heureka

Let

\(\omega\)

be a complex number such that

\(\omega^3 = 1\).
Find all possible values of

\(\dfrac{1}{1 + \omega} + \dfrac{1}{1 + \omega^2}\).

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{1 + \omega} + \dfrac{1}{1 + \omega^2}} \\\\ &=& \dfrac{1 + \omega^2+1 + \omega}{(1 + \omega)(1 + \omega^2) } \\\\ &=& \dfrac{2 + \omega + \omega^2 }{1 + \omega^2 + \omega + \omega^3 } \quad | \quad \omega^3 = 1 \\\\ &=& \dfrac{2 + \omega + \omega^2 }{1 + \omega^2 + \omega + 1 } \\\\ &=& \dfrac{2 + \omega + \omega^2 }{2 + \omega + \omega^2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

If

\(z^2 + z + 1 = 0\),

find \(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\).

\(\begin{array}{|rcll|} \hline \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z+2z^2 +2z^3 \\ \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z(z^2 + z + 1) \quad | \quad z^2 + z + 1 = 0 \\ \mathbf{0} &\mathbf{=} & \mathbf{1+z^2+z^4} \quad | \quad 0 = z^2 + z + 1 \\ z^2 + z + 1 & = & 1+z^2+z^4 \\ z & = & z^4 \quad | \quad : z \\ 1 & = & z^3 \\ \mathbf{z^3} &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline z^4 = z^3z= 1\cdot z &=& z \\ z^5 = z^3z^2= 1\cdot z^2 &=& z^2 \\ z^6 = \left(z^3\right)^2= 1^2 &=& 1 \\ \hline \end{array}\)

In general:

\(\begin{array}{|lcll|} \hline z^{0+3n} = 1 \\ z^{1+3n} = z \\ z^{2+3n} = z^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} = z^{1+3\cdot16} &=& z \\ z^{50} = z^{2+3\cdot16} &=& z^2 \\ z^{51} = z^{0+3\cdot17} &=& 1 \\ z^{52} = z^{1+3\cdot17} &=& z \\ z^{53} = z^{2+3\cdot17} &=& z^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline z^{49} + z^{50} + z^{51} + z^{52} + z^{53} &=& z+z^2+1+z+z^2 \quad | \quad z^2 + z + 1 = 0 \\ z^{49} + z^{50} + z^{51} + z^{52} + z^{53} &=& z+z^2 \quad | \quad z^2 + z = -1 \\ \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)

​complex roots

\(\begin{array}{|rcll|} \hline w &=& p+2i \\ w^{*} &=& p-2i \\\\ \left[ z-w \right] \left[z-w^{*} \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ z-(p+2i) \right] \left[z-(p-2i) \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ (z-p)-2i) \right] \left[(z-p)+2i) \right] &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2-(2i)^2 &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2+4 &=& z^2+(4+i+qi)z + 20 \\ z^2-2pz+(p^2+4) &=& z^2+\underbrace{(4+i+qi)}_{=-2p}z + \underbrace{20}_{=p^2+4} \\\\ p^2+4 &=& 20 \\ p^2 &=& 16 \\ \mathbf{p} & \mathbf{=} & \mathbf{\pm 4} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline p=4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& -8 \\ & qi &=& -12-i \\ & q &=& \dfrac{-12-i}{i} \\ & q &=& \dfrac{(-12-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(-12-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(-12-i)i}{-1} \\ & q &=& (12+i)i \\ & q &=& i^2+12i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1+12i} \\\\ & w &=& 4+2i \\ & w^* &=& 4-2i \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline p=-4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& 8 \\ & qi &=& 4-i \\ & q &=& \dfrac{4-i}{i} \\ & q &=& \dfrac{(4-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(4-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(4-i)i}{-1} \\ & q &=& (i-4)i \\ & q &=& i^2-4i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1-4i} \\\\ & w &=& -4+2i \\ & w^* &=& -4-2i \\ \hline \end{array}\)

The possible values of \(\mathbf{q}\) are \(\mathbf{-1+12i}\) and \(\mathbf{-1-4i}\)

and the quadratic equations are \(\mathbf{z^2-8z+20 = 0}\) and \(\mathbf{z^2+8z+20 = 0}\) .

Thank you, CPhill !

Thank you, Melody !

