heureka

Let $f(z)$ be a rotation by $\pi/2$ counterclockwise around $2+i$, as in the picture below:

\(\text{Let $z_0 = 2+i$}\)

\(\begin{array}{|rcll|} \hline f(z)-z_0 &=& (z-z_0)\left( \cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2} \right) \right) \\ f(z)-z_0 &=& (z-z_0)\left( 0 + i\cdot 1 \right) \\ f(z)-z_0 &=& (z-z_0)i \\ \mathbf{f(z)} & \mathbf{=}& \mathbf{z_0 + (z-z_0)i} \\\\ f(z) &=& z_0 + (z-z_0)i \quad | \quad z_0 = 2+i \\ f(z) &=& 2+i + \Big(z-(2+i)\Big)i \\ f(z) &=& 2+i + (z-2-i)i \\ f(z) &=& 2+i + iz-2i-i^2 \quad | \quad i^2 = -1 \\ f(z) &=& 2+i + iz-2i+1 \\ f(z) &=& iz+3-i \quad | \quad \text{compare with }f(z) = az + b \\ \\ \mathbf{(a,b)} & \mathbf{=} & \mathbf{(i,3-i)} \\ \hline \end{array} \)

Compute the sum

\(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)

The equation of the line through $2+3i$ which is perpendicular to the line through $0$ and $2+3i$ has equation \[az + b\overline{z} = 26,\]

where $a$ and $b$ are constant complex numbers.

Find the product $ab$ in rectangular form.

\(\begin{array}{|rcll|} \hline az + b\overline{z} &=& 26 \\ &\boxed{ z = b} \\ & \boxed{\overline{z} = a} \\ ab + ba &=& 26 \\ 2ab &=& 26 \\ \mathbf{ab} &\mathbf{=}& \mathbf{13} \\ \hline \end{array}\)

The equation of the line through $2 + 3i$ and $0$ can be written as \[az + b \overline{z} = 0\]for some complex numbers $a$ and $b$.

Find the quotient $b/a$ in rectangular form.

\(\begin{array}{|rcll|} \hline az + b \overline{z} &=& 0 \\ &&\boxed{ z = 2+3i} \\ && \boxed{\overline{z} = 2-3i} \\ a(2+3i) + b (2-3i) &=& 0 \\ b (2-3i) &=& -a(2+3i) \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)}\cdot\dfrac{(2+3i)} {(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)(2+3i)} {(2-3i)(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i+9i^2)} {4-9i^2} \quad & | \quad i^2=-1 \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i-9)} {4+9} \\\\ \dfrac{b}{a}&=& \dfrac{-(-5+12i)} {13} \\\\ \dfrac{b}{a}&=& \dfrac{5-12i} {13} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{=}& \mathbf{\dfrac{5} {13} -\dfrac{12} {13}i} \\ \hline \end{array}\)

Consider the line with equation

\( (2-i)z + (2+i)\overline{z} = 20. \)

 \[(2-i)z + (2+i)\overline{z} = 20.\]
Where does this line intersect the real axis?

On the real axis:  \(z = \overline z\)

\(\begin{array}{|rcll|} \hline (2-i)z + (2+i)\overline{z} &=& 20 \quad & | \quad z = \overline z \\ (2-i)z + (2+i)z &=& 20 \\ z\Big((2-i)+(2+i)\Big) &=& 20 \\ z(2-i+2+i) &=& 20 \\ 4z &=& 20 \\ \mathbf{ z } & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

The line intersect the real axis at 5.

Wie bestimme ich den Grenzwert ?

4c.)

\(\displaystyle \lim \limits_{x \to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \)

Formel: \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} f(x)\right) &=& \displaystyle\lim \limits_{x\to 0} \ln f(x) \\ \hline \end{array}\)
 

Wir logarithmieren:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right) &=& \displaystyle\lim \limits_{x\to 0} \ln \left( \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right)\\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \ln \left( 1+\arctan(x) \right)}{x} \qquad \text{Bernoulli / de l'Hospital} \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \Big( \ln \left( 1+\arctan(x) \right) \Big)'}{ (x)'} \\ && \Big( \ln \left( 1+\arctan(x) \right) \Big)' = \dfrac{\dfrac{1}{x^2+1}} {1+\arctan(x)} \qquad (x)' = 1 \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \dfrac{\dfrac{1}{x^2+1}} {1+\arctan(x)}}{ 1} \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{1} {(x^2+1)\Big(1+\arctan(x)\Big)} \\\\ &=& \dfrac{1} {(0^2+1)\Big(1+\arctan(0)\Big)} \\\\ &=& \dfrac{1} { 1\cdot(1+0)} \\\\ &=& \mathbf{1} \\ \hline \end{array} \)

Das Logarithmieren wird zurückgesetzt:

\(\begin{array}{|rcll|} \hline e^{\ln \left(\displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right) } &=& e^1 \\\\ \boxed{\mathbf{ \displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} = e}} \\ \hline \end{array}\)

Wie bestimme ich den Grenzwert ?

4b.)

