Commit 7769bded authored by Stefan Gehr's avatar Stefan Gehr
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Überall \today mit Datum ersetzt. Unkommitete Aufgaben in submission 12

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\subtitle{Blatt 2}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2020-11-18}
\begin{document}
\maketitle
......
......@@ -12,8 +12,7 @@
\subtitle{Blatt 4}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2020-12-07}
\begin{document}
\maketitle
......
......@@ -12,8 +12,7 @@
\subtitle{Blatt 5}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2020-12-13}
\begin{document}
\maketitle
......
......@@ -12,8 +12,7 @@
\subtitle{Blatt 7}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2021-01-10}
\begin{document}
\maketitle
......
......@@ -12,8 +12,7 @@
\subtitle{Blatt 11}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2020-02-07}
\begin{document}
\maketitle
......
......@@ -12,8 +12,7 @@
\subtitle{Blatt 12}
\author{Stefan Gehr\and Jennifer Lorenz}
\date{\today}
%\date{2020-11-08}
\date{2021-02-13}
\begin{document}
\maketitle
......@@ -69,6 +68,185 @@ Damit gilt \[
&= -1+i+i+1 +1+i+i-1
= 4i
.\end{align*}
\section*{Aufgabe 5}
\subsection*{a}
\subsubsection*{i}
\begin{align*}
\frac{1}{i}\oint_{\abs{z-1}=r} f_1(z)\dd{z}
&= \int_0^{2\pi}f_1\left(1+re^{i\phi}\right)re^{i\phi}\dd{\phi}
= r\int_0^{2\pi} \frac{\left(1+re^{i\phi}\right)^2+1+re^{i\phi}}{r^2e^{2i\phi}\left(3+re^{i\phi}\right)}e^{i\phi}\dd{\phi} \\
&= \frac{1}{r}\int_0^{2\pi} \frac{2+2re^{i\phi}+r^2e^{2i\phi}+re^{i\phi}}{e^{i\phi}(3+re^{i\phi})}\dd{\phi} \\
&= \frac{1}{r}\int_0^{2\pi} \frac{1}{3+re^{i\phi}}\left[2\left(e^{-i\phi}+r\right) + r\left(re^{i\phi}+1\right)\right]\dd{\phi}
.\end{align*}
\begin{align*}
\text{Res}(f_1, 1)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z-1}=r} f_1(z)\dd{z}
= \infty
.\end{align*}
\begin{align*}
\frac{1}{i}\oint_{\abs{z+2}=r} f_1(z)\dd{z}
&= \int_0^{2\pi}f_1\left(-2+re^{i\phi}\right)re^{i\phi}\dd{\phi}
= r\int_0^{2\pi} \frac{\left(-2+re^{i\phi}\right)^2-2+re^{i\phi}}{\left(-3+re^{i\phi}\right)^2re^{i\phi}}e^{i\phi}\dd{\phi} \\
&= \int_0^{2\pi} \frac{4-4re^{i\phi}+r^2e^{2i\phi}-2+re^{i\phi}}{9-6re^{i\phi}+r^2e^{2i\phi}} \dd{\phi}
= \int_0^{2\pi} \frac{2-3re^{i\phi}+r^2e^{2i\phi}}{9-6re^{i\phi}+r^2e^{2i\phi}}\dd{\phi}
.\end{align*}
\begin{align*}
\text{Res}(f_1, -2)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z+2}=r} f_1(z)\dd{z}
= \frac{1}{2\pi}\int_0^{2\pi}\frac{2}{9}\dd{\phi}
= \frac{2}{9}
.\end{align*}
\subsubsection*{ii}
\begin{align*}
\frac{1}{i}\oint_{\abs{z}=r}f_1(z)\dd{z}
&= \int_0^{2\pi}f_1\left(re^{i\phi}\right)re^{i\phi}\dd{\phi}
= \int_0^{2\pi}\frac{\exp\left(\frac{1}{r}e^{-i\phi}\right)}{\left(re^{i\phi}+1\right)^2}re^{i\phi}\dd{\phi} \\
&= \int_0^{2\pi}\frac{\sum_{n=0}^{\infty}\frac{1}{n!}\frac{1}{r^n}e^{-in\phi}}{\left(re^{i\phi}+1\right)^2}re^{i\phi}\dd{\phi} \\
