From d4aa3ec80fd4dc986b08eecbba0d21dce44beb82 Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:38:58 +0000
Subject: tex

---
 ClassJune26.page | 8 +++-----
 1 file changed, 3 insertions(+), 5 deletions(-)

(limited to 'ClassJune26.page')

diff --git a/ClassJune26.page b/ClassJune26.page
index 5267204..fb826d9 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -189,10 +189,8 @@ curve $\gamma$ through $z$ to a smooth curve $f\circ\gamma$ through
 $f(z)$. What happens to the tangent of $\gamma$ at $z$? Given by
 the derivative $df(z)$, a linear map taking vectors based at $z$
 to vectors based at $f(z)$. If we use rectangular coordinates  
-    $z\mapsto f(z)$
-
-    $x+iy\mapsto u(x,y)+iv(x,y)$  
-
+    $z\mapsto f(z)$  
+    $x+iy \mapsto u(x,y)+iv(x,y)$  
     $\left(\begin{array}{c}
 x\\
 y\end{array}\right)\mapsto\left(\begin{array}{c}
@@ -231,7 +229,7 @@ ie, it looks just like multiplication by the complex number $a+bi$.
 The function $f$ is conformal if its derivative acts like a nonzero
 complex number. Analytically, this condition is given by the following
 differential equations, called the **Cauchy-Riemann equations**:  
-$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$
+$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$  
 A complex function $f=u+iv$ is said to be **holomorphic** if
 $f$ satisfies the CR. We've shown that conformal $\Longrightarrow$
 holomorphic. Holomorphic functions are slightly more general, as the
-- 
cgit v1.2.3