\(\displaystyle \lim \limits_{x\searrow 0} x^x \qquad \text{von rechts annähern}\)

mit \(\begin{array}{|rcll|} \hline x &=& 0+\dfrac{1}{n} \\ &=& \dfrac{1}{n} \\ \hline \end{array}\)

erhalten wir :  \(\displaystyle \lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \)

Formel : \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to \infty} f(x)\right) &=& \displaystyle\lim \limits_{x\to \infty} \ln f(x) \\ \hline \end{array}\)

Wir logarithmieren:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right) &=& \displaystyle\lim \limits_{n\to \infty} \ln \left( \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right)\\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{1}{n}\cdot \ln \left( \dfrac{1}{n} \right)\\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{ \ln \left(\dfrac{1}{n}\right) }{n} \qquad \text{Bernoulli / de l'Hospital} \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{ \Big(\ln \left(\dfrac{1}{n}\right)\Big)' }{(n)'} \quad | \quad \Big(\ln \left(\dfrac{1}{n}\right)\Big)' = \dfrac{-\dfrac{1}{n^2}}{\dfrac{1}{n}}=-\dfrac{1}{n} \qquad (n)' = 1 \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \left( \dfrac{ -\dfrac{1}{n} }{1}\right) \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) \\ \hline \end{array}\)

Das Logarithmieren wird zurückgesetzt:

\(\begin{array}{|rcll|} \hline e^{ \ln \left(\displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right) } &=& e^{\displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) } \\\\ \displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} &=& e^{\displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) } \quad | \quad \displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) = -0 \\\\ \displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} &=& e^{-0} \\\\ &=& e^{0} \\\\ &=& \mathbf{1} \\\\ && \boxed{ \mathbf{\displaystyle \lim \limits_{x\searrow 0} x^x = 1 } } \\ \hline \end{array}\)

The ratios of the measures of the sides of a triangle is 2:10:7.
Find the length of each side if the perimeter of the triangle is 228 meters.

\(\begin{array}{|rclcl|} \hline a+b+c &=& 228 \quad & | & \quad \dfrac{c}{a} = \dfrac{7}{2} \Rightarrow \boxed{c=\dfrac{7}{2}a } \\ & & \quad & | & \quad \dfrac{a}{b} = \dfrac{2}{10} \Rightarrow \boxed{b=\dfrac{10}{2}a } \\ a+\dfrac{10}{2}a+\dfrac{7}{2}a &=& 228 \\ \dfrac{19}{2}a &=& 228 \\ a &=& \dfrac{228\cdot 2}{19} \\ \mathbf{a} & \mathbf{=} & \mathbf{24} \\\\ 24+b+c &=& 228 \\ b+c &=& 228-24 \\ b+c &=& 204 \quad & | & \quad \dfrac{b}{c} = \dfrac{10}{7} \Rightarrow \boxed{b=\dfrac{10}{7}c } \\ \dfrac{10}{7}c+c &=& 204 \\ \dfrac{17}{7}c &=& 204 \\ c &=& \dfrac{204\cdot 7}{17} \\ \mathbf{c} & \mathbf{=} & \mathbf{84} \\\\ b &=& \dfrac{10}{2}a\\ b &=& 5a\\ b &=& 5\cdot 24 \\ \mathbf{b} & \mathbf{=} & \mathbf{120} \\ \hline \end{array} \)

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 .
She turns the cards face-down, shuffles the cards until their order is random,
then pulls the top two cards off the deck.
What is the probability that at least one of those two cards shows a square number?

\(\color{red}\text{square number}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{card 2} & {\color{red}1} & 2 & 3 & {\color{red}4} & 5 & 6 \\ \hline \text{card 1} & & & & & & & \\ \hline {\color{red}1} & & - & \times & \times & \times & \times & \times \\ \hline 2 & & \times & - & & \times & & \\ \hline 3 & & \times & & - & \times & & \\ \hline {\color{red}4} & & \times & \times & \times & - & \times & \times \\ \hline 5 & & \times & & & \times & - & \\ \hline 6 & & \times & & & \times & & - \\ \hline \end{array}\)

\(\text{The probability is $\dfrac{18}{30} = \dfrac{3}{5} \quad (60\ \%) $ } \)

Meyer rolls two fair, ordinary dice with the numbers  on their sides.
What is the probability that at least one of the dice shows a square number?

\(\color{red}\text{square number} \)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{dice 2} & {\color{red}1} & 2 & 3 & {\color{red}4} & 5 & 6 \\ \hline \text{dice 1} & & & & & & & \\ \hline {\color{red}1} & & \times & \times & \times & \times & \times & \times \\ \hline 2 & & \times & & & \times & & \\ \hline 3 & & \times & & & \times & & \\ \hline {\color{red}4} & & \times & \times & \times & \times & \times & \times \\ \hline 5 & & \times & & & \times & & \\ \hline 6 & & \times & & & \times & & \\ \hline \end{array} \)

\(\text{The probability is $\dfrac{20}{36} = \dfrac{5}{9} \quad (55.6\ \%) $ } \)