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authorjoshuab <>2010-06-29 15:36:27 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-29 15:36:27 +0000
commit96374a3d0bdc16652390ff32925f67fadce39c75 (patch)
treeac64bb627ed103d86c9b5ef915f1bff868e0f282
parente46ae05267fef706fb575f773781b939dea07763 (diff)
downloadafterklein-wiki-96374a3d0bdc16652390ff32925f67fadce39c75.tar.gz
afterklein-wiki-96374a3d0bdc16652390ff32925f67fadce39c75.zip
tex
-rw-r--r--ClassJune26.page8
1 files changed, 4 insertions, 4 deletions
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 $\Longrightarrow$