\documentclass{homework}
\course{Math 5522H}
\author{Alex Li}
\input{preamble}

\begin{document}
\maketitle

\begin{inspiration}
Some of the most important results (e.g. Cauchy's theorem) are so surprising at first sight that nothing short of a proof can make them credible.
\byline{Sir Harold Jeffreys} % and where is this quotation from?
\end{inspiration}

\section{Terminology}

\begin{problem}
  What is a \textbf{Jordan curve}?
  \end{problem}
  \begin{solution}
  A Jordan curve is a simple closed piecewise smooth function $\gamma:[a,b]\mapsto \C$.
  \end{solution}
  \begin{problem}
    Define the \textbf{winding number} $n(\gamma,z)$ of a piecewise
      smooth closed curve $\gamma$ around a point $z \in \C$.
      \end{problem}
      \begin{solution}
      It's the net number of times you spin around clockwise, or
      \[n(\gamma, z) = \frac{1}{2\pi i}\int_\gamma \frac{1}{x-z}dx \]

      \end{solution}
      \section{Numericals}

      \begin{problem}\label{Fresnel integral}
      Compute the \textbf{Fresnel integrals}
        \[
            \int_0^\infty \sin \left( x^2 \right) \, dx \mbox{ and } \int_0^\infty \cos \left( x^2 \right) \, dx.
              \]
              \end{problem}
              We can do them both at once, let $f(x) = \cos(x^2) + i\sin(x^2)$. Then the real part of $\int_0^\infty f(x)dx$ is the value of the cosine integral, and the imaginary part is the value of the sin integral. Noting that $f(x) = e^{ix^2}$, we itegrate around the eighth circle of radius $R$, $\gamma$ in figure \ref{octant-tikz}.

              \begin{figure}[h]
              \begin{tikzpicture}
              \draw [<->] (-4, 0) -- (4, 0);
              \draw [<->] (0, -4) -- (0, 4);
              \node [below] at (4.24, 0) {$R$};
              \node [right] at (3, 3) {$\theta=\pi/4$};
              \begin{scope}[very thick,decoration={
                  markings,
                      mark=at position 0.5 with {\arrow{>}}}
                          ]
                          \draw[postaction={decorate}] (0,0) -- (4.24, 0);
                          \draw[postaction={decorate}] (3,3) -- (0,0);
                          \draw[postaction={decorate}] (4.24, 0) arc [radius=4.24, start angle=0, end angle=45];
                          \end{scope}
                          \end{tikzpicture}
                          \caption{The curve $\gamma$ for \ref{Fresnel integral}}
                          \label{octant-tikz}
                          \end{figure}
                          \begin{align*}
                          \int_\gamma e^{iz^2}
                          &= \int_0^R e^{iz^2}dz + \int_0^{\frac{\pi}{4}} (ie^{i\theta}) e^{i(Re^{i\theta})^2} d\theta + \int_0^R e^{5\pi/4}e^{i(ze^{i\pi/4})^2)}dz \\
                          &= \int_0^R e^{iz^2}dz + \int_0^{\frac{\pi}{4}} ie^{iR^2\cos(2\theta) - R^2\sin(2\theta) + i\theta} d\theta - e^{\pi/4} \int_0^R e^{-z^2}dz
                          \end{align*}
                          Letting $R$ go to infinity, the middle term vanishes since the real part of the exponent $R^2\sin(2\theta) + \cos(\theta)$ goes to $-\infty$. The last term becomes the classic statistics integral with value $\frac{\sqrt{\pi}}{2}$. Since $\int_\gamma f(x)$ is the integral of a holomorphic function around a closed curve, by cauchy's theorem it's value is 0 so we can deduce that
                          \[\int_0^\infty e^{iz^2}dz = e^{\pi/4} \int_0^R e^{-z^2}dz =  \frac{1+i}{\sqrt{2}}\frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{\sqrt{8}} + \frac{i\sqrt{\pi}}{\sqrt{8}}\]
                          so
                            \[
                                \int_0^\infty \sin \left( x^2 \right) \, dx = \int_0^\infty \cos \left( x^2 \right) \, dx = \frac{\sqrt{\pi}}{\sqrt{8}}
                                  \]
                                  \begin{problem}
                                    For an integrable function $f : \R \to \C$, define the
                                      \textbf{Fourier transform} of $f$, denoted $\hat{f}$, by
                                        \[
                                            {\hat {f}}(\xi ) := \int _{-\infty }^{\infty} f(x) \, e^{-2\pi ix \xi} \,dx.
                                              \]
                                                Find the Fourier transform of $f(x) = e^{-\pi x^2}$.
                                                \end{problem}
                                                \begin{solution}
                                                We solve this by completing the square in the exponent:
                                                \begin{align}\label{completed_square_fourier}
                                                -\pi x^2 - 2\pi ix\xi = -\pi(x-i\xi)^2 - \pi \xi^2
                                                \end{align}
                                                We want to do something like below,
                                                \begin{align*}
                                                {\hat {f}}(\xi ) &=  \int _{-\infty }^{\infty} e^{-\pi x^2 -2\pi ix \xi} \,dx\\
                                                &=  e^{-\xi^2\pi}\int _{-\infty }^{\infty} e^{-\pi (x-i\xi)^2} \,dx\quad \color{purple} \text{ By Eq. \ref{completed_square_fourier}}
                                                \end{align*}
                                                This integral can almost be evaluated with a $u$ subsitution of
                                                \[u = x-i\xi.\]
                                                However, the bounds of integration will be messed up and we will be going over the complex numbers. So let's consider the closed rectangular contour that goes on a straight line between the four points below.
                                                \[R \mapsto R - i\zeta \mapsto - R - i\zeta \mapsto -R\]
                                                As $R\to\infty$, the sides of this rectangle with nonzero imaginary derivative will disappear as
                                                \[
                                                \abs{\int_{-\zeta}^{0} e^{-\pi((\pm R+ix) -i\zeta)^2} dx} \leq \abs{\zeta} \abs{e^{-\pi R}}\to 0.
                                                \]
                                                Thus the integral of the bottom of the rectangle is equal to the integral going backwards along the top of the rectangle, which we can compute
                                                \begin{align*}
                                                \hat f(\zeta) &=  e^{-\xi^2\pi}\int_{-\infty}^\infty e^{-\pi x^2} \,dx\\
                                                &=  \frac{e^{-\xi^2\pi}}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-x^2} \,dx\quad \color{purple} u = \sqrt{\pi}x, du = \sqrt{\pi} dx\\
                                                &=  \frac{e^{-\xi^2\pi}}{\sqrt{\pi}} \sqrt{\pi} =  e^{-\xi^2\pi}
                                                \end{align*}

