From f03b9e8d339eb235f03498b51b9e40a74408d3ba Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:37:32 +0000
Subject: tex

---
 ClassJune26.page | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

diff --git a/ClassJune26.page b/ClassJune26.page
index 06d0305..5267204 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -189,17 +189,17 @@ 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  
-$z\mapsto f(z)$
+    $z\mapsto f(z)$
 
-$x+iy\mapsto u(x,y)+iv(x,y)$  
+    $x+iy\mapsto u(x,y)+iv(x,y)$  
 
-$\left(\begin{array}{c}
+    $\left(\begin{array}{c}
 x\\
 y\end{array}\right)\mapsto\left(\begin{array}{c}
 u(x,y)\\
-v(x,y)\end{array}\right)$
-then the derivative is 
-$df=\left(\begin{array}{cc}
+v(x,y)\end{array}\right)$  
+then the derivative is  
+    $df=\left(\begin{array}{cc}
 \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
 \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$  
 If $f$ is conformal, then this matrix had better take the (orthogonal)
-- 
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