A function defined as $f(x,y)=0$ is said to be implicitly defined by the equation. The concept of inverting $f$ is closely related to the idea of solving the equation $y=f(x)$ for x as a function of y. Some form of localization is usually necessary to obtain an implicit function. Consider for instance the problem $S=\{(x,y)|x-y^2=0\}$. The graph of S is not a function by itself, but it can have a neighborhood around the points $(x,y)$ where it becomes a function by itself. The linearization of $f$ at a particular point is used in the implicit function theorem for $f(x,y)=0$, which demands that $f$ be invertible by requesting that its Jacobian be nonsingular. For the problem I want to solve, I consider the continuously differentiable function: \begin{equation} f(x,y)=\frac{\log \left(c_3^2 y^2+c_3^2 y-c_1\right)}{2 c_3^2}-\frac{\arctan\left(\frac{c_3 (2 y+1)}{\sqrt{-c_3^2-4 c_1}}\right)}{c_3 \sqrt{-c_3^2-4 c_1}}-\frac{i x }{c_3}-c_2 \end{equation} where $y=y(x)$, $c_1$, $c_2$, and $c_3$, are constant values. As suggested by Tyma Gaidash, this equation can be rearranged as follows: \begin{equation} e^{2 \sqrt{c_{3}^2+4 c_1} (i x +c_2 c_{3})}=\frac{\left(c_{3}^2 y (y+1)-c_1\right){}^{\frac{\sqrt{c_{3}^2+4 c_1}}{c_{3}}} \left(2 c_{3} y+\sqrt{c_{3}^2+4 c_1}+c_{3}\right)}{\sqrt{c_{3}^2+4 c_1}-c_{3}-2 y c_{3}} \end{equation} I would like to explicit $y$ as a function of $x$. For $y \neq 0$, the Jacobian is non singular. Then the solution mapping $\{y|f(x,y)=0\}$ has a single-valued localization around $\bar{x}$ for $\bar{y}$ which is continuously differentiable in a neighborhood of $\bar{x}$. The above equation can be rearranged as \begin{equation} x = -\frac{i \log \left(y^2 c_3^2+y c_3^2-c_1\right)}{2 c_3}-\frac{i \text{ arctanh} \left(\frac{(2 y+1) c_3}{\sqrt{c_3^2+4 c_1}}\right)}{\sqrt{c_3^2+4 c_1}}+i c_2 c_3 \end{equation} Now \begin{equation} x = y^{-1}\big(y(x)\big) \end{equation} Is it possible to use the inverse function theorem to compute $y(x)$ analytically? If so, what conditions must be fulfilled for this theorem to apply?
Here is my idea.
Consider a function $g(y)$ of a complex variable $y$, holomorphic in a neighborhood of $y = 0$. Suppose $f(0)=0$ and $g^{'}(0) \neq 0$, so by the Inverse Function Theorem, $g(y)$ is one-to-one inside a small circle $C$ defined by $|y| = \epsilon$, and there is a unique inverse function $g(x)$ defined near $x = 0$ with $f(g(y)) = y$. Using the use of the Cauchy Residue Theorem and a new set of variables, $y = f(\zeta)$, $\zeta=g(y)$, and $d\zeta = g^{'}(y) dy$, we can obtain: \begin{equation} f(x)=\frac{1}{2 \pi i} \oint_{f(C)} \frac{f(\zeta)}{\zeta -x} d\zeta=\frac{1}{2 \pi i} \oint_{f(C)} \frac{g^{'}(y) y}{g(y)-x} dy \end{equation} So I can write \begin{equation} f(x)=\frac{1}{2 \pi i} \oint_{f(C)} \frac{\frac{\partial f(x,y)}{\partial y} y}{f(x,y)} dy \end{equation} where \begin{equation} \frac{\partial f(x,y)}{\partial y} = \frac{y}{c_{3}^2 y (y+1)-c_1} \end{equation} The integral that results by substituting $f(x,y)$ and the partial derivative $ \frac{\partial f(x,y)}{\partial y}$ into the expression above is rather complex and I do not know how to calculate it. Unfortunately, I am rather rusty as far as calculating residuals is concerned (I have some memories from when I was a student). Any suggestions are welcome.