                                                \end{solution}
                                                \section{Exploration}

                                                \begin{problem}\label{conjz_no_primitive}
                                                  Does the function $f : \C \to \C$ given by $f(z) = \conj{z}$ have a
                                                    primitive?
                                                    \end{problem}
                                                    \begin{solution}
                                                    Let $F$ be a primitive for $f$ and $\gamma: [0,1]\mapsto \C$ with $\gamma(t)=e^{it2\pi}$ be a closed curve.
                                                    Then
                                                    \begin{align*}
                                                        0 = F(\gamma(2\pi)) - F(\gamma(0)) = \int_\gamma \conj{z} = \int_\gamma \frac{1}{z} = 2\pi i,
                                                        \end{align*}
                                                        a contradiction.
                                                        \end{solution}
                                                        \begin{problem}
                                                          We proved Goursat's theorem for rectangles.  Without simply
                                                            recapitalulating the proof (e.g., your argument should not again
                                                              invoke \ref{nested-subsets-convergence}), deduce a theorem for
                                                                \textit{triangles} from our result about rectangles.
                                                                \end{problem}
                                                                \begin{solution}
                                                                We will prove the following:
                                                                \begin{theorem}\label{triangle-theorem}
                                                                Let $\gamma_{\Delta}$ be a curve around some triangle be a triangle contained in an open set $S$, and $f:S\mapsto \C$ be holomorphic. Then
                                                                \[\int_{\gamma_{\Delta}} f = 0\]
                                                                In fact, this generalizes to any polygon with no holes in it whose interior is contained in $S$ (by gluing positively oriented triangles together to make that polygon and cancelling).
                                                                \end{theorem}
                                                                Choose some side of the triangle as well as a small $\delta$. Now, draw a rectangle parallel to that side of height $\delta$ as wide as possible, and then draw a rectangle on that and so on until you can't anymore. (See Figure \ref{curve-triangle-rectangles}).
                                                                \begin{figure}[h]
                                                                \begin{tikzpicture}
                                                                \begin{scope}[very thick,decoration={
                                                                    markings,
                                                                        mark=at position 0.2 with {\arrow{>}},
                                                                            mark=at position 0.5 with {\arrow{>}},
                                                                                mark=at position 0.8 with {\arrow{>}}}
                                                                                ]
                                                                                \draw[postaction={decorate}] (0,0) -- (4, 0) -- (8, 4) -- (0,0);
                                                                                \end{scope}
                                                                                \begin{scope}[decoration={
                                                                                markings,
                                                                                mark=at position 0.2 with {\arrow{>}}}
                                                                                ]
                                                                                \draw[postaction={decorate}] (1, 0) -- (1, .5) -- (4, .5) -- (4, 0) -- (1, 0);
                                                                                \draw[postaction={decorate}] (2, .5) -- (2, 1) -- (4.5, 1) -- (4.5, .5) -- (2, .5);
                                                                                \draw[postaction={decorate}] (3, 1) -- (3, 1.5) -- (5, 1.5) -- (5, 1) -- (3, 1);
                                                                                \draw[postaction={decorate}] (4, 1.5) -- (4, 2) -- (5.5, 2) -- (5.5, 1.5) -- (4, 1.5);
                                                                                \draw[postaction={decorate}] (5, 2) -- (5, 2.5) -- (6, 2.5) -- (6, 2) -- (5, 2);
                                                                                \draw[postaction={decorate}] (6, 2.5) -- (6, 3) -- (6.5, 3) -- (6.5, 2.5) -- (6, 2.5);
                                                                                \end{scope}
                                                                                \end{tikzpicture}
                                                                                \caption{Cutting a triangle into small rectangles}
                                                                                \label{curve-triangle-rectangles}
                                                                                \end{figure}
                                                                                Since each of the rectangles have a value of 0, the integral around the triangle is equal to the sum of the integrals around the smaller triangles between the thin rectangles and the big triangle.

