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\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\im}[1]{\mbox{Im }#1}
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\date{\today{}}
\author{Thomas Kuster}
\title{UWIS, Mathematik 1, Lösung Serie 11}

\begin{document}
\markright{UWIS, Mathematik 1, Lösung Serie 11}
\pagestyle{myheadings}

\maketitle

\section{Bestimme $\re{z}$, $\im{z}$, $\abs{z}$, $\arg{z}$}
\subsection{}
\begin{eqnarray*}
\abs{\frac{1}{z}} = 3 & \Rightarrow & \abs{z} = \frac{1}{3}\\
\arg{\bar{z}} = \frac{4\pi}{3} & \Rightarrow & \arg{z} = \frac{2\pi}{3}
\end{eqnarray*}
\begin{eqnarray*}
\frac{a}{\frac{1}{3}} & = & \cos{\frac{2\pi}{3}}\\
3a & = & -\frac{1}{2}\\
a & = & -\frac{1}{6}\\
\sqrt{a^2+b^2} & = & \frac{1}{3}\\
\left(-\frac{1}{6}\right)^2 + b^2 = \frac{1}{36}+b^2 & = & \frac{1}{9}\\
b^2 & = & \frac{1}{9}-\frac{1}{36} = \frac{4-1}{36}\\
b & = & \sqrt{\frac{3}{36}} =  \sqrt{\frac{1}{12}}\\
\re{z} & = & -\frac{1}{6}\\
\im{z} & = & \frac{1}{\sqrt{12}}
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
z & = & \frac{-1 + i\sqrt{3}}{1+i\sqrt{3}} = \frac{(-1 + i\sqrt{3})(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})} = \frac{-1 + 2i\sqrt{3}-3i^2}{1-3i^2}\\
z & = & \frac{2 + 2i\sqrt{3}}{4} = \frac{1}{2}+\frac{\sqrt{3}}{2}i\\
\re{z} & = & \frac{1}{2}\\
\im{z} & = & \frac{\sqrt{3}}{2}
\end{eqnarray*}
\begin{eqnarray*}
\abs{z} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} & = & 1\\
\arg{z} = \arctan{\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}} = \arctan{\sqrt{3}} & = & \frac{\pi}{3}\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\re{(i\bar{z})} = \re{\left(i(a-ib)\right)} = \re{(ai-bi^2)} = \re{ai+b} & = & 3\\
b & = & 3\\
\abs{i\bar{z}} = \abs{i(a-ib)} = \abs{ia+b} = \sqrt{a^2+b^2} = \sqrt{a^2 + 3^2} & = & 5\\
a^2 + 9 & = & 25\\
a^2 & = & 16\\
a & = & \pm4\\
\re{z} & = & \pm4\\
\im{z} & = & 3\\
\abs{z} & = & 5\\
\arg{z} & = & \pm\arctan{\frac{3}{4}}
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
2 & = & \im{\frac{z}{i}} = \im{\frac{a+ib}{i}} = \im{\frac{(a+ib)i}{i^2}} = \im{\frac{ia-b}{-1}} = \im{(b-ia)}\\
a & = & -2\\
-\frac{\pi}{3} & = & \arg{(i\bar{z})} = \arg{i(a+ib)} = \arg{(ia-b)} = \arg{(-2i-b)}\\
\frac{\pi}{3} & = & \arg{(-b+2i)}\\
\frac{2}{-b} & = & \tan{\frac{\pi}{3}} = \sqrt{3}\\
b & = & -\frac{2}{\sqrt{3}}\\
\re{z} & = & -2\\
\im{z} & = & -\frac{2}{\sqrt{3}}\\
\abs{z} & = & \sqrt{(-2)^2+\left(-\frac{2}{\sqrt{3}}\right)^2} = \frac{4}{\sqrt{3}}\\
\arg{z} & = & \arctan{\frac{-\frac{2}{\sqrt{3}}}{-2}} = \arctan{\frac{1}{\sqrt{3}}} = \frac{\pi}{6}
\end{eqnarray*}

\section{Schreibe in der Form $z = a + ib$}
\subsection{}
\subsubsection{$z_1$}
\begin{equation*}\begin{split}
\frac{\sqrt{3}+1-i\left(\sqrt{3}+1\right)}{1+i} = \frac{\left(\sqrt{3}+1-i\sqrt{3}-i\right)(1-i)}{(1+i)(1-i)}&\\
= \frac{\sqrt{3}+1-i\sqrt{3}-i-i\sqrt{3}-i-i^2\sqrt{3}-i^2}{1-i^2} = & \frac{2\sqrt{3}+2-2\sqrt{3}i-2i}{2}\\
= & \sqrt{3}+1+i(-\sqrt{3}-1)
\end{split}\end{equation*}

