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@@ -47,3 +47,257 @@ addition, and multiplication given by dilation and rotation. Real
 numbers form the horizontal axis, and imaginary numbers form the vertical
 axis. Why does it make sense to call these arrows {}``numbers''?
 Because they satisfy all the basic rules of arithmetic:
+
+\begin{itemize}
+\item Addition is commutative.
+
+\begin{itemize}
+\item the two orders form a paralellogram, with the sum on the diagonal.
+$z+w=w+z$
+\end{itemize}
+\item Addition is associative.
+
+\begin{itemize}
+\item $v+(z+w)=(v+z)+w$ since both compute the straight-line path from
+$0$ to the end of the $\overset{w}{\rightarrow}\overset{z}{\rightarrow}\overset{v}{\rightarrow}$
+trail.
+\end{itemize}
+\item Multiplication is commutative.
+
+\begin{itemize}
+\item Multiplying $(r_{1},\theta_{1})$ by $(r_{2},\theta_{2})$ rescales
+by $r_{2}$ and rotates by $\theta_{2}$, giving $(r_{1}r_{2},\theta_{1}+\theta_{2})$,
+which is manifestly symmetric. Put another way, multiplication of
+complex numbers multiplies magnitudes and adds angles/arguments.
+\end{itemize}
+\item Multiplication is associative.
+
+\begin{itemize}
+\item exercize
+\end{itemize}
+\item Multiplication distributes over addition.
+
+\begin{itemize}
+\item Take any two complex numbers $v,w$ and add them, making a parallelogram.
+To find $z(v+w)$, rotate and scale our sum. But that's the same as
+rotating and scaling each of $v$ and $w$ individually, to $zv$
+and $zw$, then adding them together: $zv+zw=z(v+w)$. The distributive
+law for complex numbers is the fact that a rotated, rescaled paralellogram
+is still a parallelogram, or, more basically, that dilation and rotation
+of the plane preserve angles.
+\end{itemize}
+\end{itemize}
+
+\subsection*{Coordinates}
+
+We can't be constantly drawing arrows in the plane; to do anything
+quantitative, we need to introduce coordinates. Rectangular coordinates
+are easy, each vector has an $x$ and $y$-coordinate, given by its
+projections onto the real and imaginary axis. If we use column vector
+notation, we have \begin{eqnarray*}
+z & = & \left(\begin{array}{c}
+a\\
+b\end{array}\right)\\
+ & = & a\left(\begin{array}{c}
+1\\
+0\end{array}\right)+b\left(\begin{array}{c}
+0\\
+1\end{array}\right)\\
+ & = & a\cdot1+b\cdot i\\
+ & = & a+bi\end{eqnarray*}
+Put another way, we are using $1$ and $i$ as basis vectors. For
+example, \[
+\left(\begin{array}{c}
+3\\
+2\end{array}\right)=3(\rightarrow)+2(\uparrow)=3+2i\]
+and\[
+\overset{2}{\nwarrow}=\left(\begin{array}{c}
+-\sqrt{2}\\
+\sqrt{2}\end{array}\right)=-\sqrt{2}(\rightarrow)+\sqrt{2}(\uparrow)=-\sqrt{2}+\sqrt{2}i\]
+
+
+Polar coordinates represent each complex number by its magnitude and
+argument, its radius and angle (in radians). So \[
+i=\uparrow=(1,\pi/2)\]
+and\[
+\overset{2}{\nwarrow}=(2,3\pi/4)\]
+
+
+The familiar angle addition formula from trigonometry now follows
+from the distributive law applied to unit vectors.\[
+\cos(\theta+\phi)+i\sin(\theta+\phi)=(\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi+i(\cos\theta\sin\phi+\sin\theta\cos\phi)\]
+
+
+In the workshop session, you'll use a similar trick to compute triple
+and quadruple angle formulas, for $\sin$ and $\cos$, deriving De
+Moivre's theorem.
+
+
+\subsection*{Functions}
+
+The most basic transformations of the complex plane are given by multipling
+by some fixed complex number $\rho=r(\cos\theta+i\sin\theta)=a+bi$,
+which, as we saw before, is a dilation by $r$ plus a rotation by
+$\theta$. The distributive law says precisely that this is a linear
+transformation of the plane, viewed as a two-dimensional vector space.
