diff --git a/docker-compose.yaml b/docker-compose.yaml
index 0c4a399..48da0c6 100644
--- a/docker-compose.yaml
+++ b/docker-compose.yaml
@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.2
+    image: trance0/notenextra:v1.1.5
     restart: on-failure:5
     ports:
       - 13000:3000
diff --git a/pages/Math416/Math416_L14.md b/pages/Math416/Math416_L14.md
index 1f0186e..fd4a653 100644
--- a/pages/Math416/Math416_L14.md
+++ b/pages/Math416/Math416_L14.md
@@ -143,6 +143,6 @@ We proceed by contradiction. Suppose $z_n\to w\in U$, $f(z_0)=0$, $f(w)=0$. $w$
 
 QED
 
-#### Corollary 7.13.3: Identity principle
+#### Corollary 7.14: Identity principle
 
 If $f,g\in O(U)$, $U$ is a domain and $\exists$ sequence $z_0$ that converges to $w\in U$, such that $f(z_n)=g(z_n)$, then $f\equiv g$ on U$.
\ No newline at end of file
diff --git a/pages/Math416/Math416_L15.md b/pages/Math416/Math416_L15.md
new file mode 100644
index 0000000..e06caeb
--- /dev/null
+++ b/pages/Math416/Math416_L15.md
@@ -0,0 +1,124 @@
+# Math416 Lecture 15
+
+## Review on Cauchy Integrals
+
+The cauchy integral of function $\phi$ (may not be holomorphic) on curve $\Gamma$ (may not be closed) is
+
+$$
+\int_{\Gamma}\frac{\phi(\zeta)}{\zeta-z}d\zeta
+$$
+
+The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain $D$, then the integral of $f$ over any closed curve $\gamma$ in $D$ is 0.
+
+$$
+\int_{\gamma}f(z)dz = 0
+$$
+
+The Cauchy integral formula states that if $f$ is holomorphic on a simply connected domain $D$, then $f$ over any closed curve $\gamma$ in $D$ is
+
+$$
+f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}d\zeta
+$$
+
+## Continue on Cauchy Integrals (Chapter 7)
+
+### Convergence of functions
+
+#### Theorem 7.15 Weierstrass Convergence Theorem
+
+Limit of a sequence of holomorphic functions is holomorphic.
+
+Let $G$ be an open subset of $\mathbb{C}$ and let $\left(f_n\right)_{n\in\mathbb{N}}$ be a sequence of holomorphic functions on $G$ that converges locally uniformly to $f$ on $G$. Then $f$ is holomorphic on $G$.
+
+Proof:
+
+Let $z_0\in G$. There exists a neighborhood $\overline{B_r(z_0)}\subset G$ of $z_0$ such that $\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly on $\overline{B_r(z_0)}$.
+
+Let $C_r=partial B_r(z_0)$.
+
+By Cauchy integral formula, we have
+
+$$
+f_n(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f_n(\zeta)}{\zeta-z_0}d\zeta
+$$
+
+$\forall z\in B_r(z_0)$, we have $\frac{f_n(w)}{w-\zeta}$ converges uniformly on $C_r$.
+
+So $\lim_{n\to\infty}f_n(z_0) = f(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f(w)}{w-z_0}dw$
+
+So $f$ is holomorphic on $G$.
+
+QED
+
+#### Theorem 7.16 Maximum Modulus Principle
+
+If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), then $|f|$ does not attain a local maximum value on $D$.
+
+Proof:
+
+Assume at some point $z_0\in D$, $|f(z_0)|$ is a local maximum. $\exists r>0$ such that $\forall z\in \overline{B_r(z_0)}$, $|f(z)|\leq |f(z_0)|$.
+
+If $f(z_0)=0$, then $f(z)$ is identically 0 on $B_r(z_0)$. (by identity theorem)
+
+Else, we can assume that without loss of generality that $f(z_0)>0$. By mean value theorem,
+
+$$
+f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta
+$$
+
+So $f(z_0) =$
+
+/* TRACK LOST */
+
+#### Corollary  7.16.1 Minimum Modulus Principle
+
+If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), and $f$ is non zero on $D$, then $\frac{1}{f}$ does not attain a local minimum value on $D$.
+
+Proof: 
+
+Let $g(z) = \frac{1}{f(z)}$. $g$ is holomorphic on $D$.
+
+QED
+
+#### Theorem 7.17 Schwarz Lemma
+
+Let $f$ be a holomorphic map of the open unit disk $D$ into itself, and $f(0)=0$. Then $\forall z\in D$, $|f(z)|\leq |z|$ and $|f'(0)|\leq 1$.
+
+And the equality holds if and only if $f$ is a rotation, that is, $f(z)=e^{i\theta}z$ for some $\theta\in\mathbb{R}$.
+
+Proof:
+
+Let
+
+$$
+g(z) = \begin{cases}
+\frac{f(z)}{z} & z\neq 0 \\
+f'(0) & z=0
+\end{cases}
+$$
+
+We claim that $g$ is holomorphic on $D$.
+
+For $z\neq 0$, $g(z)$ is holomorphic since $f$ is holomorphic on $D$.
+
+For $z=0$, $g(z)$ is holomorphic since $f$'s power series expansion has $c_0=f(0)=0$. $g'(0)=f'(0)=c_1+c_2z+c_3z^2+\cdots$.
+
+So $g$ is (analytic) thus holomorphic on $D$.
+
+On the boundary of $D$, $|g(z)|\leq\frac{1}{r} \cdot 1$. By maximum modulus principle, $|g(z)|\leq 1$ on $D$.
+
+So $|f(z)|\leq |z|$ on $D$.
+
+And $|f'(0)|\leq 1$.
+
+QED
+
+#### Schwarz-Pick Lemma
+
+Let $f$ be a holomorphic map of the open unit disk $D$ into itself, then for any $z,w\in D$,
+
+$$
+\frac{|f(z)-f(w)|}{|1-\overline{f(w)}f(z)|}\leq\frac{|z-w|}{|1-\overline{w}z|}=\rho(z,w)
+$$
+
+Prove after spring break.
diff --git a/pages/Math416/_meta.js b/pages/Math416/_meta.js
index 683a868..4b4d3d8 100644
--- a/pages/Math416/_meta.js
+++ b/pages/Math416/_meta.js
@@ -18,4 +18,5 @@ export default {
     Math416_L12: "Complex Variables (Lecture 12)",
     Math416_L13: "Complex Variables (Lecture 13)",
     Math416_L14: "Complex Variables (Lecture 14)",
+    Math416_L15: "Complex Variables (Lecture 15)",
 }