\[ \overline z=\sqrt{3}-i,\quad |z|=\sqrt{3+1}=\sqrt{4}=2. \]

\[ \overline z=7,\quad |z|=7. \]

\[ z=(1+i)(1-i)=1-i+i-i^2=1-i^2=1+1=2 \]

\[ \overline z=2,\quad |z|=2. \]

\[ z=\frac{1}{i}=-i \]

\[ \overline z=i,\quad |z|=1. \]

\[ z=\frac{1}{1+i}=\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2} \]

\[ \overline z=\frac{1}{2}+\frac{i}{2}\,,\quad |z|=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\,. \]

\[ z=\frac{1+2i}{3-i}=\frac{(1+2i)(3+i)}{(3-i)(3+i)}=\frac{3+i+6i+i^2}{9-i^2}=\frac{2+7i}{10}=\frac{1}{5}+\frac{7i}{10} \]

\[ \overline z=\frac{1}{5}-\frac{7i}{10}\,,\quad |z|=\sqrt{\frac{1}{25}+\frac{49}{100}}=\sqrt{\frac{53}{100}}=\frac{\sqrt{53}}{10}\,. \] Damit sind alle gesuchten Größen ermittelt.\( \qquad\Box \)

\begin{align} i^0 &= 0, \\[0.6ex] i^1 &= i, \\[0.6ex] i^2 &= -1, \\[0.6ex] i^3 &= (-1)\cdot i=-i, \\[0.6ex] i^4 &= i^3\cdot i=(-i)\cdot i=-i^2=1, \\[0.6ex] i^5 &= i^4\cdot i=1\cdot i=i, \\[0.6ex] i^6 &= i^5\cdot i=i\cdot i=-1, \\[0.6ex] i^7 &= i^6\cdot i=(-1)\cdot i=-i, \\[0.6ex] i^8 &= i^7\cdot i=(-i)\cdot i=-i^2=1. \end{align}

\[ i^n =\left\{ \begin{array}{rl} 1, & \mbox{falls}\ 4|n \\[0.6ex] i, & \mbox{falls}\ 4|(n-1) \\[0.6ex] -1, & \mbox{falls}\ 4|(n-2) \\[0.6ex] -i, & \mbox{falls}\ 4|(n-3) \\[0.6ex] \end{array} \right.. \]

\[ z=2+5i+8i^2+9i^3+6i^7=2+5i-8-9i-6i=-6-10i. \]

\[ z=i^{10}+i^{14}-i^{18}+(-i)^{23}=-1-1+1-i^{23}=-1+i. \]

\[ z=\left(\frac{1+i}{1-i}\right)^k =\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^k =\left(\frac{1+2i+i^2}{1-i^2}\right)^k =\left(\frac{2i}{2}\right)^k =i^k \]

Damit sind alle gesuchten Größen bestimmt.\( \qquad\Box \)

\[ z_{1/2}=5\pm\sqrt{-15}=5\pm\sqrt{15}\,i. \]

\[ z_1=5-\sqrt{15}\,i,\quad z_2=5+\sqrt{15}\,i. \]

\[ z_{1/2}=\frac{5}{2}\pm\sqrt{\frac{25}{4}-17}=\frac{5}{2}\pm\sqrt{\frac{25}{4}-\frac{68}{4}}=\frac{5}{2}\pm\frac{1}{2}\,\sqrt{-43}=\frac{5}{2}\pm\frac{1}{2}\,\sqrt{43}\,i. \]

\[ z_1=\frac{5}{2}-\frac{1}{2}\,\sqrt{43}\,i,\quad z_2=\frac{5}{2}+\frac{1}{2}\,\sqrt{43}\,i. \] Damit sind alle gesuchten Größen bestimmt.\( \qquad\Box \)

Wir schätzen zunächst wie folgt ab \begin{align} \sum_{k=1}^{2^n-1}\frac{1}{k^2} &= \frac{1}{1^2}+\left(\frac{1}{2^2}+\frac{1}{3^2}\right)+\left(\frac{1}{4^2}+\ldots+\frac{1}{7^2}\right)+\ldots+\left(\frac{1}{(2^{n-1})^2}+\ldots+\frac{1}{(2^n-1)^2}\right) \\[2ex] &\le \frac{1}{1^2}+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\left(\frac{1}{4^2}+\ldots+\frac{1}{4^2}\right)+\ldots+\left(\frac{1}{(2^{n-1})^2}+\ldots+\frac{1}{(2^{n-1})^2}\right) \\[2ex] &= \frac{1}{1^2}+\frac{2}{2^2}+\frac{4}{4^2}+\ldots+\frac{2^{n-1}}{(2^{n-1})^2} \\[2ex] &= \frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^{n-1}} \\[2ex] &= \sum_{k=0}^{n-1}\frac{1}{2^k}\le\sum_{k=0}^\infty\frac{1}{2^k}=\frac{1}{1-\frac{1}{2}}=2\lt\infty\,. \end{align} Also gilt im Grenzfall \[ \sum_{k=1}^\infty\frac{1}{k^2}\lt\infty\,. \] Das war zu zeigen.\( \qquad\Box \)

\[ \sum_{k=1}^n\frac{1}{k^2+1}\le\sum_{k=1}^n\frac{1}{k^2} \]

\[ \sum_{k=1}^\infty\frac{1}{k^2+1}\le\sum_{k=1}^n\frac{1}{k^2}\lt\infty\,. \]

