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Questions 17
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 #1
avatar+26319 
+3

The equation of the line joining the complex numbers
\(-5+4i\) and \(7+2i\) can be expressed in the form
\(az + b \overline{z} = 38\)
for some complex numbers a and b. Find the product ab
.


\(\text{Let $z_1=-5+4i$ and $\overline{z_1}=-5-4i$} \\ \text{Let $z_2=7+2i$ and $\overline{z_2}=7-2i$} \\ \text{Let $az_1 + b \overline{z_1} = 38$} \\ \text{Let $az_2 + b \overline{z_2} = 38$}\)

 

\(\begin{array}{|rcll|} \hline az_2 + b \overline{z_2} &=& 38 \\ az_2 &=& 38- b \overline{z_2} \\ \mathbf{a} &=& \mathbf{\dfrac{38- b \overline{z_2}}{z_2}} \\ \hline az_1 + b \overline{z_1} &=& 38 \\\\ \mathbf{\dfrac{38- b \overline{z_2}}{z_2}}z_1 + b \overline{z_1} &=& 38 \quad | \quad *z_2 \\\\ (38- b \overline{z_2})z_1 + b \overline{z_1}z_2 &=& 38z_2 \\ 38z_1- b \overline{z_2}z_1+ b \overline{z_1}z_2 &=& 38z_2 \\ b(\overline{z_1}z_2-\overline{z_2}z_1) &=& 38(z_2-z_1) \\\\ \mathbf{b} &=& \mathbf{ \dfrac{38(z_2-z_1)}{ (\overline{z_1}z_2-\overline{z_2}z_1) }} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \overline{z_1}z_2-\overline{z_2}z_1\\ &=& (-5-4i)(7+2i)-(7-2i)(-5+4i) \\ &=& \mathbf{-76i} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{b} &=& \mathbf{ \dfrac{38(z_2-z_1)}{\overline{z_1}z_2-\overline{z_2}z_1}} \\\\ &=& \dfrac{38\Big(7+2i-(-5+4i) \Big)}{-76i} \\\\ &=& \dfrac{38(7+2i+5-4i)}{-76i} \\\\ &=& \dfrac{7+2i+5-4i}{-2i} \\\\ &=& \dfrac{12-2i}{-2i} \\\\ &=& \dfrac{-6+i}{i} \\\\ &=& \dfrac{-6}{i}+1 \\\\ &=& \dfrac{-6i}{i^2}+1 \\\\ &=& \dfrac{-6i}{-1}+1 \\\\ &=& 6i+1 \\ \mathbf{b} &=& \mathbf{1+6i} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline az_2 + b \overline{z_2} &=& 38 \\\\ az_2 + \mathbf{ \dfrac{38(z_2-z_1)}{(\overline{z_1}z_2-\overline{z_2}z_1)}} \overline{z_2} &=& 38 \\ az_2 &=& 38-\dfrac{38(z_2-z_1)\overline{z_2}}{(\overline{z_1}z_2-\overline{z_2}z_1)} \quad | \quad : z_2 \\\\ a &=& \dfrac{38}{z_2}-\dfrac{38(z_2-z_1)\overline{z_2}}{(\overline{z_1}z_2-\overline{z_2}z_1)z_2} \\\\ a &=& \dfrac{38(\overline{z_1}z_2-\overline{z_2}z_1)-38(z_2-z_1)\overline{z_2}}{z_2(\overline{z_1}z_2-\overline{z_2}z_1)} \\\\ a &=& \dfrac{38 \Big(\overline{z_1}z_2-\overline{z_2}z_1-(z_2-z_1)\overline{z_2} \Big) } {z_2(\overline{z_1}z_2-\overline{z_2}z_1)} \\\\ a &=& \dfrac{38 (\overline{z_1}z_2-\overline{z_2}z_1-z_2\overline{z_2}+z_1\overline{z_2}) } {z_2(\overline{z_1}z_2-\overline{z_2}z_1)} \\\\ a &=& \dfrac{38 (\overline{z_1}z_2-z_2\overline{z_2}) } {z_2(\overline{z_1}z_2-\overline{z_2}z_1)} \\\\ a &=& \dfrac{38z_2 (\overline{z_1}-\overline{z_2}) } {z_2(\overline{z_1}z_2-\overline{z_2}z_1)} \\\\ \mathbf{a} &=& \mathbf{ \dfrac{38(\overline{z_1}-\overline{z_2}) } {(\overline{z_1}z_2-\overline{z_2}z_1)} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{ \dfrac{38(\overline{z_1}-\overline{z_2}) } {(\overline{z_1}z_2-\overline{z_2}z_1)} } \\\\ &=& \dfrac{38\Big(-5-4i-(7-2i) \Big)}{-76i} \\\\ &=& \dfrac{38(-7+2i-5-4i)}{-76i} \\\\ &=& \dfrac{-7+2i-5-4i}{-2i} \\\\ &=& \dfrac{-12-2i}{-2i} \\\\ &=& \dfrac{6+i}{i} \\\\ &=& \dfrac{6}{i}+1 \\\\ &=& \dfrac{6i}{i^2}+1 \\\\ &=& \dfrac{6i}{-1}+1 \\\\ &=& -6i+1 \\ \mathbf{a} &=& \mathbf{1-6i} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline ab &=& (1+6i)(1-6i) \\ ab &=& 1+36 \\ \mathbf{ab} &=& \mathbf{37} \\ \hline \end{array}\)

