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Algrebra Help

a) Find the square roots of \( 3+4i\).

\(\begin{array}{rcll} z= a+bi &=& 3+4i \\ |z| &=& \sqrt{a^2+b^2} \\ &=& \sqrt{3^2+4^2} \\ &=& 5 \\ \sin(\varphi) &=& \frac{b}{|z|} \\ &=& \frac{4}{5} \\ &=& 0.8 \\ \varphi &=& \arcsin(0.8) +2k\pi \\\\ z &=& |z|e^{i\varphi} \\ \mathbf{z} & \mathbf{=} & \mathbf{5e^{i\cdot ( \arcsin(0.8)+2k\pi) }} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \sqrt{3+4i} \\ &=& (3+4i)^{\frac12} \\ &=& 5^{\frac12} \left( e^{i\cdot ( \arcsin(0.8)+2k\pi) }\right)^{\frac12} \\ &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)+2k\pi}{2}\right)} \\ &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}+k\pi\right)} \quad & | \quad k =(0,1) \\\\ \sqrt{3+4i} &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}\right)} \quad & | \quad k = 0 \\ &=& \sqrt{5} \cdot \left( \cos( \frac{\arcsin(0.8)}{2} ) + i \cdot \sin(\frac{\arcsin(0.8)}{2}) \right) \\ &=& 2.23606797750 \cdot \left( 0.89442719100 + i \cdot 0.44721359550 \right) \\ &=& 2+i\cdot 1 \\\\ \sqrt{3+4i} &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}+\pi\right)} \quad & | \quad k = 1 \\ &=& \sqrt{5} \cdot \left( \cos( \frac{\arcsin(0.8)}{2}+\pi ) + i \cdot \sin(\frac{\arcsin(0.8)}{2}+\pi) \right) \\ &=& \sqrt{5} \cdot \left( -\cos( \frac{\arcsin(0.8)}{2}) - i \cdot \sin(\frac{\arcsin(0.8)}{2}) \right) \\ &=& 2.23606797750 \cdot \left( -0.89442719100 - i \cdot 0.44721359550 \right) \\ &=& -2-i\cdot 1 \\\\ \hline \end{array}\)

All 2nd roots of \(3+4 i\) :

\(2+i \\ -2-i \)

Compute $1+i+i^2+i^3+i^4+\cdots+i^{2009}$.

Compute \(1+i+i^2+i^3+i^4+\cdots+i^{2009}.\)

geometric sequence: \(a = 1,~ r = i\)

The sum is \(\begin{array}{|rcll|} \hline s &=& 1\cdot \dfrac{i^{2010}-1}{i-1} \\ \hline \end{array}\)

\(\begin{array}{|rclcrcl|} \hline s &=& 1\cdot \left( \dfrac{i^{2010}-1}{i-1} \right) \\ && & i^{2010} &=& i^{2\cdot1005} \\ && & &=& (i^2)^{1005} \quad & | \quad i^2 = -1 \\ && & &=& (-1)^{1005} \\ && & &=& -1 \\ s &=& \left(\dfrac{-1-1}{i-1}\right)\cdot \left(\dfrac{i+1}{i+1}\right) \\\\ &=& \dfrac{(-1-1)(i+1)}{(i-1)(i+1)} \\\\ &=& \dfrac{(-1-1)(i+1)}{(i^2-1)} \quad & | \quad i^2 = -1 \\\\ &=& \dfrac{(-1-1)(i+1)}{(-1-1)} \\\\ &=& i+1 \\ \hline \end{array}\)

\( \mathbf{1+i+i^2+i^3+i^4+\cdots+i^{2009} = 1+i} \)

Find a complex number $z$ such that the real part and imaginary part of $z$ are both integers, and such that $$z\overline z = 89.$$ The z with a line over it would equal to a-bi when z=a+bi

\(\begin{array}{|rcl|rcl|} \hline z\overline z &=& 89 & z\overline z &=& (a+bi)(a-bi)\\ && & &=& a^2+b^2 \\ a^2+b^2 &=& 89 \\ \hline \end{array}\)

The factorisation of \(89\) is \(89^1\), because \(89\) is a prime number.