                                                                                Now, because $f$ is continous on the compact set that is the interior of $\gamma_{\Delta}$ it is uniformly continuous, so we can choose $\delta$ such that the value of $f$ varies by less than $\epsilon$ at any two points on the same small cut up piece. As the pieces get smaller, the top strangely shaped piece will have a very small arc length and so the value of the integral around it goes to 0, so we can ignore it. Next we can bound the integral around any other triangle $\gamma_{i}$. If $f$ is a constant, then the integral around the triangle is 0, so
                                                                                \[\abs{\int_{\gamma_i} f dz} = \abs{\int_{\gamma_i} f dz- \inf_{\gamma_i}(f)} \leq |\gamma_i|\epsilon\].

                                                                                Now, any given small triangle has it's longest side as part of the contour $\gamma_{\Delta}$, so the sum of the perimeters of all of the triangles is bounded above by $3\abs{\gamma_{\Delta}}$. Thus we can bound the integral around $\gamma_\Delta$ by choosing $\delta$ so as to make $\epsilon$ tiny:
                                                                                \[\abs{\int_{\gamma_\Delta} f}\leq 3\abs{\gamma_{\Delta}}\epsilon\to 0\]
                                                                                \end{solution}

                                                                                \begin{problem}
                                                                                Suppose $f, g : \C \to \C$ are holomorphic and agree on the unit
                                                                                circle.  How do $f$ and $g$ relate?  Morally, this problem is
                                                                                related to \ref{identity-theorem}.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                Let's consider $f-g$, it's holomorphic and equal to 0 on the unit circle. Call the positively oriented curve going around the unit circle $\gamma$. By Cauchy's Integral formula,
                                                                                \[(f-g)(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{(f-g)(w)}{z-w}dw\]
                                                                                And this is $0$ since the numerator is $0$.
                                                                                \end{solution}

                                                                                \begin{problem}
                                                                                Suppose $f : \C \to \C$ is holomorphic, and consider the circle
                                                                                $\gamma$ with center $z_0$.  How does the real part of the average
                                                                                value of $f$ on the circle $\gamma$ relate to $f(z_0)$?
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                By Cauchy's integral formula,
                                                                                \begin{align*}
                                                                                f(z_0) &= \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z_0 - z}dz\\
                                                                                &= \frac{1}{2\pi i}\int_{0}^{2\pi} \frac{f(z_0 + e^{it})}{e^{it}}ie^{it} dz\\
                                                                                &= \frac{1}{2\pi}\int_{0}^{2\pi} f(z_0 + e^{it}) dz
                                                                                \end{align*}
                                                                                And this last expression is the average value of $f$ on the circle $\gamma$, so $f(z_0)$ is is equal to average value of $f$ on the circle $\gamma$.

                                                                                \end{solution}
                                                                                \begin{problem}
                                                                                Recalling \ref{harmonic-conjugate}, suppose $u, v : \C \to \R$ are
                                                                                harmonic functions and $f = u + iv$ is holomorphic, so $u$ and $v$
                                                                                are harmonic conjugates.  Find the \textbf{Poisson kernel} for the
                                                                                unit disc, i.e., find a function $P_r(\theta)$ so that
                                                                                \[
                                                                                u(re^{i\theta}) = \frac {1}{2\pi} \int_{-\pi }^{\pi } P_{r}(\theta -t) \, u(e^{it}) \, dt
                                                                                \]
                                                                                for $r < 1$.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                We can recover the value at $re^{i\theta}$ by an integral of $u$ over the unit circle using Cauchy's integral formula
                                                                                \begin{align*}
                                                                                u(re^{i\theta}) &= \frac{1}{2\pi i}\int_{\gamma} \frac{u(e^{it})}{re^{i\theta} - e^{it}}\\
                                                                                &= \frac{1}{2\pi i}\int_{-\pi}^{\pi} \frac{u(e^{it})ie^{it}}{re^{i\theta} - e^{it}}\\
                                                                                &= \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{u(e^{it})}{re^{i(\theta - t)} - 1}\\
                                                                                \end{align*}
                                                                                Thus
                                                                                \[P_r(\theta) = \frac{1}{re^{i\theta} - 1}\]
                                                                                is the Poisson kernel.

                                                                                \end{solution}
                                                                                \section{Prove or Disprove and Salvage if Possible}

