heureka

Thank you, OldTimer !

Consider the graph of the equation \(z^2 + (\overline{z})^2 = 2\).

For each complex number in the following list,

1      0    1+i     \( 2 - i \sqrt{3}\)     2-i    -1      i.

figure out whether each one is on the graph.

\(\text{Let $z=a+ib$} \\ \text{Let $\overline{z}=a-ib$}\)

\(\begin{array}{|rcll|} \hline \mathbf{z^2 + (\overline{z})^2} &=& \mathbf{2} \\\\ (a+ib)^2 +(a-ib)^2 &=& 2 \\ a^2+2abi-b^2 +a^2-2abi-b^2 &=& 2 \\ 2a^2-2b^2 &=& 2 \quad | \quad : 2 \\ \mathbf{a^2-b^2} &=& \mathbf{1} \\ \hline \end{array} \)

\(\begin{array}{|r||r|r|l|} \hline \text{list} && & & \\ z=a+ib && a & b & \mathbf{a^2-b^2=1}\ ? \\ \hline \color{red}1 && 1 & 0 & 1^2 -0^2= 1\ \checkmark \\ \hline 0 && 0 & 0 & 0^2 -0^2\ne 1 \\ \hline 1+i && 1 & 1 & 1^2 -1^2\ne 1 \\ \hline \color{red}2-i\sqrt{3} && 2 & \sqrt{3} & 2^2 - \left(\sqrt{3}\right)^2= 1\ \checkmark \\ \hline 2-i && 2 & -1 & 2^2 -\left(-1\right)^2\ne 1 \\ \hline \color{red}-1 && -1 & 0 & \left(-1\right)^2 -0^2= 1\ \checkmark \\ \hline i && 0 & 1 & 0^2 - 1^2\ne 1 \\ \hline \end{array} \)

Thank you, Melody !

 Trigonometry

\(\begin{array}{|rcll|} \hline \mathbf{\tan{\theta}} &=& \mathbf{2\ \tan(60^\circ-\theta)} \quad | \quad \lambda = 2,\ A=60^\circ \\\\ (2-1)\sin{60^\circ} &=& (2+1)\sin(2\theta-60^\circ) \\ \sin{60^\circ} &=& 3\sin(2\theta-60^\circ) \quad | \quad \sin(60^\circ)=\dfrac{\sqrt{3} }{2} \\ \dfrac{\sqrt{3} }{2} &=& 3\sin(2\theta-60^\circ) \\ \mathbf{ \sin(2\theta-60^\circ)} &=& \mathbf{\dfrac{\sqrt{3}}{6}} \\ 2\theta-60^\circ &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 60^\circ +16.7786548810^\circ +360^\circ n \\ 2\theta &=& 76.7786548810^\circ +360^\circ n \\ \theta &=& 38.3893274405^\circ +180^\circ n \\ \\ \mathbf{\theta_1} &=& \mathbf{38.3893274405^\circ} \\\\ \theta_2 &=& 38.3893274405^\circ+180^\circ \\ \mathbf{\theta_2} &=& \mathbf{218.389327440^\circ} \\ \hline \sin(2\theta-60^\circ) &=& \sin\Big(180^\circ-(2\theta-60^\circ)\Big) \\ &=& \sin(180^\circ-2\theta+60^\circ) \\ &=& \sin(240^\circ-2\theta ) \\\\ \mathbf{ \sin(240^\circ-2\theta )} &=& \mathbf{ \dfrac{\sqrt{3}}{6} } \\ 240^\circ-2\theta &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 240^\circ-16.7786548810^\circ - 360^\circ n \\ 2\theta &=& 223.221345119^\circ - 360^\circ n \\ \theta &=& 111.610672560^\circ - 180^\circ n \\ \\ \mathbf{\theta_3} &=& \mathbf{111.610672560^\circ} \\\\ \theta_4 &=& 111.610672560^\circ+180^\circ \\ \mathbf{\theta_4} &=& \mathbf{291.610672560^\circ} \\ \hline \end{array}\)