Solve for

\( 0 \text{ to } 2\pi\)

\(\sin(3x)+\sin(4x)=0 \)

\(\begin{array}{|lrcll|} \hline & \sin(3x)+\sin(4x) &=& 0 \quad | \quad -\sin(3x) \\ & \sin(4x) &=& -\sin(3x) \quad | \quad -\sin(3x) = \sin(-3x) \\ & \sin(4x) &=& \sin(-3x) \quad | \quad \arcsin() \\\\ (1) & 4x + 2\pi n &=& -3x + 2\pi m \quad | \quad +3x \\ & 7x &=& 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{2\pi}{7} z } \qquad z \in \mathbb{Z} \\\\ (2) & \sin(4x) &=& \sin(-3x) \quad | \quad \sin(-3x) = \sin(\pi -(-3x) ) \\ & \sin(4x ) &=& \sin(\pi + 3x ) \quad | \quad \arcsin() \\ & 4x +2\pi n &=& \pi+3x + 2\pi m \quad | \quad -3x \\ & x &=& \pi + 2\pi z \\ & \mathbf{x} &\mathbf{=}& \mathbf{ \pi + 2\pi z } \qquad z \in \mathbb{Z} \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|} \hline & 0\le x\le2\pi & 0\le x\le2\pi \\ z & x=\dfrac{2\pi}{7} z & x = \pi + 2\pi z \\ \hline 0 & 0 & \pi \\ 1 & \dfrac{2\pi}{7} & \\ 2 & \dfrac{4\pi}{7} & \\ 3 & \dfrac{6\pi}{7} & \\ 4 & \dfrac{8\pi}{7} & \\ 5 & \dfrac{10\pi}{7} & \\ 6 & \dfrac{12\pi}{7} & \\ 7 & \dfrac{14\pi}{7}=2\pi & \\ \hline \end{array}\)

\(x = \left\{ 0,\ \dfrac{2\pi}{7},\ \dfrac{4\pi}{7},\ \dfrac{6\pi}{7},\ \pi ,\ \dfrac{8\pi}{7},\ \dfrac{10\pi}{7},\ \dfrac{12\pi}{7},\ 2\pi \right\}\)

Thank you, CPhill !

Find all solutions to

\(\large \sqrt[3]{x + 57} - \sqrt[3]{x - 57} = \sqrt[3]{6}\).

\(\begin{array}{|rcll|} \hline \sqrt[3]{x + 57} - \sqrt[3]{x - 57} &=& \sqrt[3]{6} \\ \left( \sqrt[3]{x + 57} - \sqrt[3]{x - 57} \right)^3 &=& 6 \\ (x + 57)-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2}-(x-57) &=& 6 \\ 114-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2} &=& 6 \\ 108-3 \sqrt[3]{(x + 57)^2(x - 57)} + 3 \sqrt[3]{(x + 57)(x - 57)^2} &=& 0 \quad | \quad : 3\\ 36- \sqrt[3]{(x + 57)^2(x - 57)} + \sqrt[3]{(x + 57)(x - 57)^2} &=& 0 \\ 36- \sqrt[3]{(x^2 - 57^2)(x + 57)} + \sqrt[3]{(x^2 - 57^2)(x - 57)} &=& 0 \\ 36- \sqrt[3]{x^2 - 57^2} \underbrace{\left(\sqrt[3]{x + 57} - \sqrt[3]{x - 57}\right)}_{=\sqrt[3]{6}} &=& 0 \\ \sqrt[3]{x^2 - 57^2}\sqrt[3]{6} &=& 36 \\ (x^2 - 57^2)\cdot 6 &=& 36^3 \\ x^2 - 57^2 &=& 7776 \\ x^2 &=& 7776 +3249 \\ x^2 &=& 11025 \\ \mathbf{x} &\mathbf{=}& \mathbf{\pm 105} \\ \hline \end{array}\)

check \(x=105\):

\(\begin{array}{rcll} \sqrt[3]{105 + 57} - \sqrt[3]{105 - 57} &\overset{?}{=}& \sqrt[3]{6} \\ \sqrt[3]{162} - \sqrt[3]{48} &\overset{?}{=}& 1.81712059283 \\ 5.45136177850 - 3.63424118566 &\overset{?}{=}& 1.81712059283 \\ 1.81712059283 & = & 1.81712059283 \quad \checkmark \\ \end{array}\)

check \(x=-105\):