                                                                                \begin{problem}
                                                                                Suppose $D = \{ z \in \C : \abs{z} < 1 \}$ and $f : D \to \C$ is
                                                                                continuous and $\gamma : [a,b] \to D$ is a piecewise smooth curve.
                                                                                Then $\displaystyle\int_\gamma f\, dz = 0$.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                No, let $f(x)=1$ and $\gamma$ be any non-closed curve. Even if it's closed, if $f$ is not holomorphic, then going around a rectangle may be a problem (for example consider the computation in \ref{conjz_no_primitive}). So we really want to suppose that $\gamma$ is closed and $f$ is holomorphic. We will prove the following.
                                                                                \begin{theorem}\label{general-curve-theorem}
                                                                                Let $\gamma$ be a piecewise smooth curve in an open set $S$, and $f:S\mapsto \C$ be holomorphic. If the interior of $\gamma$ is contained in $S$, then \[\int_{\gamma} fdz = 0\].
                                                                                \end{theorem}
                                                                                Theorem \ref{triangle-theorem} tells use this is true if $\gamma$ is the curve of a polygon. Since $\gamma$ is smooth, it is rectifiable, so we can draw a polygonal path $\gamma(t_0), \gamma(t_1), \dots, \gamma(t_{n-1}), \gamma(t_n)$ with $\gamma(t_n)=\gamma(t_0)$ that approximates $\gamma$. Since $S$ is open we can ensure this path in contained inside of it. By rectifiablity, we can choose the path such that for any $\delta > 0$,
                                                                                \[\sum_{i=0}^{n-1}\abs{\gamma(t_i) - \gamma(t_{i+1})} - \int_{t_i}^{t_{i+1}} f(\gamma(t)) dt \leq \delta \]
                                                                                Let $\gamma_{i}'$ be the integral on the line directly from $t_i$ to $t_{i+1}$ and then back from $t_{i+1}$ to $t_{i}$ along the curve $\gamma$. We can choose $t_i$ to be close enough to each other that any two points on $\gamma_{i}'$ are at most $\delta$ away from each other, and since the polygon is compact and $f$ is continuous, for any $\epsilon > 0$ we can choose $\delta$ such that
                                                                                \[x - y < \delta \implies f(x) - f(y) < \epsilon\].

                                                                                Now, since the line integral of interior of the polygon is 0 by Theorem \ref{triangle-theorem}, the value of the line integral around $\gamma$ is bounded by the sum of the integrals of the segments between the approximating polygon and the curve. Then
                                                                                \begin{align*}
                                                                                \abs{\int_{\gamma} fdz} &\leq \sum_{i=0}^{n-1}\abs{\int_{\gamma_{i}'} f dt}\\
                                                                                &= \sum_{i=0}^{n-1}\abs{\int_{\gamma_{i}'} f - \inf_{\gamma_{i}'}f dt}\color{purple}\text{ closed line integral of constant is 0}\\
                                                                                &\leq \sum_{i=0}^{n-1}\abs{\gamma_{i}'}(\sup_{\gamma_{i}'}f - \inf_{\gamma_{i}'}f)  dt\\
                                                                                &\leq \sum_{i=0}^{n-1}\abs{\gamma_{i}'}\epsilon dt\\
                                                                                &\leq \epsilon((2\int_{\gamma(t)}dt +\delta)\to 0
                                                                                \end{align*}
                                                                                Thus the line integral around $\gamma$ is 0.
                                                                                \end{solution}

                                                                                \begin{problem}
                                                                                Suppose $U$ is an open set and $f : U \to \C$ is analytic and
                                                                                $\gamma : [a,b] \to U$ is a piecewise smooth curve.  Then
                                                                                $\displaystyle\int_\gamma f \, dz = 0$.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                False- again, $\gamma$ must be closed, and the interior of $\gamma$ should be contained in $U$. In this case we can apply \ref{general-curve-theorem}.
                                                                                Otherwise a counter-example can be given by noting that $1/z$ is defined on the open set $\C/\{0\}$. For $\gamma$ the unit circle, $\int_\gamma 1/z=2\pi i\neq 0$.
                                                                                \end{solution}

