$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$
In this Chapter we consider linear ODEs with analytic coefficients in the complex domain: \begin{equation} a_n (z)u^{(n)}(z)+ a_{n-1} (z)u^{(n-1)}(z)+\ldots+a_1 (z)u'(z)+a_0(z)u(z)=0 \label{eq-3.1.1} \end{equation} where $a_k(z)$ are analytic functions. After division by $a_n(z)$ we get the similar equation albeit with the leading coefficient equal to $1$: \begin{equation} u^{(n)}(z)+ p_{n-1} (z)u^{(n-1)}(z)+\ldots+p_1 (z)u'(z)+p_0(z)u(z)=0 \label{eq-3.1.2} \end{equation} where $p_k(z)$ are also analytic functions.
Remark 1.
Definition 1. Point $z_0\ne\infty $ is an ordinary point of equation (\ref{eq-3.1.2}) if $p_k(z)$ are analytic at $z_0$ for all $k=0,1,\ldots,n-1$.
We are going to prove in Section 3.2 that then solution $u(z)$ is analytic at $z_0$ and moreover the radius of convergence of its Taylor expansion is at least the distance to the nearest singularity.
Example 1. Consider toy-model: Constant coefficient equations. Check ODE textbook.
Definition 2. Point $z_0\ne\infty $ (which is not an ordinary point in the sense of Definition 1) is a regular singular point of equation (\ref{eq-3.1.2}) if $p_k(z)(z-z_0)^{n-k}$ are analytic at $z_0$ for all $k=0,1,\ldots,n-1$.
In other words: $z_0$ could be a pole of degree not exceeding $n-k$.
Example 2. Consider toy-model: Euler equation \begin{equation} z^n u^{(n)}(z)+ z^{n-1} q_{n-1} u^{(n-1)}(z)+\ldots+zq_1 u'(z)+q_0 u(z)=0 \label{eq-3.1.3} \end{equation} with constant $q_{n-1},\ldots, q_0$ has solutions of the form $z^\alpha$ where $\alpha$ is an incidical exponent--a root of the incidical equation \begin{equation} \alpha(\alpha-1)\cdots (\alpha -n+1) + q_{n-1}\alpha(\alpha-1)\cdots (\alpha -n+2)+\ldots +q_1 \alpha +q_0=0; \label{eq-3.1.4} \end{equation} if the multiplicity of this root is $r\ge 2$ then there are also solutions $z^\alpha (\ln z)^j$, $j=1,\ldots,r-1$.
The general solution is a linear combination of solutions described above.
Remark 2.
Remark 3. In the general case assuming that $\alpha+1$,\ \alpha+2,\ \ldots$ are not incidical exponents^{1} we get solutions of the form \begin{equation} \sum|_{0\le k\le j} A_k(z)(z-z_0)^\alpha (\log (z-z_0)^k,\qquad \text{with }\ A_j(z_0)\ne 0 \label{eq-3.1.5} \end{equation} where $A_k(z)$ are analytic functions at $z_0$. Here $j=0,\ldots,r-1$.
The Taylor series of $A_k$, expanded about $z_0$, has a radius of convergence at least as large as the distance to the next nearest singularity.
Definition 3. Point $z_0\ne\infty $ is an irregular singular point of if it is neither an ordinary point nor a regular singular point.
Remark 4. There is no comprehensive theory for irregular singular points. What can be said is the following:
Recall that in the theory of Complex Variables $z=\infty\in \bC^*$ (extended complex plane) is treated as an ordinary point $\zeta=0$ after substitution $z=1/\zeta$.
Definition 4. The point $z_0 = \infty$ is:
$\Leftarrow$ $\Uparrow$ $\Rightarrow$
Without this assumption we get more complicated decomposition.↩