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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! |