                                                                                \begin{problem}
                                                                                Define $\gamma_r : [0,2\pi] \to \C$ by
                                                                                $\gamma_r(\theta) = r e^{i \theta}$, and for real numbers
                                                                                $R > r > 0$, define the annulus
                                                                                \[
                                                                                A(r,R) := \{ z \in \C : r < |z| < R \}
                                                                                \]
                                                                                and suppose $f : A(r/2,2R) \to \C$ is holomorphic.  Then
                                                                                \[
                                                                                \int_{\gamma_r} f \, dz = \int_{\gamma_R} f \, dz.
                                                                                \]
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                True. Consider integrating along the 2 curves in Fig. \ref{annulus-curve}. The two curves are closed and can be compressed to a point along the holomorphic region, so the integral along them both is 0. The two horizontal lines will cancel out, so we have the claim.
                                                                                \begin{figure}[h]
                                                                                \begin{tikzpicture}
                                                                                \begin{scope}[very thick,decoration={
                                                                                markings,
                                                                                mark=at position 0.3 with {\arrow{>}}}
                                                                                ]
                                                                                \draw[postaction={decorate}] (-1, 0) arc [radius=1, start angle=180, end angle=0];
                                                                                \draw[postaction={decorate}] (2, 0) arc [radius=2, start angle=0, end angle=180];
                                                                                \draw[postaction={decorate}] (-2, 0) arc [radius=2, start angle=180, end angle=360];
                                                                                \draw[postaction={decorate}] (1, 0) arc [radius=1, start angle=360, end angle=180];
                                                                                \draw[postaction={decorate}] (1,0) -- (2,0);
                                                                                \draw[postaction={decorate}] (2,0) -- (1,0);
                                                                                \draw[postaction={decorate}] (-1,0) -- (-2,0);
                                                                                \draw[postaction={decorate}] (-2,0) -- (-1,0);
                                                                                \end{scope}
                                                                                \end{tikzpicture}
                                                                                \caption{Annulus Curves, one is on the top half and the other is on the bottom half.}
                                                                                \label{annulus-curve}
                                                                                \end{figure}
                                                                                \end{solution}
                                                                                \pagebreak
                                                                                \begin{problem}\label{cauchy-integral-formula}
                                                                                Define $B_R(0) := \{ z \in C : \abs{z} < R \}$ and suppose
                                                                                $f : B_R(0) \to \C$ is holomorphic and let $\gamma$ be the
                                                                                positively-oriented circle of positive radius $r < R$.  Then
                                                                                \[
                                                                                f(a) = \int_\gamma \frac{f(z)}{z-a} \, dz.
                                                                                \]
                                                                                and moreover
                                                                                \[
                                                                                f'(a) = \int_\gamma \frac{f(z)}{(z-a)^2} \, dz.
                                                                                \]
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                This is true up to a factor of $2\pi i$ for any $a$ in the interior of $\gamma$. For the first claim,
                                                                                \begin{align*}
                                                                                \int_\gamma \frac{f(z)}{z-a}dz &= \int_\gamma \frac{f(z) - f(a)}{z-a}dz + f(a)\int_\gamma \frac{1}{z-a}dz
                                                                                \end{align*}
                                                                                The rightmost integral is the integral of $\frac{1}{z}$ around a circle containing the origin shifted by $a$, so it is $2\pi i$.
                                                                                \begin{align}
                                                                                \label{cauchys-integral-formula-really}
                                                                                \int_\gamma \frac{f(z)}{z-a}dz &= \int_\gamma \frac{f(z) - f(a)}{z-a}dz + f(a)2\pi i
                                                                                \end{align}
                                                                                Now, we can draw a super small circle of radius $\delta$ around $a$ contained in $B_R(0)$. Splitting this circle up into 2 curves in the spirit of the image \ref{annulus-curve}, and noting that the two curves have line integral 0, we see that the integral around $B_R(0)$ is equivalent to the integral around the small circle centered at $a$. Letting $\delta$ go to 0,
                                                                                \begin{align*}
                                                                                \lim_{\delta\to 0} \int_{\gamma_{\delta}} \frac{f(z)-f(a)}{z-a} dz &= \lim_{\delta\to 0} \int_{\gamma_{\delta}} \frac{f(a+h)-f(a)}{h} dz \color{purple}\quad |h| = \delta\\
                                                                                &\leq \frac{|2\gamma_{\delta}|}{\delta}\sup{f(a+h) - f(a)}\\
                                                                                &\leq \frac{2\pi\delta}{\delta}\epsilon \leq 0
                                                                                \end{align*}
                                                                                Here we can use the continuity of $f$ to ensure that we can choose a $\delta$ such that $a+h - a<\delta \implies f(a+h) - f(a) < \epsilon$. Thus by Eq. \ref{cauchys-integral-formula-really}
                                                                                \begin{align}\label{cauchys-integral-formula-clean}
                                                                                \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}dz &= f(a)
                                                                                \end{align}
                                                                                For the second equation, let's take the derivative of both sides with respect to $a$ with the quotient rule
                                                                                \begin{align}\label{cauchys-integral-formula-clean-d}
                                                                                \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z-a)^2}dz &= f'(a)
                                                                                \end{align}
                                                                                \end{solution}

                                                                                \begin{problem}\label{cauchy-inequalities}If $f : U \to \C$ is
                                                                                holomorphic and $U \supset B_r(z_0)$, then
                                                                                \[
                                                                                \abs{f(z_0)} \leq \sup_{z \in \partial B_r(z_0)} \abs{f(z)}
                                                                                \]
                                                                                and
                                                                                \[
                                                                                \abs{f'(z_0)} \leq \sup_{z \in \partial B_r(z_0)} \abs{f(z)}. % missing (1/r) factor
                                                                                \]
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                WLOG let $z_0=0.$ By Eq. \ref{cauchys-integral-formula-clean},
                                                                                \begin{align*}
                                                                                f(0) &= \frac{1}{2\pi i}\int_0^{2\pi} \frac{f(re^{it})ire^{it}}{re^{it}}dt\\
                                                                                &= \frac{1}{2\pi}\int_0^{2\pi} f(re^{it}) dt
                                                                                \end{align*}
                                                                                So $f(0)$ is the average value on the boundary, and thus $\abs{f(z_0)} \leq \displaystyle\sup_{z \in \partial B_r(z_0)} |f(z)|$.
                                                                                By Eq. \ref{cauchys-integral-formula-clean-d},
                                                                                \begin{align*}
                                                                                \abs{f'(0)} &= \abs{\frac{1}{2\pi i}\int_0^{2\pi} \frac{f(re^{it})ire^{it}}{(re^{it})^2}dt}\\
                                                                                &\leq \frac{1}{2\pi r}\int_0^{2\pi} \frac{\abs{f(re^{it})}}{\abs{e^{it}}} dt\\
                                                                                &\leq \frac{1}{r}\left(\frac{1}{2\pi}\int_0^{2\pi} \abs{f(re^{it})} dt\right)
                                                                                \end{align*}
                                                                                Thus
                                                                                \begin{align}\label{derivative-dies-holomorphic}
                                                                                \abs{f'(z_0)} \leq \frac{1}{r}\displaystyle\sup_{z \in \partial B_r(z_0)} |f(z)|
                                                                                \end{align}
                                                                                \end{solution}

                                                                                \begin{problem}\label{liouville-theorem}Recall that a function $f : U \to \C$ is
                                                                                \textbf{bounded} if there exists $M > 0$ so that for all $z \in U$
                                                                                we have $\abs{f(z)} \leq M$.  A bounded holomorphic function
                                                                                $f : U \to \C$ is constant.
                                                                                \end{problem}
                                                                                \begin{solution}
                                                                                False, consider $f:B_1(0)\mapsto \C, f(z) = z$. This is bounded, since $\abs{f(z)}\leq 1$. But it's clearly not constant. However, we can prove it if we take $U=\C$.