\(\begin{array}{rcll} \sqrt[3]{-105 + 57} - \sqrt[3]{-105 - 57} &\overset{?}{=}& \sqrt[3]{6} \\ \sqrt[3]{-48} - \sqrt[3]{-162} &\overset{?}{=}& 1.81712059283 \\ - 3.63424118566 -(-5.45136177850 ) &\overset{?}{=}& 1.81712059283 \\ - 3.63424118566 + 5.45136177850 &\overset{?}{=}& 1.81712059283 \\ 1.81712059283 & = & 1.81712059283 \quad \checkmark \\ \end{array}\)

Find positive integers \((a,b)\) so that \(\sqrt{37 + 20 \sqrt{3}} = a + b \sqrt{3}\).

\(\begin{array}{|rcll|} \hline \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{a + b \sqrt{3}} \\\\ 37 + 20 \sqrt{3} &=& \left(a + b \sqrt{3}\right)^2 \\ 37 + 20 \sqrt{3} &=& a^2+2ab\sqrt{3} +3b^2 \\ 37 + 20 \sqrt{3} &=& \underbrace{a^2+3b^2}_{=37}+\underbrace{2ab}_{=20} \sqrt{3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & 2ab &=& 20 \\ & ab &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{10}{a}} \\ \hline (2) & a^2+3b^2 &=& 37 \\ & a^2+3\dfrac{10^2}{a^2} &=& 37 \\ & a^2+ \dfrac{300}{a^2} &=& 37 \quad |\quad \cdot a^2 \\ & a^4+ 300 &=& 37a^2 \\ & a^4-37a^2+ 300 &=& 0 \\\\ & a^2 &=& \dfrac{37 \pm \sqrt{37^2-4\cdot 300} }{2} \\ & a^2 &=& \dfrac{37 \pm \sqrt{169} }{2} \\ & a^2 &=& \dfrac{37 \pm 13 }{2} \\\\ & a^2 = 25 &\text{or}& a^2 = 12 \\ & a = \pm 5 &\text{or}& a = \pm 2\sqrt{3} \quad |\quad \text{positive integers } (a,b) \\ & a = 5 &\text{or}& a = 2\sqrt{3} \quad |\quad a \text{ without } \sqrt{3} \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b &=& \dfrac{10}{a} \\ & b &=& \dfrac{10}{5} \\ & \mathbf{b} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

\(\begin{array}{rcll} \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{5 + 2 \sqrt{3}} \\ \end{array}\)

Compute 

\(1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb\)

\(\text{Let $x = \dfrac{1}{2} \quad|x|<1$ } \)

Now we have:  \(\displaystyle s= \sum \limits_{n=1}^{\infty} nx^n \)

Let the progression to be summed be put equal to s:

\(s = x+2x^2+3x^3+4x^4+\cdots + nx^n +\cdots\)

It is divided by \(x\) and multiplied by \(dx\) then

\(\dfrac{s\ dx}{x} = dx +2x\ dx+3x^2\ dx+4x^3\ dx+\cdots + nx^{n-1}\ dx +\cdots \)

and with the integrals taken this equation is found
\(\displaystyle \int \dfrac{s\ dx}{x} = x +x^2+x^3+x^4 +\cdots + x^n + \cdots = \dfrac{x}{1-x} \quad \text{ infinite geometric progression }\)

Therefore from the equation:
\(\displaystyle \int \dfrac{s\ dx}{x} = \dfrac{x}{1-x}\)

on differentiation s is found. For the equation becomes:
\(\begin{array}{rcll} \displaystyle \dfrac{s}{x} &=& \dfrac{x}{1-x}\left( \dfrac{1}{x}- \dfrac{(-1)}{1-x} \right) \\\\ &=& \dfrac{1}{1-x} + x(-1)(1-x)^{-2}(-1) \\\\ &=& \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\\\ &=& \dfrac{1-x+x}{(1-x)^2} \\\\ &=& \dfrac{1 }{(1-x)^2} \end{array}\)

thus there is produced:
\(\displaystyle s = \dfrac{x}{(1-x)^2} \\\)

So  \(\text{let $x = \dfrac{1}{2} $}:\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{ \dfrac{1}{2} } {\left(1-\dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ \dfrac{1}{2} } {\left( \dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ 1 } { \dfrac{1}{2} } \\\\ \mathbf{s} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

finally
\(\displaystyle 1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb = 2\)