 

laugh

6 juin 2022
 #1
avatar+26319 
+3

What is the remainder when the sum

\(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\) 

is divided by 17?

 

\(\begin{array}{|rcll|} \hline 1^2 \pmod {17} &=& 1 \\ 2^2 \pmod {17} &=& 4 \\ 3^2 \pmod {17} &=& 9 \\ 4^2 \pmod {17} &=& 16 \\ 5^2 \pmod {17} &=& 8 \\ 6^2 \pmod {17} &=& 2 \\ 7^2 \pmod {17} &=& 15 \\ 8^2 \pmod {17} &=& 13 \\ 9^2 \pmod {17} &=& 13 \\ 10^2 \pmod {17} &=& 15 \\ 11^2 \pmod {17} &=& 2 \\ 12^2 \pmod {17} &=& 8 \\ 13^2 \pmod {17} &=& 16 \\ 14^2 \pmod {17} &=& 9 \\ 15^2 \pmod {17} &=& 4 \\ 16^2 \pmod {17} &=& 1 \\ 17^2 \pmod {17} &=& 0 \\ \hline \end{array}\)

cycle mod 17: \((1, 4, 9, 16, 8, 2, 15, 13, 13, 15, 2, 8, 16, 9, 4, 1, 0)\)

 

circle mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9+ 4+ 1+ 0\pmod {17} \\ &=& 136 \pmod {17} \\ &=& \mathbf{0 \pmod {17}} \\ \hline \end{array}\)

 

2020/17 = 118 cycles remainder 14

 

remainder 14, this is the sum in the cycle from 1 to 14  mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9\pmod {17} \\ &=& 131 \pmod {17} \\ &=&\mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

 

\( \begin{array}{|rcll|} \hline && 1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\pmod {17} \\ &=& \underbrace{0*118 \pmod {17}}_{118 \text{ cycles}} + \underbrace{\mathbf{ 12 \pmod {17}}}_{\text{remainder 14}} \\ &=& \mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

 

The remainder when the sum \(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2 \) is divided by 17 is \(\mathbf{12}\)

 

laugh

27 mai 2022
 #1
avatar+26319 
+2

Find all complex solutions such that z^4 = -4
Note: All solutions should be expressed in the form a+bi,
where a and b are real numbers help!