                                                                                Choose a point $z_0$, and construct a ball of a massive radius $r$ around it. Then by \ref{derivative-dies-holomorphic},
                                                                                \[\abs{f'(z_0)} \leq \frac{1}{r}\sup_{z \in \partial B_r(z_0)} \abs{f(z)} \leq rM \to 0\]
                                                                                Thus the derivative of $f$ is 0 everywhere, so $f$ is constant.
                                                                                \end{solution}
                                                                                \begin{problem}
                                                                                For every polynomial $p \in \C[z]$ there is $z \in \C$ so that $p(z) = 0$.
                                                                                \end{problem}
                                                                                \begin{proof}
                                                                                False. Consider the polynomial $p(z)=1$. Instead we prove the following:
                                                                                \begin{theorem}
                                                                                If a polynomial $p \in \C[z]$ has no zeros, it is constant.
                                                                                \end{theorem}
                                                                                \[
                                                                                \abs{p(z)} = \abs{\sum_{n=0}^d a_nz^n} \geq \abs{z}^d - \sum_{n=0}^{d-1} \abs{a_n}\abs{z}^n
                                                                                \]
                                                                                Once $\abs{z}$ gets sufficently big (say $\abs{z}>M$), the leading term will dominate and we can say that, for some $M_2$ $\abs{p(z)}>M_2$.

                                                                                Suppose that $p(z)$ never takes on the value $0$. Then $\frac{1}{p(z)}$ is holomorphic, and outside of the ball of radius $M$, it's value is at most $\frac{1}{M_2}$. And it's holomorphic hence continuous and thus also bounded inside the compact set that is the ball of radius $M$. Thus $\frac{1}{p(z)}$ is bounded, and by \ref{liouville-theorem}, it is constant.
                                                                                \end{proof}
                                                                                \end{document}


\chapter*{Notation} \addcontentsline{toc}{chapter}{Notation}

Following notation is used throughout this text:

\bigskip

\begin{tabular}{l p{0.8\textwidth}}
	\(\mathbb{N}\) & set of natural numbers excluding zero \\
	\(\mathbb{N}_0\) & set of natural numbers including zero \\
	\(\mathbb{R}\) & set of real numbers \\
	\(t\) & discrete time moment; \(t \in \mathbb{N}_0\) \\
	\(a_t\) & value of quantity \(a\) at time \(t\); \(a_t \in \mathbb{R}^n, n \in \mathbb{N}\) \\
	\(a_{t|t'}\) & quantity with two indices: \(t\) and \(t'\) \\
		& there is no implicit link between \(a_{t}\), \(a_{t|t'}\) and \(a_{t'}\) \\
	\(a_{t:t'}\) & sequence of quantities \((a_t, a_{t+1}, \dotsc, a_{t'-1}, a_{t'})\) \\
	\(p(a_t)\) & probability density function{\footnotemark[1]} of quantity \(a\) at time \(t\) (unless
		noted otherwise) \\
	\(p(a_t|b_{t'})\) & conditional {\pdf} of quantity \(a\) at time \(t\) given value of quantity
		\(b\) at time \(t'\) \\
	\(\delta(a)\) & Dirac delta function; used exclusively in context of {\pdfs} to denote discrete
		distribution within framework of continuous distributions{\footnotemark[2]} \\
	\(\mathcal{N}(\mu, \Sigma)\) & multivariate normal (Gaussian) {\pdf} with mean vector \(\mu\)
		and covariance matrix \(\Sigma\) \\
\end{tabular}

% \footnote does not work in tabular environment, - this can be worked around with this
\footnotetext[1]{for the purpose of this text, {\pdf} \(p\) is multivariate non-negative function
\(\mathbb{R}^n \rightarrow \mathbb{R}; \; \int_{\supp p} p(x_1, x_2, \dotsc, x_n) \; \dx_1 \dx_2 \dotsb \dx_n = 1\)}

\footnotetext[2]{so that \(\int_{-\infty}^{\infty} f(x) \delta(x - \mu) \; \dx = f(\mu)\) and more
complex expressions can be derived using integral linearity and Fubini's theorem.}


\subsection{Functions of rational numbers}

\subsubsection{Rational addition}

Then we can define addition as:

\((a,b)+(c,d)=(a.d+b.c,b.d)\)

\(a+b=c\)

\(c_1=a_1b_2+a_2b_1\)

\(c_1=a_2b_2\)

\subsubsection{Rational subtraction}

\(a-b=c\)

\(c_1=a_1b_2-a_2b_1\)

\(c_1=a_2b_2\)

\subsubsection{Rational multiplication}

Similarly, multiplication can be defined as:

\((a,b).(c,d)=(a.c, b.d)\)

\(ab=c\)

\(c_1=a_1b_1\)

\(c_1=a_2b_2\)

\subsubsection{Rational division}

\(\dfrac{a}{b}=c\)

\(c_1=a_1b_2\)

\(c_1=a_2b_1\)