&= \int_0^{2\pi}\frac{1}{\left(re^{i\phi}+1\right)^2}\left[re^i{\phi}+\sum_{n=1}^{\infty}\frac{1}{n!}\frac{1}{r^{n-1}}e^{-i(n-1)\phi}\right]\dd{\phi} \\
&= \int_0^{2\pi}\frac{1}{\left(re^{i\phi}+1\right)^2}\left[re^{i\phi}+\sum_{n=0}^{\infty}\frac{1}{(n+1)!}\frac{1}{r^n}e^{-in\phi}\right]\dd{\phi} \\
&= \int_0^{2\pi}\frac{re^{i\phi}}{\left(re^{i\phi}+1\right)^2} \dd{\phi}
+ \sum_{n=0}^{\infty}\frac{1}{(n+1)!r^n}\int_0^{2\pi}\frac{e^{-in\phi}}{\left(re^{i\phi}+1\right)^2}\dd{\phi}
.\end{align*}
\begin{align*}
\text{Res}(f_2, 0)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z}=r} f_2(z)\dd{z}
= \infty
.\end{align*}
\begin{align*}
\frac{1}{i}\oint_{\abs{z+1}=r}f_2(z)\dd{z}
&= r\int_0^{2\pi}\frac{\exp\left(\frac{1}{-1+re^{i\phi}}\right)}{r^2e^{2i\phi}}e^{i\phi}\dd{\phi}
= \frac{1}{r}\int_0^{2\pi}\exp\left(\frac{1}{-1+re^{i\phi}}\right)e^{-i\phi}\dd{\phi}
.\end{align*}
\begin{align*}
\text{Res}(f_2, -1)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z+1}=r} f_2(z)\dd{z}
= \infty
.\end{align*}
\subsubsection*{iii}
\begin{align*}
f_3(z)
&= \frac{z+3}{z^2-3z+2}
= \frac{z+3}{\left(z-1\right)\left(z-2\right)}
.\end{align*}
\begin{align*}
\frac{1}{i}\oint_{\abs{z-1}=r} f_3(z)\dd{z}
&= r\int_0^{2\pi}f_3\left(1+re^{i\phi}\right)e^{i\phi}\dd{\phi}
= r\int_0^{2\pi}\frac{4+re^{i\phi}}{re^{i\phi}\left(-1+re^{i\phi}\right)}e^{i\phi}\dd{\phi} \\
&= \int_0^{2\pi} \frac{4 + re^{i\phi}}{-1+re^{i\phi}}
.\end{align*}
\begin{align*}
\text{Res}(f_3, 1)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z-1}=r} f_3(z)\dd{z}
= \frac{1}{2\pi}\int_0^{2\pi} (-4)\dd{\phi}
= -4
.\end{align*}
\begin{align*}
\frac{1}{i}\oint_{\abs{z-2}=r} f_3(z)\dd{z}
&= r\int_0^{2\pi}f_3\left(2+re^{i\phi}\right)e^{i\phi}\dd{\phi}
= r\int_0^{2\pi}\frac{5+re^{i\phi}}{\left(1+re^{i\phi}\right)re^{i\phi}}e^{i\phi}\dd{\phi} \\
&= \int_0^{2\pi} \frac{5 + re^{i\phi}}{1+re^{i\phi}}
.\end{align*}
\begin{align*}
\text{Res}(f_3, 2)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z-2}=r} f_3(z)\dd{z}
= \frac{1}{2\pi}\int_0^{2\pi} 5\dd{\phi}
= 5
.\end{align*}
\subsubsection*{iv}
Für \(n\in \mathbb{Z}\) liegen die Polstellen von \(f_4\) bei \(z_n = i\pi \left(1 + 2n\right)\).
\begin{align*}
\frac{1}{i}\oint_{\abs{z+z_n}=r}f_4(z)\dd{z}
&= r\int_0^{2\pi} f_4\left(z_n + re^{i\phi}\right)e^{i\phi}\dd{\phi}
= r\int_0^{2\pi} \left[1 + \exp\left(z_n + re^{i\phi}\right)\right]^{-1}e^{i\phi}\dd{\phi} \\
&= r\int_0^{2\pi} \left[1 + \exp(z_n) \exp\left(re^{i\phi}\right)\right]^{-1}e^{i\phi}\dd{\phi}
= r\int_0^{2\pi} \left[1 - \exp\left(re^{i\phi}\right)\right]^{-1}e^{i\phi}\dd{\phi} \\
&= r\int_0^{2\pi} \frac{e^{i\phi}}{\sum_{n=1}^{\infty}\frac{1}{n!}r^ne^{in\phi}} \dd{\phi}
= \int_0^{2\pi} \left[\sum_{n=1}^{\infty}\frac{1}{n!}r^{n-1}e^{i(n-1)\phi}\right]^{-1}\dd{\phi} \\
&= \int_0^{2\pi} \left[\sum_{n=0}^{\infty}\frac{1}{(n+1)!}r^ne^{in\phi}\right]^{-1}
.\end{align*}
\begin{align*}
\text{Res}(f_4, z_n)
&= \lim_{r\searrow 0} \frac{1}{2\pi i}\oint_{\abs{z+z_n}=r} f_4(z)\dd{z}