\subsubsection{$z_2$}
\begin{equation*}\begin{split}
\left(\sqrt{3}+i\right)^6(1-i) = \left(\sqrt{\sqrt{3}^2+1^2}\quad e^{\arctan{\left(\frac{1}{\sqrt{3}}\right)}}\right)^6(1-i)\\
= 2^6 \underbrace{e^{\frac{\pi}{6}6}}_{=-1}(1-i) = -64(1-i) = -64 +i64
\end{split}\end{equation*}

\subsubsection{$z_3$}
$e^{-i\frac{\pi}{4}} \Rightarrow$ gleichschenkliges ($a=b$), rechtwinkliges Dreieck wobei die Hypotenuse die Länge 1 hat.
\begin{eqnarray*}
1^2 = a^2 + a^2\\
\sqrt{\frac{1}{2}} & = & a\\
a + ib & = & \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}
\end{eqnarray*}

\subsubsection{$z_4$}
\begin{eqnarray*}
\frac{1}{e^{i\frac{2\pi}{3}}} & = & e^{-i\frac{2\pi}{3}} = e^{i\frac{4\pi}{3}}\\
\im{z} & = & sin{\frac{4\pi}{3}} = -\frac{\sqrt{3}}{2}\\
\re{z} & = & cos{\frac{4\pi}{3}} = -\frac{1}{2}\\
a + ib & = & -\frac{\sqrt{3}}{2} -i\frac{1}{2}
\end{eqnarray*}

\subsubsection{$z_5$}
\begin{eqnarray*}
i\underbrace{e^{-i\frac{\pi}{6}}}_{z_1}\\
\im{z_*} & = & sin{-\frac{\pi}{6}} = -\frac{1}{2}\\
\re{z_*} & = & cos{-\frac{\pi}{6}} = \frac{\sqrt{3}}{2}\\
i\left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) & = & i\frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}
\end{eqnarray*}

\subsubsection{$z_6$}
\begin{displaymath}
i^i = \left(e^{i\frac{\pi}{2}}\right)^i = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}} \approx 0.209
\end{displaymath}

\subsection{}
\subsubsection{}
\begin{displaymath}
\arg{z_1} = \arctan{\left(\frac{-\sqrt{3}-1}{\sqrt{3}+1}\right)} = \arctan{-1} = \frac{\pi}{4}
\end{displaymath}

\subsubsection{}
\begin{equation*}\begin{split}
{z_3}^2z_4 = \left(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right)^2\left(-\frac{\sqrt{3}}{2} -i\frac{1}{2}\right) = \left(\frac{1}{2} -2i\frac{1}{2}+i^2\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2} -i\frac{1}{2}\right)&\\
 = -i\left(-\frac{\sqrt{3}}{2} -i\frac{1}{2}\right) = i\frac{\sqrt{3}}{2} +i^2\frac{1}{2} =\-\frac{1}{2} + i\frac{\sqrt{3}}{2}&
\end{split}\end{equation*}

\subsubsection{}
\begin{equation*}\begin{split}
\frac{z_4}{z_5} & = \frac{-\frac{\sqrt{3}}{2} -i\frac{1}{2}}{\frac{1}{2} + i\frac{\sqrt{3}}{2}} = -\frac{\left(\frac{\sqrt{3}}{2} +i\frac{1}{2}\right)\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}\\
& = \frac{\frac{\sqrt{3}}{4}-i\frac{3}{4}+i\frac{1}{4}+\frac{\sqrt{3}}{4}}{\frac{1}{4}+\frac{3}{4}} = \frac{2\sqrt{3}}{4}-i\frac{2}{4} = \frac{\sqrt{3}}{2}-i\frac{1}{2}
\end{split}\end{equation*}

\section{Bestimme alle komplexen Zahlen $z$}
\subsection{}
Polynom zweiten Grades $\Rightarrow$ 2 zwei Lösungen.
\begin{eqnarray*}
z^2 = (a+ib)^2 & = & i\\
a^2+2iab-b^2 & = & i\\
a^2-b^2 & = & 0 \Rightarrow a = \pm b\\
2iab & = & i\\
ab & = & \frac{1}{2} \mbox{ da } a = \pm b \Rightarrow a = b\\
a^2 & = & \frac{1}{2}\\
a = b & = & \pm \frac{1}{\sqrt{2}}
\end{eqnarray*}
Lösungen: $\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$ und $-\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$

