heureka

Suppose that f is a Mobius transformation such that f(1)=i, f(i)=-1, and f(-1)=1.
Find the value of f(-i).

Suppose that f is a Mobius transformation such that \(f(1)=i\), \(f(i)=-1\), and \(f(-1)=1\).
Find the value of \(f(-i)\).

Möbius transformation: \(w={\dfrac {az+b}{cz+d}}\) of one complex variable z; here the coefficients a, b, c, d are complex number.

\( a=\det {\begin{pmatrix}z_{1}w_{1}&w_{1}&1\\z_{2}w_{2}&w_{2}&1\\z_{3}w_{3}&w_{3}&1\end{pmatrix}}\, \\ b=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&w_{1}\\z_{2}w_{2}&z_{2}&w_{2}\\z_{3}w_{3}&z_{3}&w_{3}\end{pmatrix}}\, \\ c=\det {\begin{pmatrix}z_{1}&w_{1}&1\\z_{2}&w_{2}&1\\z_{3}&w_{3}&1\end{pmatrix}}\, \\ d=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&1\\z_{2}w_{2}&z_{2}&1\\z_{3}w_{3}&z_{3}&1\end{pmatrix}}\)

Adult tickets to a basketball game cost $5.
Student tickets cost $1.
A total of $2,783 was collected on the sale of 1,235 tickets.
How many of each type of ticket were sold?

\(\text{Let number of adult tickets $=a$} \\ \text{Let number of student tickets $=s$}\)

\(\begin{array}{|rcll|} \hline a+s &=& 1235 \\ \mathbf{s} &=& \mathbf{1235-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 5a+s &=& 2783 \\ 5a + (1235-a) &=& 2783 \\ 5a + 1235-a &=& 2783 \\ 4a + 1235 &=& 2783 \\ 4a &=& 2783-1235 \\ 4a &=& 1548 \quad | \quad : 4 \\ \mathbf{a} &=& \mathbf{387} \\\\ \mathbf{s} &=& \mathbf{1235-a} \\ s &=& 1235-387 \\ \mathbf{s} &=& \mathbf{848} \\ \hline \end{array}\)

The number of adult tickets is 387
The number of student tickets is 848

How many two digit positive integers in the base-four representation are
twice the two digit positive integers in the base-three representation?

I assume:

\(\begin{array}{|rl|c|c|c|c|c|c|l|} \hline & & 10 & 11 & 12 & 20 & 21 & 22 & \text{base}~3 \\ & & =3 & =4 & =5 & =6 & =7 & =8 & \text{base}~10 \\ \text{base}~4 & \text{base}~10 & 6 & 8 & 10 & 12 & 14 & 16 & 2x \\ \hline 10 & = 4 & & & & & & & \\ 11 & = 5 & & & & & & & \\ 12 & = 6 &6=6\checkmark& & & & & & \\ 13 & = 7 & & & & & & & \\ 20 & = 8 & &8=8\checkmark& & & & & \\ 21 & = 9 & & & & & & & \\ 22 & = 10 & & &10=10\checkmark& & & & \\ 23 & = 11 & & & & & & & \\ 30 & = 12 & & & &12=12\checkmark& & & \\ 31 & = 13 & & & & & & & \\ 32 & = 14 & & & & &14=14\checkmark& & \\ 33 & = 15 & & & & & & & \\ \hline \end{array}\)

The answer is 5.

\(\begin{array}{|ll|} \hline 1. &12_4 = 2\times 10_3 \\ 2. &20_4 = 2\times 11_3 \\ 3. &22_4 = 2\times 12_3 \\ 4. &30_4 = 2\times 20_3 \\ 5. &32_4 = 2\times 21_3 \\ \hline \end{array}\)

Suppose we have a square
with vertices at \((0),~ (-6+13i),~ (-19+7i) ,~~\text{and}~ (-13-6i)\).
Suppose that we want to multiply these points
by a single complex number a+bi to get a square
with vertices \((0),~ (8+2i),~ (6+10i),~ ~\text{and}~ (-2+8i)\).
What is (a,b)?

I assume:

We have a black square with vertices: \(\mathbf{z =|z|e^{i\varphi}}\)

We have a blue square with vertices: \(\mathbf{w =|w|e^{i\omega}}\)

We have a single complex number: \(\mathbf{x=a+bi}\)

Formula: \(\mathbf{w = x\cdot z \qquad \text{with}\qquad x = \dfrac{|w|}{|z|}e^{i(\omega-\varphi)} }\)

\(\text{For example we set $z=(-6+13i) \quad \text{and} \quad w=(8+2i)$ }\)

\(\begin{array}{|rcll|} \hline |z| &=& \sqrt{ \left(-6\right)^2+13^2 } \\ &=& \sqrt{36+169} \\ &=& \sqrt{205} \\ \mathbf{|z|} &=& \mathbf{14.3178210633}\\\\ |w| &=& \sqrt{8^2+2^2} \\ &=& \sqrt{68} \\ \mathbf{|w|} &=& \mathbf{8.2462112512}\\\\ \dfrac{|w|}{|z|}&=& \dfrac{8.2462112512}{14.3178210633} \\\\ \mathbf{\dfrac{|w|}{|z|}}&=& \mathbf{0.5759403763} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \omega-\varphi &=& \arctan\left(\dfrac{2}{8}\right) -\arctan\left(\dfrac{13}{-6}\right) \\ &=& 14.03624346791^\circ - (180^\circ-65.2248594312^\circ) \\ &=& 14.03624346791^\circ - 114.775140569^\circ \\ \mathbf{\omega-\varphi} &=& \mathbf{-100.738897101^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{|w|}{|z|}*e^{i(\omega-\varphi)}} \\\\ x &=& 0.5759403763*e^{i(-100.738897101^\circ)} \\\\ x &=& 0.5759403763* \Big( \cos(-100.738897101^\circ)+i*\sin(-100.738897101^\circ) \Big) \\ x &=& 0.5759403763*(-0.1863336512-0.9824865243i) \\ x &=&-0.1073170732-0.5658536585i \\\\ \mathbf{a} &=& \mathbf{-0.1073170732} \\ \mathbf{b} &=& \mathbf{-0.5658536585} \\ \hline \end{array}\)

