From e46ae05267fef706fb575f773781b939dea07763 Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:35:11 +0000
Subject: tex

---
 ClassJune26.page | 72 +++++++++++++++++++++++++++-----------------------------
 1 file changed, 35 insertions(+), 37 deletions(-)

(limited to 'ClassJune26.page')

diff --git a/ClassJune26.page b/ClassJune26.page
index 15c603a..27b30b9 100644
--- 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!
 
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