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diff --git a/ClassJune26.page b/ClassJune26.page index 60c2e22..81cd2eb 100644 --- a/ClassJune26.page +++ b/ClassJune26.page @@ -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! + |