Because \(89 \equiv 1 \pmod 4\) there are \(4\cdot ( \underbrace{1}_{\text{exponent of } 89} + 1 ) = 8 \) solutions,

\(\begin{array}{|r|r|r|r|r|} \hline & & a & b & z \\ \hline 1 & 8^2 + 5^2 = 89 & 8 & 5 & 8+5i \\ 2 & 5^2 + 8^2 = 89 & 5 & 8 & 5+8i \\ 3 & (-8)^2 + 5^2 = 89 & -8 & 5 & -8+5i \\ 4 & 5^2 + (-8)^2 = 89 & 5 & -8 & 5-8i \\ 5 & 8^2 + (-5)^2 = 89 & 8 & -5 & 8-5i \\ 6 & (-5)^2 + 8^2 = 89 & -5 & 8 & -5+8i \\ 7 & (-8)^2 + (-5)^2 = 89 & -8 & -5 & -8-5i \\ 8 & (-5)^2 + (-8)^2 = 89 & -5 & -8 & -5-8i \\ \hline \end{array} \)

The sum of 18 consecutive odd integers is a perfect fifth power of n.

If p is the smallest first number of the series,

then the product np=?

arithmetic progression:

\(\begin{array}{|lcll|} \hline p + 0 \cdot 2,~ p + 1 \cdot 2,~ p+ 2 \cdot 2,~ p+ 3 \cdot 2,~\ldots,~ p+ 17 \cdot 2 \\ p,~ p +2,~ p+ 4,~ p+ 6,~ \ldots,~ p+ 34 \\ \hline \end{array}\)

sum of a arithmetic progression:

\(\begin{array}{|rcll|} \hline && a_1 = p \qquad a_m = p+34 \qquad d = 2 \\\\ s_m &=& \left( \frac{a_1+a_m}{2} \right) \cdot m \\ \hline \end{array} \)

\(\begin{array}{rcll} s_{18} &=& \left( \frac{p+(p+34)}{2} \right) \cdot 18 \\ s_{18} &=& \left( \frac{2p+34}{2} \right) \cdot 18 \\ s_{18} &=& (p+17) \cdot 18 \\ s_{18} &=& 18p+17\cdot 18 \quad & | \quad s_{18} = n^5 \\ n^5 &=& 18p+17\cdot 18 \\ n^5 - 17\cdot 18 &=& 18p \quad & | \quad : 18 \\ \frac{n^5}{18} - 17 &=& p \\ p&=& \dfrac{n^5}{18} - 17 \quad & | \quad 18 = 2^1 \cdot 3^2 \\\\ p&=& \dfrac{n^5}{2^1 \cdot 3^2} - 17 \quad & | \quad n = 2^i\cdot 3^j \\\\ p&=& \dfrac{(2^i\cdot 3^j)^5}{2^1 \cdot 3^2} - 17 \\\\ p&=& \dfrac{2^{5i}\cdot 3^{5j}}{2^1 \cdot 3^2} - 17 \\\\ p&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \\\\ p_{\text{min}}&=& 2^{5i-1}\cdot 3^{5j-2} - 17 \quad & | \quad p_{\text{min}} \text{ if }i = 1 \text{ and } j = 1 \\\\ p_{\text{min}}&=& 2^{4}\cdot 3^{3} - 17 \\ p_{\text{min}}&=& 16\cdot 27 - 17 \\ \mathbf{p_{\text{min}}}& \mathbf{=}& \mathbf{415} \\\\ n &=& 2^i\cdot 3^j \quad & | \quad i = 1 \text{ and } j = 1 \\ n &=& 2^1\cdot 3^1 \\ \mathbf{n}& \mathbf{=}& \mathbf{6} \\\\ n\cdot p &=& 6\cdot 415 \\ &=& 2490 \end{array}\)

In how many ways can you write 20 as a sum of three counting numbers?

One Way: 1+18+1

Not a Way: 1+1+1+17

Anybody now how to solve this?

