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authorjoshuab <>2010-06-29 15:35:11 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-29 15:35:11 +0000
commite46ae05267fef706fb575f773781b939dea07763 (patch)
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parent45a4bb100ca5aea14554f265ffc09e981d79a36c (diff)
downloadafterklein-wiki-e46ae05267fef706fb575f773781b939dea07763.tar.gz
afterklein-wiki-e46ae05267fef706fb575f773781b939dea07763.zip
tex
-rw-r--r--ClassJune26.page72
1 files changed, 35 insertions, 37 deletions
diff --git a/ClassJune26.page b/ClassJune26.page
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--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -178,35 +178,35 @@ between them (actually, the angles between their tangent vectors at
the intersection point) stays the same.
-\subsection*{Conformality, Holomorphicity}
+## Conformality, Holomorphicity
-A map $\mathbb{C}\rightarrow\mathbb{C}$ is \textbf{conformal} if
-it preserves oriented angles. We (meaning me \emph{and} you) will
+A map $\mathbb{C}\rightarrow\mathbb{C}$ is **conformal** if
+it preserves oriented angles. We (meaning me *and* you) will
show that polynomials, exponentials, etc are conformal (almost everywhere).
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\[
-z\mapsto f(z)\]
-\[
-x+iy\mapsto u(x,y)+iv(x,y)\]
-\[
-\left(\begin{array}{c}
+to vectors based at $f(z)$. If we use rectangular coordinates
+$z\mapsto f(z)$
+
+$x+iy\mapsto u(x,y)+iv(x,y)$
+
+$\left(\begin{array}{c}
x\\
y\end{array}\right)\mapsto\left(\begin{array}{c}
u(x,y)\\
-v(x,y)\end{array}\right)\]
-then the derivative is \[
-df=\left(\begin{array}{cc}
+v(x,y)\end{array}\right)$
+then the derivative is
+$df=\left(\begin{array}{cc}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
-\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).\]
+\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
angle, and $df$ is linear, its image must bisect too, that is, the
-little square must become a little \emph{square}. Numerically, this
+little square must become a little *square*. Numerically, this
means that $\left(\begin{array}{c}
\frac{\partial u}{\partial y}\\
\frac{\partial u}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c}
@@ -216,33 +216,33 @@ means that $\left(\begin{array}{c}
a\\
b\end{array}\right)=\left(\begin{array}{c}
\frac{\partial u}{\partial x}\\
-\frac{\partial v}{\partial x}\end{array}\right)\]
-, then \[
+\frac{\partial v}{\partial x}\end{array}\right)$
+, then $
\left(\begin{array}{c}
\frac{\partial u}{\partial y}\\
\frac{\partial u}{\partial y}\end{array}\right)=\left(\begin{array}{c}
-b\\
-a\end{array}\right).\]
-Put another way, the linear transformation $df(z)$ has the form \[
-\left(\begin{array}{cc}
+a\end{array}\right).$
+Put another way, the linear transformation $df(z)$ has the form
+$\left(\begin{array}{cc}
a & -b\\
-b & a\end{array}\right),\]
+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 \textbf{is} a nonzero
+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:\[
-\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.\]
-A complex function $f=u+iv$ is said to be \textbf{holomorphic} if
+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 $\Longrightarrow$
holomorphic. Holomorphic functions are slightly more general, as the
Jacobian can vanish; $df$ can be the complex number $0$. This is
what happens at the origin for $f=z^{n}$. We can now check that $e^{z}$
is conformal, and that $z^{n}$ is holomorphic, so conformal away
from $0$.
-\begin{example}
-$z^{2}$ is holomorphic. $z^{2}=(x+iy)^{2}=x^{2}-y^{2}+i(2xy).$ So
+**Example:** $z^{2}$ is holomorphic. $z^{2}=(x+iy)^{2}=x^{2}-y^{2}+i(2xy).$ So
$u=x^{2}-y^{2}$ and $v=2xy$.
-\end{example}
+
+
This class is, essentially, the study of holomorphic functions, which
includes all angle-preserving transformations of the plane. The lesson
here is that complex analysis is geometry; complex numbers are dilations
@@ -254,19 +254,17 @@ Why study complex numbers, holomorphic functions, and conformal maps?
Because of the beautiful and deep structure to which they lead.
Mathematically,
-\begin{itemize}
-\item fundamental theorem of algebra
-\item difficult integrals ($\int_{-\infty}^{\infty}\frac{1}{(x^{2}+1)^{2}}dx=\frac{\pi}{2}$)
-\item theory of surfaces
-\item hyperbolic geometry
-\end{itemize}
+
+- fundamental theorem of algebra
+- difficult integrals ($\int_{-\infty}^{\infty}\frac{1}{(x^{2}+1)^{2}}dx=\frac{\pi}{2}$)
+- theory of surfaces
+- hyperbolic geometry
Applicationally,
-\begin{itemize}
-\item conformal maps preserve maxwell's equations (in 2D) and incompressible,
+- conformal maps preserve maxwell's equations (in 2D) and incompressible,
irrotational fluid flow.
-\item fourier transform, and its generalization, the laplace transform for
+- fourier transform, and its generalization, the laplace transform for
understanding dynamical signals and systems.
-\end{itemize}
+
We'll do all that and more, in the next six weeks. Hold on, dig in,
and enjoy!