\gotosection{1}{1}
\subsection{Points and vectors}

\begin{exercise}{1}
  All sketches are hand-drawn in the separate sketch paper.

  \begin{enumerate}
  \item $\begin{bmatrix}1\\3\end{bmatrix} +
         \begin{bmatrix}2\\1\end{bmatrix} =
         \begin{bmatrix}1+2\\3+1\end{bmatrix} =
         \begin{bmatrix}3\\4\end{bmatrix}$
  \item $2\begin{bmatrix}2\\4\end{bmatrix} =
         \begin{bmatrix}2 \times 2\\2 \times 4\end{bmatrix} =
         \begin{bmatrix}4\\8\end{bmatrix}$

  \item $\begin{bmatrix}1\\3\end{bmatrix} -
         \begin{bmatrix}2\\1\end{bmatrix} =
         \begin{bmatrix}1-2\\3-1\end{bmatrix} =
         \begin{bmatrix}-1\\2\end{bmatrix}$

  \item $\begin{bmatrix}3\\2\end{bmatrix} + \vec{\mathbf{e}}_1 =
         \begin{bmatrix}3\\2\end{bmatrix} +
         \begin{bmatrix}1\\0\end{bmatrix} =
         \begin{bmatrix}3+1\\2+0\end{bmatrix}=
         \begin{bmatrix}4\\2\end{bmatrix}$
  \end{enumerate}
\end{exercise}

\begin{exercise}{4}
  \begin{enumerate}
    \item The first trivial subspace is $V = \{\vec{\mathbf{0}}\}$, a singleton set containing only the zero       vector. The second one is $V = \mathbb{R}^n$, namely the whole vector space itself.

    \item Such two elements exist, which can be proven geometrically, as shown in the figure. Picking two vectors $\Vect{x} = \Vect{OA}$ and $\Vect{y} = \Vect{OB}$ from $S_1$, it is obvious that $\Vect{x} + \Vect{y} = \Vect{OC}$ is the diagonal of the parallelogram OACB. If $\Vect{x} + \Vect{y} \in S_1$, the sum vector must has the same length as $\Vect{x}$ and $\vec{y}$, being the radius of the unit circle. Therefore, triangle ACO and triangle COB must be equilateral triangles. In other words, as long as angle AOB is 120 degrees, which is evidently possible, the sum $\Vect{x} + \Vect{y} \in S_1$. \QED

    \begin{figure}[h]
      \begin{center}
        \begin{tikzpicture}
          \tkzInit[xmax=2.1,ymax=2.1,xmin=-2.1,ymin=-2.1]
          \tkzDrawX[noticks]
          \tkzDrawY[noticks]
          \tkzDefPoint(0,0){O}
          \tkzDefPoint[label=above:$A$](-1,1.732){A}
          \tkzCalcLength[cm](O,A)
          \tkzGetLength{rOA}
          \tkzDefPoint(\rOA,0){B}
          \tkzDefPoint[label=above:$C$]({-1 + \rOA}, 1.732){C}
          \tkzDrawCircle[R](O,\rOA cm)
          \draw[thick,->] (O) -- (A);
          \draw[thick,->] (O) -- (B);
          \draw[thick,->] (O) -- (C);
          \tkzDrawSegments[postaction={decorate},dashed](A,C)
          \tkzDrawSegments[postaction={decorate},dashed](B,C)
          \tkzDrawPoints(O,A,B,C)
          \tkzLabelPoints(O,B)
        \end{tikzpicture}
      \end{center}
    \end{figure}
  \end{enumerate}
\end{exercise}

\begin{exercise}{5}
  $$
    \begin{bmatrix}1\\1\\\vdots\\1\\1\end{bmatrix} = \sum\limits_{i=1}^n \Vect{e}_i
    \quad\quad\quad
    \begin{bmatrix}1\\2\\\vdots\\n-1\\n\end{bmatrix} = \sum\limits_{i=1}^n i\Vect{e}_i
    \quad\quad\quad
    \begin{bmatrix}0\\0\\3\\4\\\vdots\\n-1\\n\end{bmatrix} = \sum\limits_{i=3}^n \Vect{e}_i
  $$
\end{exercise}

\begin{exercise}{6}
  All sketches are hand-drawn in the separate sketch paper.
\end{exercise}

\begin{exercise}{8}
  \begin{enumerate}
    \item Suppose the axis of the pipe is the point set $\{\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}: z_0 \in \mathbb{R}\}$.
    Let $\vec{F}$ be the vector field. It's obvious that outside the pipe the water speed is 0, namely for all $p \in \{\begin{pmatrix}x\\y\\z\end{pmatrix} : (x-x_0)^2 + (y-y_0)^2 > r^2, z \in \mathbb{R}\}$, $\vec{F}(p) = 0$.
    \footnote{More precisely, $\vec{F}(p)$ is meaningless for those $p$, since there is no water. However, since vector field is a function whose domain is the whole vector space, I have to give it a default value. When grading please take into account that I have considered this.}

