Given is the implicit equation
$$ c(\zeta) = \frac{\alpha}{1 - \zeta q}\left(2 + \frac{1}{2}c(\zeta)\left[1- \cot\left(\frac{1}{2}c(\zeta)\right)\right]\right), $$ where $\cot$ is the cotangent function and $\alpha, q>0$.
I would like to show that $c$ as a function of $\zeta$ is (complex) analytic around the origin and I would like to bound the derivatives of this $c$ (likely using Cauchy's integral formula).
The implicit function theorem gives us the condition that $$ 1 \neq \frac{\alpha}{2(1 - \zeta q)}\left[ 1 - \cot(\frac{1}{2}c(\zeta)) + \frac{1}{2}c(\zeta)\frac{1}{\sin^2(\frac{1}{2}c(\zeta))}\right]. $$ I have tried to play around with this equation, but cannot find any solutions/estimates on where $c(\zeta)$ is complex analytic or find any bounds on the derivatives of $c(\zeta)$.
Origin of implicit equation
I found the implicit equation for $c(\zeta)$ from a generating function $c(\zeta) = \sum_{j \geq 1} \zeta^{j-1} \chi_j$, where $\chi_j$ the coefficients and know that it satisfies $$ c(\zeta) = \frac{\alpha}{1- \zeta q}\left(1 + \sum_{i \geq 1}\frac{|\mathcal{B}_i|}{i!} c(\zeta)^i \right)\\ = \frac{\alpha}{1- \zeta q}\left(1 + \sum_{i \geq 1}\frac{|\mathcal{B}_i|}{i!} \sum_{k_1 + \dots + k_i \geq i}\zeta^{k_1}\chi_{k_1} \dots \zeta^{k_i} \chi_{k_i} \right) $$ where $\mathcal{B}_i$ are the Bernoulli numbers (with $\mathcal{B}_1 = -\frac{1}{2}$).
Define the function
$$ f(x) := 2 \!+\! \frac{x}2 \!-\! \frac{x}2\cot\bigg(\frac{x}2\bigg) = 1 \!+\! \frac{x}2 \!+\! \sum_{n=1}^\infty B_{2n}\,\frac{(-x^2)^n}{(2n)!}. $$
Suppose the function $\,c(\zeta)\,$ is required to satisfy the equation
$$ c(\zeta) = \frac{\alpha}{1-\zeta q} f(c(\zeta)). \tag1 $$
Substitute $\,\zeta=0\,$ into the equation to get $\, c(0) = \alpha\,f(c(0)). \,$ Thus, for any $\,\alpha>0,\,$ the value of $\,c(0)\,$ must be a fixed point of the function $\, g(x) := \alpha\,f(x). \,$ For $\,0<\alpha< k:=.88650312\,$ this fixed point $\,c(0)>\alpha\,$ exists and is the limit of the iteration of $\,g(x)\,$ starting from $\,x=\alpha.\,$ For $\,\alpha>k\,$ the iterations are chaotic unless $\,f(\alpha)=1\,$ and $\,c(0)=\alpha.$ In this case, there is an infinite discrete set of solutions to $\,f(\alpha)=1.$
In the case that $\,\alpha\,$ is any solution to $\,f(\alpha)=1\,$ then define the constant $$\beta := 2-2\alpha-\alpha^2 \qquad \text{ and } \qquad z := \zeta q. $$ Then the power series for $\,c(\zeta)\,$ is $$ c(\zeta) = c_0 + c_1 z + c_2 z^2 + c_3 z^3 +\cdots \tag2 $$ where $$ c_0 = \alpha, \; c_1 = 2\alpha/\beta, \; c_2 = 2\alpha(4-6\alpha-4\alpha^2-\alpha^3)/\beta^3, \; \dots $$ Differentiate equation $(1)$ w.r.t. $\,\zeta\,$ to get $$ c'(\zeta) = \frac{\alpha q}{(1-z)^2} f(c(\zeta)) + \frac{\alpha}{1-\zeta q} f'(c(\zeta))c'(\zeta) \tag3 $$ which can be used to solve for $\,c'(\zeta)\,$ in terms of $\,c(\zeta).$ This can then be used to determine the coefficients of the power series in equation $(2)$.
In the case $\,0<\alpha<k\,$ the power series coefficients are not so simple, but equation $(3)$ can still be used to compute the coefficients given $\,c_0 := c(0).\,$