I assume in your case, there is only a recursive method:

Let P(n,k) is the number of partitions of a positive integer n into exactly k parts:

For instance, \(7 = 5+1+1=4+2+1=3+2+2\), so \(p(7,3) = 4\)

Formula:

\(\begin{array}{|rcll|} \hline P(0,0) &=& 1 \\ P(n,0) &=& 0 \qquad n\ge 1 \\ P(n,1 )&=& 1 \\ P(n,n) &=& 1 \\ P(n,k) &=& P(n-k,k)+P(n-1,k-1) \qquad \text{ or } \qquad P(n+k,k) = \sum \limits_{j=1}^{k}P(n,j) \\ \hline \end{array} \)

In how many ways can you write 20 as a sum of three counting numbers?

p(20,3) = 33

P(n,k):

p(n,k):

n = 1 -------------------------

p(1,1) = 1

n = 2 -------------------------

p(2,1) = 1

p(2,2) = 1

n = 3 -------------------------

p(3,1) = 1

p(3,2) = 1

p(3,3) = 1

n = 4 -------------------------

p(4,1) = 1

p(4,2) = 2

p(4,3) = 1

p(4,4) = 1

n = 5 -------------------------

p(5,1) = 1

p(5,2) = 2

p(5,3) = 2

p(5,4) = 1

p(5,5) = 1

n = 6 -------------------------

p(6,1) = 1

p(6,2) = 3

p(6,3) = 3

p(6,4) = 2

p(6,5) = 1

p(6,6) = 1

n = 7 -------------------------

p(7,1) = 1

p(7,2) = 3

p(7,3) = 4

p(7,4) = 3

p(7,5) = 2

p(7,6) = 1

p(7,7) = 1

n = 8 -------------------------

p(8,1) = 1

p(8,2) = 4

p(8,3) = 5

p(8,4) = 5

p(8,5) = 3

p(8,6) = 2

p(8,7) = 1

p(8,8) = 1

n = 9 -------------------------

p(9,1) = 1

p(9,2) = 4

p(9,3) = 7

p(9,4) = 6

p(9,5) = 5

p(9,6) = 3

p(9,7) = 2

p(9,8) = 1

p(9,9) = 1

n = 10 -------------------------

p(10,1) = 1

p(10,2) = 5

p(10,3) = 8

p(10,4) = 9

p(10,5) = 7

p(10,6) = 5

p(10,7) = 3

p(10,8) = 2

p(10,9) = 1

p(10,10) = 1

n = 11 -------------------------

p(11,1) = 1

p(11,2) = 5

p(11,3) = 10

p(11,4) = 11

p(11,5) = 10

p(11,6) = 7

p(11,7) = 5

p(11,8) = 3

p(11,9) = 2

p(11,10) = 1

p(11,11) = 1

n = 12 -------------------------

p(12,1) = 1

p(12,2) = 6

p(12,3) = 12

p(12,4) = 15

p(12,5) = 13

p(12,6) = 11

p(12,7) = 7

p(12,8) = 5

p(12,9) = 3

p(12,10) = 2

p(12,11) = 1

p(12,12) = 1

n = 13 -------------------------

p(13,1) = 1

p(13,2) = 6

p(13,3) = 14

p(13,4) = 18

p(13,5) = 18

p(13,6) = 14

p(13,7) = 11

p(13,8) = 7

p(13,9) = 5

p(13,10) = 3

p(13,11) = 2

p(13,12) = 1

p(13,13) = 1

n = 14 -------------------------

p(14,1) = 1

p(14,2) = 7

p(14,3) = 16

p(14,4) = 23

p(14,5) = 23

p(14,6) = 20

p(14,7) = 15

p(14,8) = 11

p(14,9) = 7

p(14,10) = 5

p(14,11) = 3

p(14,12) = 2

p(14,13) = 1

p(14,14) = 1

n = 15 -------------------------

p(15,1) = 1

p(15,2) = 7

p(15,3) = 19

p(15,4) = 27

p(15,5) = 30

p(15,6) = 26

p(15,7) = 21

p(15,8) = 15

p(15,9) = 11

p(15,10) = 7

p(15,11) = 5

p(15,12) = 3

p(15,13) = 2

p(15,14) = 1

p(15,15) = 1

n = 16 -------------------------

p(16,1) = 1

p(16,2) = 8

p(16,3) = 21

p(16,4) = 34

p(16,5) = 37

p(16,6) = 35