\subsection{}
Eine Lösung von $z^3 = -1$ kann erraten werden: $(-1)^3 = -1$.Die anderen drei Lösungen sind gleichmässig auf dem Einheitskreis verteilt da $-1 = e^{i\pi}$ ist sind die zwei Lösungen: $e^{i\frac{\pi}{3}}$ und $e^{i\frac{2\pi}{3}}$.
\begin{eqnarray*}
z_1 & = & -1\\
z_2 & = & \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}} = \frac{1}{2} + i\frac{\sqrt{3}}{2}\\
z_3 & = & \frac{1}{2} - i\frac{\sqrt{3}}{2}
\end{eqnarray*}

\subsection{}
\begin{equation*}\begin{split}
z^4 = & \frac{81\left(\sqrt{3}+i\right)}{-\sqrt{3}+i} = \frac{81\left(\sqrt{3}+i\right)\left(-\sqrt{3}-i\right)}{\left(-\sqrt{3}+i\right)\left(-\sqrt{3}-i\right)} = \frac{81\left(-3 -2i\sqrt{3}+1\right)}{3+1}\\
= & \frac{81\cdot-2\left(1+i\sqrt{3}\right)}{4} = -\frac{81}{2}-\frac{i81\sqrt{3}}{2}
\end{split}\end{equation*}
\begin{eqnarray*}
r & = & \sqrt{\frac{81^2}{4}+\frac{3\cdot81^2}{4}} = \sqrt{\frac{81^2(1+3)}{4}} = 81\\
\varphi & = & \arctan{\frac{-\frac{81\sqrt{3}}{2}}{-\frac{81}{2}}} = \arctan{\sqrt{3}} = \frac{\pi}{3}
\end{eqnarray*}
\begin{eqnarray*}
z^4 & = & 81 e^{i\left(\frac{\pi}{3}+2k\pi\right)}\\
z & = & 3 e^{i\left(\frac{\pi}{3\cdot4}+\frac{2k\pi}{4}\right)} = 3 e^{i\left(\frac{\pi}{12}+\frac{6k\pi}{12}\right)}\\
z_1 & = & 3 e^{i\frac{\pi}{12}}\\
z_2 & = & 3 e^{i\frac{7\pi}{12}}\\
z_3 & = & 3 e^{i\frac{13\pi}{12}}\\
z_4 & = & 3 e^{i\frac{19\pi}{12}}\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
4z\bar{z} + (z-\bar{z})^2 & = & 3\\
4(a+ib)(a-ib) + \left((a+ib)-(a-ib)\right)^2 & = & 3\\
4(a^2+b^2)(2ib)^2 & = & 3\\
4a^2+4b^2+4i^2b^2 & = & 3\\
4a^2 + 4b^2 -4b^2 & = & 3\\
4a^2 & = & 3\\
a & = & \pm\frac{\sqrt{3}}{2}\\
b & = & \in \mathbb{R}
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
iz^2+2z+\sqrt{3} = 0\\
z_1 = \frac{-2+\sqrt{2^2-4i\sqrt{3}}}{2i}\\
z_2 = \frac{-2-\sqrt{2^2-4i\sqrt{3}}}{2i}\\
\end{eqnarray*}

\section{Teich mit Forellen}
\subsection{}
\begin{eqnarray*}
V_0 & = & 17280 \mathrm{l} \mbox{zur Zeit } t_0\\
N(t) & = & \mbox{Anzahl Forellen im Teich zur Zeit } t_0\\
N(0) & = & 1864 \mbox{ Forellen zur Zeit } t_0\\
z & = & 12 \mathrm{\frac{l}{min}}\\
a & = & 8 \mathrm{\frac{l}{min}}\\
z_F & = & \frac{2}{40}\mathrm{\frac{Forellen}{l}}\\
k & = & 12 \frac{2}{40} = \frac{3}{5} \mathrm{\frac{Forellen}{min}} \mbox{ Zufluss Forellen}\\
\alpha & = & \mbox{Proportionalitätsfaktor}
\end{eqnarray*}
Wassermenge im Teich ist egal, da die Anzahl abfliessenden Forellen proportinal zur Anzahl Forellen im Teich ist.
\begin{eqnarray*}
N(0+\Delta t) & = & N(0) + \Delta t k - \Delta t N(0) \alpha\\
\frac{N(0+\Delta t) - N(0)}{\Delta t} & = & k - N(0) \alpha\\
\dot{N}(t) & = & k - N(t) \alpha\\
\end{eqnarray*}

\begin{eqnarray*}
N(t) & = & ce^{-\alpha t} + k\\
N(0) = 1864 & = & c\underbrace{e^{-\alpha 0}}_{=1} + \frac{3}{5}\\
1864 & = & c + \frac{3}{5}\\
1864 - \frac{3}{5} & = & c\\
c & = & \frac{9317}{5} = 1863.4\\
N(t) & = & 1863.4e^{-\alpha t} + \frac{3}{5}
\end{eqnarray*}

\subsection{}


\end{document}
