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authorjoshuab <>2010-06-29 15:37:32 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-29 15:37:32 +0000
commitf03b9e8d339eb235f03498b51b9e40a74408d3ba (patch)
treeb6986f7e586dcc62609ac8e45178f397ae04d6f8
parent96374a3d0bdc16652390ff32925f67fadce39c75 (diff)
downloadafterklein-wiki-f03b9e8d339eb235f03498b51b9e40a74408d3ba.tar.gz
afterklein-wiki-f03b9e8d339eb235f03498b51b9e40a74408d3ba.zip
tex
-rw-r--r--ClassJune26.page12
1 files 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)