p(16,7) = 28

p(16,8) = 22

p(16,9) = 15

p(16,10) = 11

p(16,11) = 7

p(16,12) = 5

p(16,13) = 3

p(16,14) = 2

p(16,15) = 1

p(16,16) = 1

n = 17 -------------------------

p(17,1) = 1

p(17,2) = 8

p(17,3) = 24

p(17,4) = 39

p(17,5) = 47

p(17,6) = 44

p(17,7) = 38

p(17,8) = 29

p(17,9) = 22

p(17,10) = 15

p(17,11) = 11

p(17,12) = 7

p(17,13) = 5

p(17,14) = 3

p(17,15) = 2

p(17,16) = 1

p(17,17) = 1

n = 18 -------------------------

p(18,1) = 1

p(18,2) = 9

p(18,3) = 27

p(18,4) = 47

p(18,5) = 57

p(18,6) = 58

p(18,7) = 49

p(18,8) = 40

p(18,9) = 30

p(18,10) = 22

p(18,11) = 15

p(18,12) = 11

p(18,13) = 7

p(18,14) = 5

p(18,15) = 3

p(18,16) = 2

p(18,17) = 1

p(18,18) = 1

n = 19 -------------------------

p(19,1) = 1

p(19,2) = 9

p(19,3) = 30

p(19,4) = 54

p(19,5) = 70

p(19,6) = 71

p(19,7) = 65

p(19,8) = 52

p(19,9) = 41

p(19,10) = 30

p(19,11) = 22

p(19,12) = 15

p(19,13) = 11

p(19,14) = 7

p(19,15) = 5

p(19,16) = 3

p(19,17) = 2

p(19,18) = 1

p(19,19) = 1

n = 20 -------------------------

p(20,1) = 1

p(20,2) = 10

p(20,3) = 33

p(20,4) = 64

p(20,5) = 84

p(20,6) = 90

p(20,7) = 82

p(20,8) = 70

p(20,9) = 54

p(20,10) = 42

p(20,11) = 30

p(20,12) = 22

p(20,13) = 15

p(20,14) = 11

p(20,15) = 7

p(20,16) = 5

p(20,17) = 3

p(20,18) = 2

p(20,19) = 1

p(20,20) = 1

...

In how many ways can you write 20 as a sum of three counting numbers?

p(20,3) = 33

\(\begin{array}{|r|ll|} \hline & 20 = \\ \hline 1 & 1+1+18 \\ 2 & 1+2+17 \\ 3 & 1+3+16 \\ 4 & 1+4+15 \\ 5 & 1+5+14 \\ 6 & 1+6+13 \\ 7 & 1+7+12 \\ 8 & 1+8+11 \\ 9 & 1+9+10 \\ \hline 10 & 2+2+16 \\ 11 & 2+3+15 \\ 12 & 2+4+14 \\ 13 & 2+5+13 \\ 14 & 2+6+12 \\ 15 & 2+7+11 \\ 16 & 2+8+10 \\ 17 & 2+9+9 \\ \hline 18 & 3+3+14 \\ 19 & 3+4+13 \\ 20 & 3+5+12 \\ 21 & 3+6+11 \\ 22 & 3+7+10 \\ 23 & 3+8+9 \\ \hline 24 & 4+4+12 \\ 25 & 4+5+11 \\ 26 & 4+6+10 \\ 27 & 4+7+9 \\ 28 & 4+8+8 \\ \hline 29 & 5+5+10 \\ 30 & 5+6+9 \\ 31 & 5+7+8 \\ \hline 32 & 6+6+8 \\ 33 & 6+7+7 \\ \hline \end{array}\)

x^5+3x^4+8:3x^3-x-1:3 Funktion fünften Grades Nullstellen errechnen

Brüche entfernen:

\(\begin{array}{|rcll|} \hline x^5+3x^4+\frac{8}{3}x^3-x-\frac{1}{3} &=& 0 \quad | \quad \cdot 3\\ \mathbf{3x^5+9x^4+8x^3-3x-1} &=& \mathbf{0} \\ \hline \end{array}\)

Vermutete Lösung: \(x_0 =\pm1\)

\(\begin{array}{rcrrcl} x_0 &=& 1: & 3+9+8-3-1 &\ne& 0 \\ x_0 &=& -1: & -3+9-8+3-1 &=& 0 ~ \checkmark \\ \end{array}\)

1. Lösung: \(x_0=-1\)

weitere Lösungen:

\(\begin{array}{rcll} (3x^5+9x^4+8x^3-3x-1):(x-x_0) \quad | \quad x_0=-1 \\ \end{array}\)

\(x^5+3x^4+\frac{8}{3}x^3-x-\frac{1}{3} = \frac13\cdot(x+1)(3x^4+6x^3+2x^2-2x-1) \)

\(\begin{array}{|rcll|} \hline \mathbf{3x^4+6x^3+2x^2-2x-1} &=& \mathbf{0} \\ \hline \end{array}\)

