From 96374a3d0bdc16652390ff32925f67fadce39c75 Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:36:27 +0000
Subject: tex

---
 ClassJune26.page | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

(limited to 'ClassJune26.page')

diff --git a/ClassJune26.page b/ClassJune26.page
index 27b30b9..06d0305 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -188,10 +188,10 @@ Consider a smooth map $f$ from the plane to itself; it takes a smooth
 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
+to vectors based at $f(z)$. If we use rectangular coordinates  
 $z\mapsto f(z)$
 
-$x+iy\mapsto u(x,y)+iv(x,y)$
+$x+iy\mapsto u(x,y)+iv(x,y)$  
 
 $\left(\begin{array}{c}
 x\\
@@ -201,7 +201,7 @@ v(x,y)\end{array}\right)$
 then the derivative is 
 $df=\left(\begin{array}{cc}
 \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
-\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$
+\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$  
 If $f$ is conformal, then this matrix had better take the (orthogonal)
 standard basis to orthogonal vectors; the little square becomes a
 little rectangle. Since the diagonal of the square bisects the right
@@ -230,7 +230,7 @@ b & a\end{array}\right),$
 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**:
+differential equations, called the **Cauchy-Riemann equations**:  
 $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$
 A complex function $f=u+iv$ is said to be **holomorphic** if
 $f$ satisfies the CR. We've shown that conformal $\Longrightarrow$
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