\((a,~b) = (-0.1073170732,~-0.5658536585)\)

Four houses in a row are each to be painted with one of the colors red, blue, green, and yellow.

In how many different ways can the houses be painted so that no two adjacent houses are of the same color?

\(\begin{array}{|c|c|c|} \hline \text{House} & \text{selectable colors}& \text{number of selectable colors} \\ \hline A & red,~ blue,~ green,~ yellow & 4 \\ \hline B & not~A & 3 \\ \hline C & not~B & 3 \\ \hline D & not~C & 3 \\ \hline && \text{product}~ 4\times 3 \times 3 \times 3 = 108 \\ \hline \end{array} \)

The houses can be painted in 108 different ways.

At an all-you-can-eat barbequefundraiser that you are sponsoring adults pay $6 for a dinner and children pay $4 for a dinner.
212 people attend and you raise $1128.
What is the total number of adults and the total number of children attending?

\(\text{Let number of adults $=a$} \\ \text{Let number of children $=c$}\)

\(\begin{array}{|rcll|} \hline a+c &=& 212 \\ \mathbf{c} &=& \mathbf{212-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 6a+4c &=& 1128 \\ 6a +4(212-a) &=& 1128 \\ 6a + 4*212-4a &=& 1128 \\ 6a + 848-4a &=& 1128 \\ 2a &=& 1128-848 \\ 2a &=& 280 \quad | \quad : 2 \\ \mathbf{a} &=& \mathbf{140} \\\\ \mathbf{c} &=& \mathbf{212-a} \\ c &=& 212-140 \\ \mathbf{c} &=& \mathbf{72} \\ \hline \end{array}\)

The total number of adults is 140
The total number of children is 72

Five times the second of three consecutive even integers is six more than twice the sum of the first and third integer.

Find the middle even integer.

I assume:

\(\text{Let the first even integer $=2(n-1)$ } \\ \text{Let the middle even integer $=2n$ } \\ \text{Let the third even integer $=2(n+1)$ }\)

\(\begin{array}{|rcll|} \hline 5(2n) &=& 6+2\Big( 2(n-1)+2(n+1) \Big) \\ 10n &=& 6+2( 2n-2+2n+2 ) \\ 10n &=& 6+2*4n \\ 10n &=& 6+8n \\ 10n-8n &=& 6 \\ \mathbf{2n} &=& \mathbf{6} \\ \hline \end{array}\)

The middle even integer is 6

Jason has 43 stamps. Some are worth 15 cents and some are worth 20 cents.
If their total value is $7.50, how many of each kind does he have?

\(\text{Let stamps with worth $20$ cents $= x$ }\\ \text{Let stamps with worth $15$ cents $= y$ } \)

\(\begin{array}{|rcll|} \hline x+y &=& 43 \\ \mathbf{y} &=& \mathbf{43-x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 20x+15y &=& 750 \quad | \quad \mathbf{y=43-x} \\ 20x+15(43-x)&=& 750 \\ 20x+15*43-15x &=& 750 \\ 20x+645-15x &=& 750 \\ 5x +645 &=& 750 \\ 5x &=& 750 - 645 \\ 5x &=&105 \quad | \quad : 5 \\ \mathbf{x} &=& \mathbf{21} \\\\ \mathbf{y} &=& \mathbf{43-x} \\ y &=& 43-21 \\ \mathbf{y} &=& \mathbf{22} \\ \hline \end{array}\)

Jason does have 21 stamps with worth of 20 cents and
he does have 22 stamps with worth of 15 cents.

The rectangle ABCD has vertices at A = (1, 2, 3), B = (3, 6, −2), and C = (0, 2, −6).

Determine the coordinates of vertex D.

\(\begin{array}{|rcll|} \hline \mathbf{\vec{B}-\vec{A}} &=& \mathbf{\vec{C}-\vec{D}} \\\\ \left( \begin{array}{r} 3 \\ 6 \\ -2 \end{array}\right) - \left( \begin{array}{r} 1 \\ 2 \\ 3 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} 3-1 \\ 6-2 \\ -2-3 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} 2 \\ 4 \\ -5 \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0 \\ 2 \\ -6 \end{array}\right) - \left( \begin{array}{r} 2 \\ 4 \\ -5 \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0-2 \\ 2-4 \\ -6-(-5) \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} 0-2 \\ 2-4 \\ -6+5 \end{array}\right) \\\\ \left( \begin{array}{r} x_D \\ y_D \\ z_D \end{array}\right) &=& \left( \begin{array}{r} -2 \\ -2 \\ -1 \end{array}\right) \\ \hline \end{array}\)

\(\mathbf{\vec{D} =} \left( \begin{array}{r} \mathbf{-2} \\ \mathbf{-2} \\ \mathbf{-1} \end{array}\right)\)