\[ \sum_{k=1}^n\frac{k+1}{k^3+2k^2+k+1} \le\sum_{k=1}^n\frac{2k}{k^3+2k^2+k+1} \le\sum_{k=1}^n\frac{2k}{k^3} =\sum_{k=1}^n\frac{2}{k^2} =2\,\sum_{k=1}^n\frac{1}{k^2} \]

\[ \sum_{k=1}^\infty\frac{k+1}{k^3+2k^2+k+1}\le 2\,\sum_{k=1}^\infty\frac{1}{k^2}\lt\infty\,. \]

\[ \frac{1}{3k^4+k^3+7k^2-k-1}\le\frac{1}{3k^4+1}\quad\mbox{für alle}\ k\in\mathbb N, \]

\[ \begin{array}{ll} & 7\ge 2 \\[0.6ex] \Longrightarrow & k+7\ge 2+k \\[0.6ex] \Longrightarrow & k^3+7k^2\ge 2+k \\[0.6ex] \Longrightarrow & k^3+7k^2-k-1\ge 1 \\[0.6ex] \Longrightarrow & 3k^4+k^3+7k^2-k-1\ge 3k^4+1 \end{array} \]

\[ \frac{1}{3k^4+1}\le\frac{1}{3k^4}\le\frac{1}{k^4}\le\frac{1}{k^2}\quad\mbox{für alle}\ k\in\mathbb N, \]

\[ \sum_{k=1}^n\frac{1}{3k^4+k^3+7k^2-k-1}\le\sum_{k=1}^n\frac{1}{k^2} \]

\[ \sum_{k=1}^\infty\frac{1}{3k^4+k^3+7k^2-k-1}\le\sum_{k=1}^\infty\frac{1}{k^2}\lt\infty\,. \]

\[ \sum_{k=1}^n\frac{1}{k^\alpha}\le\sum_{k=1}^n\frac{1}{k^2} \]

\[ \sum_{k=1}^\infty\frac{1}{k^\alpha}\le\sum_{k=1}^\infty\frac{1}{k^2}\lt\infty\,. \] Damit ist alles gezeigt.\( \qquad\Box \)

\[ \frac{1}{3k}\le\frac{1}{3k-1}\quad\mbox{für alle}\ k\in\mathbb N \]

\[ \sum_{k=1}^n\frac{1}{3k-1}\ge\sum_{k=1}^n\frac{1}{3k}\longrightarrow\infty\quad\mbox{für}\ n\to\infty\,. \]

\begin{align} \sum_{k=1}^n\frac{3k}{k+2} &\ge \sum_{k=1}^n\frac{k}{k+2} =\frac{1}{1+2}+\sum_{k=2}^n\frac{k}{k+2}+\frac{1}{3}+\sum_{k=2}^n\frac{1}{k+k} \\[2ex] &= \frac{1}{3}+\sum_{k=2}^n\frac{1}{2k} =\frac{1}{3}+\frac{1}{2}\,\sum_{k=1}^n\frac{1}{k} \end{align}

\[ \sum_{k=1}^n\frac{3k}{k+2}\longrightarrow\infty\quad\mbox{für}\ n\to\infty\,. \]

\[ \frac{k+1}{k(k+1)}=\frac{1}{k}\quad\mbox{für alle}\ k=1,2,\ldots \]

\[ \sum_{k=1}^n\frac{k+1}{k(k+1)}=\sum_{k=1}^n\frac{1}{k}\longrightarrow\infty\quad\mbox{für}\ n\to\infty\,. \]

\[ \frac{1}{\sqrt{k(k+1)}}\ge\frac{1}{k+1}\,, \]

\[ \frac{1}{k\cdot(k+1)}\ge\frac{1}{(k+1)\cdot(k+1)}\,. \]

\[ \sum_{k=1}^n\frac{1}{\sqrt{k(k+1)}} \ge\sum_{k=1}^n\frac{1}{k+1} =\sum_{k=2}^{n+1}\frac{1}{k} =\sum_{k=1}^{n+1}\frac{1}{k}-1 \longrightarrow\infty\quad\mbox{für}\ n\to\infty\,. \]

\begin{align} -k+\sqrt{k^2+1} &= \frac{(-k+\sqrt{k^2+1})(k+\sqrt{k^2+1})}{k+\sqrt{k^2+1}} =\frac{-k^2+k^2+1}{k+\sqrt{k^2+1}} \\[2ex] &= \frac{1}{k+\sqrt{k^2+1}} \ge\frac{1}{k+k+1} =\frac{1}{2k+1} \ge\frac{1}{2k+2} =\frac{1}{2}\cdot\frac{1}{k+1}\,, \end{align}

\[ \begin{array}{ll} & 0\le 2k \\[0.6ex] \Longrightarrow & k^2+1\le k^2+2k+1=(k+1)^2 \\[0.6ex] \Longrightarrow & \sqrt{k^2+1}\le k+1 \\[0.6ex] \Longrightarrow & k+\sqrt{k^2+1}\le k+k+1 \end{array} \]

\[ \sum_{k=1}^n(-k+\sqrt{k^2+1}) \ge\frac{1}{2}\,\sum_{k=1}^n\frac{1}{k+1} \longrightarrow\infty\quad\mbox{für}\ n\to\infty\,. \] Damit ist alles gezeigt.\( \qquad\Box \)