    For $p \in \{\begin{pmatrix}x\\y\\z\end{pmatrix}\} : (x-x_0)^2 + (y-y_0)^2 \leq r^2, z \in \mathbb{R}$, let $p = \begin{pmatrix}x\\y\\z\end{pmatrix}$, the distance between the axis and $p$ is $a = \sqrt{(x - x_0)^2 + (y - y_0)^2}$. The water is flowing in the direction of the $z$-axis, so $z$ equals the water speed. Therefore:

    $$\vec{F}(\begin{pmatrix}x\\y\\z\end{pmatrix}) =
      \begin{bmatrix}0\\0\\r^2 - a^2\end{bmatrix} =
      \begin{bmatrix}0\\0\\1 - a^2\end{bmatrix} =
      \begin{bmatrix}0\\0\\1 - (x-x_0)^2 - (y-y_0)^2\end{bmatrix}$$

    \item Assuming that the flow's direction is anticlockwise. Since the axis of
    the pipe is the unit circle $S_2$ in the $(x, y)$-plane, the water speed in
    the direction of the $z$-axis is 0, and the direction of the flow is tangent
    to the pipe. For any point $p = \begin{pmatrix}x\\y\\0\end{pmatrix} \in S_2$
    , we can have a plane $P_p$ intersecting $p$ and the origin point. Let the
    point set of the pipe be $D$, and $P_p' = D \cap P_p$, then we have:

    \begin{enumerate}
      \item Since $\begin{bmatrix}x\\y\\0\end{bmatrix}$ is a vector on $P_p$ and perpendicular
      to $\vec{dir} = \begin{bmatrix}-y\\x\\0\end{bmatrix}$, $P_p$ is also perpendicular to
      $\vec{dir}$. That is to say, for all $\begin{pmatrix}x'\\y'\\z'\end{pmatrix} \in P_p$,
      $\begin{pmatrix}x'\\y'\\z'\end{pmatrix}\begin{bmatrix}-y\\x\\0\end{bmatrix} = 0$.

      \item $\begin{bmatrix}-y\\x\\0\end{bmatrix}$ is also the direction of water flow for points
      in $P_p'$, since the direction of flow is tangent to the pipe, which also means that
      it is tangent to the unit circle. Therefore, the water velocity for a point
      $q \in P_p'$, $\vec{F}(q) = \begin{bmatrix}v_x\\v_y\\0\end{bmatrix}$, is parallel to
      $\begin{bmatrix}-y\\x\\0\end{bmatrix}$, or $\frac{v_x}{v_y} = \frac{-y}{x}$.
    \end{enumerate}

    $D$ is the union of all sets $Pq'$ where $q$ is a point of the unit circle $S_2$:

    $$D = \bigcup\limits_{(x, y, 0) \in S_2}
            \{(x_1, y_1, z_1): (x_1 - x)^2 + (y_1 - y)^2 + z_1^2 \leq r^2,
                               \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}
                               \begin{bmatrix}-y\\x\\0\end{bmatrix} = 0\}$$

    For every point $p \in D$, let $p = \begin{pmatrix}x\\y\\z\end{pmatrix}$ and
    $\vec{F}(p) = \begin{bmatrix}v_x\\v_y\\0\end{bmatrix}$. From (b)
    we know that:
    \begin{align*}
      \frac{v_x}{v_y} &= \frac{-y}{x} \\
                  v_x &= -\frac{y}{x}v_y
    \end{align*}

    Next, we need to figure out what $a$ is. One way to do it is to omit the $z$
    -component for now, and figure the distance $a'$ between the axis and $p$'s
    projection $p'$ on $(x, y)$-plane, which is also the difference between the
    distance from the axis to the origin point and that from $p'$ to the origin
    point: $a' = \sqrt{x^2 + y^2} - 1$. Obviously
    $$a^2 = a'^2 + z^2 = (\sqrt{x^2 + y^2} - 1)^2 + z^2$$

    \def \tmp{(\sqrt{x^2 + y^2} - 1)^2 - z^2}
    Then,
    \begin{align*}
                         |\vec{F}(p)| &= r^2 - a^2 \\
                 \sqrt{v_x^2 + v_y^2} &= r^2 - \tmp \\
      \sqrt{\frac{y^2+x^2}{x^2}v_y^2} &= r^2 - \tmp \\
        \frac{\sqrt{y^2 + x^2}}{x}v_y &= \pm (r^2 - \tmp) \\
      v_y &= \pm \frac{x(r^2 - \tmp)}{\sqrt{y^2 + x^2}} \\ & \\
      v_x &= \mp \frac{y(r^2 - \tmp)}{\sqrt{y^2 + x^2}}
    \end{align*}

    With a bit simplification, for $p = \begin{pmatrix}x\\y\\z\end{pmatrix}$, we have

    $$\vec{F}(p) =
      \left\{
      \begin{array}{lcr}
        0, & \quad & p \notin D \\ & & \\
        \frac{r^2 - \tmp}{\sqrt{y^2 + x^2}}
        \begin{bmatrix}-y\\x\\0\end{bmatrix},
           & \quad & p \in D
      \end{array}
      \right.$$

    One should note the assumption that the direction of flow is anticlockwise, which means
    $v_y$ is negative and $v_y$ is positive when both $x$ and $y$ is positive. If the flow
    is clockwise, the result should be ``inversed":

    $$\vec{F}(p) =
      \left\{
      \begin{array}{lcr}
        0, & \quad & p \notin D \\ & & \\
        \frac{r^2 - \tmp}{\sqrt{y^2 + x^2}}
        \begin{bmatrix}y\\-x\\0\end{bmatrix},
           & \quad & p \in D
      \end{array}
      \right.$$

  \end{enumerate}
\end{exercise}