diff --git a/pages/Math416/Math416_L19.md b/pages/Math416/Math416_L19.md
index 45b4936..fa5ad16 100644
--- a/pages/Math416/Math416_L19.md
+++ b/pages/Math416/Math416_L19.md
@@ -72,3 +72,19 @@ the singularity at $z=3$ is a simple pole with order 1 $f(z)=\frac{z+1}{z-3}=\fr
 
 there are three poles at $z=1,5,8$, the order of the poles are 2, 6, 1 respectively.
 
+If $f$ has a pole of order $m$ at $z_0$,
+
+$$
+f(z) = \sum_{n=-m}^{\infty} a_n (z-z_0)^n
+$$
+
+then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$. Value of holomorphic extension of $(z-z_0)^m f(z)$ at $z_0$ is $a_{-m}$.
+
+- $f$ is given by a power series in $A(z_0;0,R)$
+- $f=(z-z_0)^{-m} g(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$, $\frac{1}{f}=(z-z_0)^m \frac{1}{g(z)}$ has a pole of order $m$ at $z_0$. So $f$ has a pole of order $m$ at $z_0$ if and only if $\frac{1}{f}$ has a zero of order $m$ at $z_0$
+
+$e^{1/z}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ has an essential singularity at $z=0$ since it has infinitely many terms with negative powers of $z$.
+
+
+
+