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authorjoshuab <>2010-06-29 15:38:58 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-29 15:38:58 +0000
commitd4aa3ec80fd4dc986b08eecbba0d21dce44beb82 (patch)
treea649419892ba890ff4c83c178a05899bda087d3a /ClassJune26.page
parentf03b9e8d339eb235f03498b51b9e40a74408d3ba (diff)
downloadafterklein-wiki-d4aa3ec80fd4dc986b08eecbba0d21dce44beb82.tar.gz
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tex
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@@ -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