n} = \dfrac{1 - r^2\rho^2 +\rho^{2/n}}{2\rho^{1/n}}\\[4pt] k \in\mathbb Z, \end{cases}\tag{11}$$ or $$\begin{cases} \varphi = \pi m + (-1)^m \arccos\dfrac{-1 - r^2\rho^2 + \rho^{2/n}}{2r\rho}-\alpha\\[4pt] \pi m + (-1)^m\left(\arccos\dfrac{-1 - r^2\rho^2 + \rho^{2/n}}{2r\rho}\right)-\alpha = n\left(\pi j+(-1)^j\arccos\dfrac{1 - r^2\rho^2 +\rho^{2/n}}{2\rho^{1/n}}\right)\\[4pt] jn\in\{m-1, m, m+1\}\\[4pt] (m, j) \in \mathbb Z^2, \end{cases}\tag{12}$$ looks more usable.