= \infty
.\end{align*}
\subsection*{b}
\begin{align*}
\oint_{\abs{z}=r} \frac{z+3}{z^2-3z+2}\dd{z}
&= ri\int_0^{2\pi} \frac{re^{i\phi}+3}{r^2e^{2i\phi}-3re^{i\phi}+2}e^{i\phi}\dd{\phi} \\
&= ri\int_0^{2\pi} \left[\frac{re^{2i\phi}}{r^2e^{2i\phi}-3re^{i\phi}+2}
+ \frac{3e^{i\phi}}{r^2e^{2i\phi}-3re^{i\phi}+2}\right]\dd{\phi}
.\end{align*}
Wolframalpha sagt nein. Aber für \(r\searrow 0\) erhält man
\begin{align*}
\lim_{r\searrow 0}\oint_{\abs{z}=r} \frac{z+3}{z^2-3z+2}\dd{z}
&= 0
.\end{align*}
\section*{Aufgabe 6}
\subsection*{a}
\begin{align*}
\int_{\gamma_{2,1}} \frac{z^7}{z^2(z^4+1)}\dd{z}
&= 0
.\end{align*}
\subsection*{b}
\begin{align*}
\int_{\gamma_{0,\frac{1}{2}}} \frac{e^{1-z}}{z^3(1-z)}\dd{z}
&= \int_0^{2\pi} \frac{\exp\left(1-\frac{1}{2}e^{it}\right)}{\frac{1}{8}e^{3it}\left(1-\frac{1}{2}e^{it}\right)}\frac{1}{2}ie^{it}\dd{t}
= 4i\int_0^{2\pi} \frac{\exp\left(1-\frac{1}{2}e^{it}\right)}{e^{2it}\left(1-\frac{1}{2}e^{it}\right)}\dd{t} \\
&= 4i\int_0^{2\pi} e^{-2it} \frac{1}{1-\frac{1}{2}e^{it}}
\sum_{n=0}^{\infty}\frac{1}{n!}\left(1-\frac{1}{2}e^{it}\right)^n \\
&= 4i\int_0^{2\pi} e^{-2it}\left[\frac{1}{1-\frac{1}{2}e^{it}} +
\sum_{n=0}^{\infty}\frac{1}{(n+1)!}\left(1-\frac{1}{2}e^{it}\right)^n\right]
= 0
.\end{align*}
\subsection*{c}
\begin{align*}
\oint_{\gamma_{1,\frac{1}{2}}} \frac{e^{1-z}}{z^3(1-z)}\dd{z}
&= \frac{1}{2}i\int_0^{2\pi} \frac{\exp\left(\frac{1}{2}e^{it}\right)}{\left(1+\frac{1}{2}e^{it}\right)^3\frac{1}{2}e^{it}}
e^{it}\dd{t}
= i \int_0^{2\pi} \frac{\exp\left(\frac{1}{2}e^{it}\right)}{\left(1+\frac{1}{2}e^{it}\right)^3}\dd{t} \\
= 2\pi i
.\end{align*}
Das sagt zumindest Wolframalpha. Keine Ahnung wie das sonst gehen sollte?
\section*{Aufgabe 7}
\[
f(x) = \frac{1}{x^4+4}
.\]
\begin{align*}
\int_{-\infty}^{\infty}f(x)\dd{x}
&= \lim_{r\to\infty}\left[\int_{\gamma_r}f(z)\dd{z} - \int_{\gamma_{2,r}}f(z)\dd{z}\right]
,\end{align*}
wobei \(\gamma\) eine Hintereinanderverkettung von \(\gamma_{1,r}\) und \(\gamma_{2,r}\) ist und
\begin{align*}
\gamma_{1,r}: [-r,r] &\longrightarrow \mathbb{C} \\
z &\longmapsto \gamma_{1,r}(z) = z
\end{align*}
\begin{align*}
\gamma_{2,r}: [0,\pi] &\longrightarrow \mathbb{C} \\
t &\longmapsto \gamma_{2,r}(t) = re^{it}
.\end{align*}
Die Polstellen im von \(\gamma_r\) eingeschlossenen Bereich lauten
\begin{align*}
z_1 = 1+i
\qquad z_2 = 1-i
.\end{align*}
Beide davon werden umkreist, also
\begin{align*}
\int_{\gamma_r}f(z)\dd{z}
&= 4\pi i
.\end{align*}
Es bleibt noch
\begin{align*}
\lim_{r\to\infty}\int_{\gamma_{2,r}}f(z)\dd{z}
&= \lim_{r\to\infty}ri\int_0^{\pi} \frac{e^{i\phi}}{r^4e^{4i\phi}+4}\dd{\phi}
= \lim_{r\to\infty}i\int_0^{\pi} \frac{re^{i\phi}}{r^4e^{4i\phi}+4}\dd{\phi}
= 0
.\end{align*}
Also bleibt nur
\begin{align*}
\int_{-\infty}^{\infty} \frac{1}{x^4+4}\dd{x}
&= 4\pi i
.\end{align*}
Ja das stimmt nicht. Irgendwas hab ich da falsch verstanden.
\end{document}
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