 

\(\huge{z^2 = \pm 2i}\)

 

\(\begin{array}{|rcll|} \hline z^2 &=& \pm 2i \quad | \quad z = a+bi \\ \left(a+bi\right)^2 &=& \pm 2i \\ a^2+2abi+b^2i^2 &=& \pm 2i \quad | \quad i^2 = -1 \\ a^2+2abi-b^2 &=& \pm 2i \\ \mathbf{a^2-b^2+2abi} &=& \mathbf{\pm 2i} \\ \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{+2i}\\ a^2-b^2+2abi &=& 0+2i \quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& 2 \\ \mathbf{b} &=& \dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_1 &=& 1 \\ a_2 &=& -1 \\ b_1 &=& \dfrac1a_1 \\ b_1 &=& 1 \\ b_2 &=& \dfrac1a_2 \\ b_2 &=& -1 \\ \hline \end{array}\\ \hline \end{array} &\text{or}& \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{-2i}\\ a^2-b^2+2abi &=& 0-2i\quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& -2 \\ \mathbf{b} &=& -\dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(-\dfrac1a\right)^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_3 &=& 1 \\ a_4 &=& -1 \\ b_3 &=& -\dfrac1a_3 \\ b_3 &=& -1 \\ b_4 &=& -\dfrac1a_4 \\ b_4 &=& 1 \\ \hline \end{array}\\ \hline \end{array} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline z&=& a_1+b_1i = 1+i\\ z&=& a_2+b_2i = -1-i\\ z&=& a_3+b_3i = 1 -i\\ z&=& a_4+b_4i = -1+i\\ \hline \end{array}\)

 

laugh

26 mai 2022
 #1
avatar+26319 
+2

Compute

 

\(\mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10}}\)

 

\(\begin{array}{|rcll|} \hline s&=& \dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} \\\\ &=& \dfrac{1}{2(2^2-1)} + \dfrac{1}{3(3^2-1)} + \dfrac{1}{4(4^2-1)} + \dots + \dfrac{1}{10(10^2-19} \\\\ &=& \dfrac{1}{(2-1)2(2+1)} + \dfrac{1}{(3-1)3(3+1)} + \dfrac{1}{(4-1)4(4+1)} + \dots + \dfrac{1}{(10-1)10(10+1)} + \dots + \dfrac{1}{(n-1)n(n+1)} \\\\ \hline s &=&\sum \limits_{n=2}^{10}\dfrac{1}{(n-1)n(n+1)} \\ \\ &&\boxed{ \frac{1}{(n-1)n} =\frac{1}{n-1}-\frac{1}{n} \\ \frac{1}{(n+1)n} =\frac{1}{n}-\frac{1}{n+1}\\\ldots\\ \frac{1}{(n-1)n(n+1)} = \frac12\left( \frac{1}{n-1}-\frac{2}{n} +\frac{1}{n+1} \right) } \\\\ s &=&\dfrac12\sum \limits_{n=2}^{10}\left( \dfrac{1}{n-1}-\dfrac{2}{n} +\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=2}^{10}\dfrac{1}{n-1}-\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=2}^{10}\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=1}^{9}\dfrac{1}{n} -\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=3}^{11}\dfrac{1}{n} \right) \\ \\ &&\boxed{ \sum \limits_{n=1}^{9}\frac{1}{n} = \frac11 + \frac12 + \sum \limits_{n=3}^{9}\frac{1}{n} \\ -\sum \limits_{n=2}^{10}\frac{2}{n} = -\frac{2}{2}-\frac{2}{10}-2\sum \limits_{n=3}^{9}\frac{1}{n} \\ \sum \limits_{n=3}^{11}\frac{1}{n} = \frac{1}{10}+\frac{1}{11}+\sum \limits_{n=3}^{9}\frac{1}{n} } \\\\ s &=&\dfrac12\left( \dfrac11 + \dfrac12 + \sum \limits_{n=3}^{9}\dfrac{1}{n} -\dfrac{2}{2}-\dfrac{2}{10}-2\sum \limits_{n=3}^{9}\dfrac{1}{n} +\dfrac{1}{10}+\dfrac{1}{11}+\sum \limits_{n=3}^{9}\dfrac{1}{n}\right) \\\\ &=&\dfrac12\left( \dfrac11 + \dfrac12 -\dfrac{2}{2}-\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac14 -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{5}{20} -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{4}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{1}{5}+\dfrac{1}{22} \\ \\ &=&\dfrac{22+5}{5*22}\\ \\ \mathbf{s}&=&\mathbf{\dfrac{27}{110}}\\ \hline \end{array}\)

 

\(\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} = \mathbf{\dfrac{27}{110}}\)

 

laugh

22 mai 2022