Vermutete Lösung: \( x_0 =\pm1\)

\(\begin{array}{rcrrcl} x_0 &=& 1: & 1+6+2-2-1 &\ne& 0 \\ x_0 &=& -1: & 3-6+2+2-1 &=& 0 ~ \checkmark \\ \end{array}\)

2. Lösung: \(x_0=-1\)

weitere Lösungen:
\(\begin{array}{rcll} (3x^4+6x^3+2x^2-2x-1):(x-x_0) \quad | \quad x_0=-1 \\ \end{array}\)

\(x^5+3x^4+\frac{8}{3}x^3-x-\frac{1}{3} = \frac13\cdot(x+1)\cdot(x+1)(3x^3+3x^2-x-1) \)

\(\begin{array}{|rcll|} \hline \mathbf{3x^3+3x^2-x-1} &=& \mathbf{0} \\ \hline \end{array}\)

Vermutete Lösung: \(x_0 =\pm1\)

\(\begin{array}{rcrrcl} x_0 &=& 1: & 3+3-1-1 &\ne& 0 \\ x_0 &=& -1: & -3+3+1-1 &=& 0 ~ \checkmark \\ \end{array}\)

3. Lösung: \(x_0=-1\)

weitere Lösungen:

\(\begin{array}{rcll} (3x^3+3x^2-x-1):(x-x_0) \quad | \quad x_0=-1 \\ \end{array}\)

\(x^5+3x^4+\frac{8}{3}x^3-x-\frac{1}{3} = \frac13\cdot(x+1)\cdot(x+1)\cdot(x+1)(3x^2-1) \)

\(\begin{array}{|rcll|} \hline \mathbf{3x^2-1} &=& \mathbf{0} \\ 3x^2 &=& 1 \\ x^2 &=& \frac13 \\ x &=& \pm \sqrt{\frac13} \\ \hline \end{array} \)

4. Lösung: \(x_0= \sqrt{\frac13}\)
5. Lösung: \(x_0=- \sqrt{\frac13}\)

Die 5 Lösungen lauten:

\(x_1 = -1 \qquad x_2 = -1 \qquad x_3 = -1 \qquad x_4 = \sqrt{\frac13} \qquad x_5 = -\sqrt{\frac13} \)

\(x^5+3x^4+\frac{8}{3}x^3-x-\frac{1}{3} = (x+1)^3(x-\sqrt{\frac13})(x+\sqrt{\frac13}) \)

1504   2807   0310   1811   2008

    15       10       20       20       13

3004   0708   2310       ?     0209

Welche Zahl ist das Fragezeichen?

Also ich käme für das ? auf:
\(6e65b61d030e5ac6492802c6f619a63006e5b0e2cafc0170ab92a68083162528_{sha256} - 100_{10} \\ = c0d7c54022345b39c9be73e19b30ccb040ba03b1fefb2566a9f510a09aba4796_{sha256}\)

Ich habe eine logische Lösung gefunden, aber ob das tatsächlich die einfachste Logik zu diesem Rätsel ist, weiß ich nicht.

What is the value of X?

\(\begin{array}{rcll} 11x-3 &=& 9x+23 \\ 2x &=& 26 \\ x &=& 13 \\ \end{array}\)

X and Y are natural numbers.

X = 1; Y = 1

X = 2; Y = 2

X = 3; Y = 3

X = 4; Y = 4

X = 5; Y = 3

X = 6; Y = 2

X = 7; Y = 1

X = 8; Y = 0

How do I express this with a formula?

\(\begin{array}{|rcll|} \hline Y = \min(X,8-X) \\ \hline \end{array}\)

finding angle b in question 1?

\(\begin{array}{|rcll|} \hline \sin(B) &=& \frac{3.6}{4.2} \times \sin(110^{\circ}) \\ \sin(B) &=& 0.85714285714 \times 0.93969262079 \\ \sin(B) &=& 0.80545081782 \\\\ B &=& \arcsin(0.80545081782) \\ B &=& 53.6538157268^{\circ} \\\\ \mathbf{B} &\mathbf{=}& \mathbf{53.7^{\circ}} \\\\ A &=& 180^{\circ}- 110^{\circ} - 53.7^{\circ} \\ \mathbf{A} &\mathbf{=}& \mathbf{16.3^{\circ}} \\\\ \hline \end{array}\)