+And linear maps are given in rectangular coordinates by matrices:\[
+\left(\begin{array}{c}
+x\\
+y\end{array}\right)\rightsquigarrow\left(\begin{array}{cc}
+a & c\\
+b & d\end{array}\right)\left(\begin{array}{c}
+x\\
+y\end{array}\right)=\left(\begin{array}{c}
+ax+cy\\
+bx+dy\end{array}\right).\]
+What is the matrix corresponding to multiplication by $\rho$? Well,
+the first column is the image of $\left(\begin{array}{c}
+1\\
+0\end{array}\right)$, ie, of $1$, which just $v=\left(\begin{array}{c}
+a\\
+b\end{array}\right)$ itself, and the second column is the image of $i=\left(\begin{array}{c}
+0\\
+1\end{array}\right)$, which is $\rho$ rotated by $90^{\circ}$, which has coordinates
+$\left(\begin{array}{c}
+-b\\
+a\end{array}\right)$. So complex number $a+bi$ is identified with the matrix \[
+\left(\begin{array}{cc}
+a & -b\\
+b & a\end{array}\right).\]
+As a sanity check, note that $i$ corresponds to $\left(\begin{array}{cc}
+0 & -1\\
+1 & 0\end{array}\right)$, which squares to $\left(\begin{array}{cc}
+-1 & 0\\
+0 & -1\end{array}\right)\sim-1$.
+
+Another natural transformation of the complex plane is given by squaring,
+sending $z\rightsquigarrow z^{2}$. This squares the length of each
+vector, and doubles its angle. PICTURE. What does $z\rightsquigarrow z^{n}$
+look like? PICTURE.
+
+Another key example is the exponential map. Recall the power series
+for $e^{z}$:
+
+\[
+e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\frac{z^{5}}{5!}+\cdots\]
+We can raise complex numbers to powers, divide by the real denominators,
+and add them up just fine, so we can exponentiate complex values of
+$z$. We know what happens to real values, what happens to pure imaginary
+ones? Let $y\in\mathbb{R}$. Then \begin{eqnarray*}
+e^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
+ & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
+ & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
+ & = & \cos y+i\sin y\end{eqnarray*}
+
+
+Substituting $y=\pi$, we recover Euler's famous identity,\[
+e^{i\pi}=-1.\]
+ Given a real argument $y$, the $e^{iy}$ gives the unit vector with
+that argument. Given an arbitrary complex number $z=x+iy$, it's exponnt
+$e^{z}$ is the complex number with magnitude $e^{x}$ and angle $y$
+radians.
+
+What does $z\rightsquigarrow e^{z}$ look like? Periodic in the $i$
+direction with period $2\pi$. Takes a horizontal strip and wraps
+it around, forming an annulus. PICTURE.
+
+Let's take another look at these functions. What do you notice?
+
+Images of grid lines still intersect orthogonally! In fact, all angles
+are preserved: these two curves go to those two curves, and the angle
+between them (actually, the angles between their tangent vectors at
+the intersection point) stays the same.
+
+
+\subsection*{Conformality, Holomorphicity}
+
+A map $\mathbb{C}\rightarrow\mathbb{C}$ is \textbf{conformal} if
+it preserves oriented angles. We (meaning me \emph{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\[
+z\mapsto f(z)\]
+\[
+x+iy\mapsto u(x,y)+iv(x,y)\]
+\[
+\left(\begin{array}{c}
+x\\
+y\end{array}\right)\mapsto\left(\begin{array}{c}
+u(x,y)\\
+v(x,y)\end{array}\right)\]
+then the derivative is \[
+df=\left(\begin{array}{cc}
+\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
+\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).\]
+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
+means that $\left(\begin{array}{c}
+\frac{\partial u}{\partial y}\\
+\frac{\partial u}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c}
+\frac{\partial u}{\partial x}\\
+\frac{\partial v}{\partial x}\end{array}\right)$ rotated by $\pi/2$. If we write \[
+\left(\begin{array}{c}
+a\\
+b\end{array}\right)=\left(\begin{array}{c}
+\frac{\partial u}{\partial x}\\
+\frac{\partial v}{\partial x}\end{array}\right)\]
+, then \[
+\left(\begin{array}{c}
+\frac{\partial u}{\partial y}\\
+\frac{\partial u}{\partial y}\end{array}\right)=\left(\begin{array}{c}
+-b\\
+a\end{array}\right).\]
+Put another way, the linear transformation $df(z)$ has the form \[
+\left(\begin{array}{cc}
+a & -b\\
+b & a\end{array}\right),\]
+ie, it looks just like multiplication by the complex number $a+bi$.
+The function $f$ is conformal if its derivative \textbf{is} a nonzero
+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
+$f$ 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
+$u=x^{2}-y^{2}$ 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
+and rotations, and holomorphic functions (defined in terms of differential
+equations) are (almost everywhere) angle-preserving (if they're not
+constant).
+
+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}
+Applicationally,
+\begin{itemize}
+\item conformal maps preserve maxwell's equations (in 2D) and incompressible,
+irrotational fluid flow.
+\item 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!
+