Trigonometry

\(\begin{array}{|rcll|} \hline \tan{\theta} &=& \lambda\ \tan(A-\theta) \\ \boxed{\tan{\theta}=\dfrac{\sin(\theta)}{\cos(\theta)} \\ \tan(A-\theta) = \dfrac{\sin(A-\theta)}{\cos(A-\theta)} } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A-\theta)}{\cos(A-\theta)} \right) \\ \boxed{ \sin(A-\theta) = \sin(A)\cos(\theta)-\cos(A)\sin(\theta) \\ \cos(A-\theta) = \cos(A)\cos(\theta)+\sin(A)\sin(\theta) } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A)\cos(\theta)-\cos(A)\sin(\theta)} {\cos(A)\cos(\theta)+\sin(A)\sin(\theta)} \right) \\ \sin(\theta) \Big(\cos(A)\cos(\theta)+\sin(A)\sin(\theta) \Big) &=& \lambda\ \cos(\theta) \Big(\sin(A)\cos(\theta)-\cos(A)\sin(\theta) \Big) \\ \cos(A)\sin(\theta)\cos(\theta)+ \sin(A)\sin^2(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\lambda\ \cos(A)\sin(\theta)\cos(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\sin(A)\sin^2(\theta) \\ \boxed{ \cos^2(\theta) = 1-\sin^2(\theta) } \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \Big(1-\sin^2(\theta) \Big) \sin(A)-\sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \sin(A)-(\lambda+1) \sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta)+(\lambda+1) \sin(A)\sin^2(\theta) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(\theta)\cos(\theta)+\sin(A)\sin^2(\theta) \Big) &=& \lambda\ \sin(A) \\ \boxed{\sin(\theta)\cos(\theta)=\dfrac{\sin(2\theta)}{2} \\ \sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2} } \\ (\lambda+1)\Big( \cos(A)\dfrac{\sin(2\theta)}{2}+\sin(A) \Big(\dfrac{1-\cos(2\theta)}{2}\Big) \Big) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)+\sin(A) \Big( 1-\cos(2\theta)\Big) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ \boxed{ \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) = \sin(2\theta-A) } \\ (\lambda+1)\Big( \sin(2\theta-A) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) +(\lambda+1)\sin(A) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-(\lambda+1)\sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-\lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& \lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& (\lambda -1) \sin(A)\\ \mathbf{(\lambda -1) \sin(A)} &=& \mathbf{(\lambda+1)\sin(2\theta-A)} \\ \hline \end{array}\)

Show that from any five integers,

one can always choose three of these integers such that their sum is divisible by 3.

Five integers have a remainder of 0 or 1 or 2 if they become divisible by 3:

\(\begin{array}{|c|c|c|c|} \hline & & & \text{one choose three} \\ \text{remainder 0} & \text{remainder 1} & \text{remainder 2} & \text{of these is divisible by 3} \\ 0 \pmod{3} & 1 \pmod{3} & 2 \pmod{3} & \text{sum of the remainder} \\ \hline 0 & 0 & 5 & 2+2+2 \\ 0 & 5 & 0 & 1+1+1 \\ 0 & 4 & 1 & 1+1+1 \\ 0 & 3 & 2 & 1+1+1 \\ 0 & 2 & 3 & 2+2+2 \\ 0 & 1 & 4 & 2+2+2 \\ \hline 1 & 0 & 4 & 2+2+2 \\ 1 & 4 & 0 & 1+1+1 \\ 1 & 3 & 1 & 1+1+1 \\ 1 & 2 & 2 & 0+1+2 \\ 1 & 1 & 3 & 2+2+2 \\ \hline 2 & 0 & 3 & 2+2+2 \\ 2 & 3 & 0 & 1+1+1 \\ 2 & 2 & 1 & 0+1+2 \\ 2 & 1 & 2 & 0+1+2 \\ \hline 3 & 0 & 2 & 0+0+0 \\ 3 & 2 & 0 & 0+0+0 \\ 3 & 1 & 1 & 0+0+0 \\ \hline 4 & 0 & 1 & 0+0+0 \\ 4 & 1 & 0 & 0+0+0 \\ \hline 5 & 0 & 0 & 0+0+0 \\ \hline \end{array}\)

Find the value of 

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} } \\\\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\\\ \dfrac{1}{x-1} &=& 2+ \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}} \\\\ \dfrac{1}{x-1} &=& 2+ (x-1) \\\\ 1 &=& (x-1)(2+x-1) \\ 1 &=& (x-1)(x+1) \\ 1 &=& x^2-1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)

A)

Let \(f(x)\) be a polynomial of degree \(4\)  with rational coefficients which has \(1+2\sqrt{3}\) and \(3-\sqrt{2}\)
as roots, and such that \(f(0) = -154\).
Find \(f(1)\).

Vieta:

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ax^4+bx^3+cx^2+dx+e} \quad | \quad \text{polynomial of degree 4} \\\\ \dfrac{e}{a} &=& x_1x_2x_3x_4 \quad | \quad \text{Vieta-theorem},\ f(0) =e \\\\ \dfrac{f(0)}{a} &=& x_1x_2x_3x_4 \\\\ \mathbf{a} &=& \mathbf{\dfrac{f(0)}{x_1x_2x_3x_4}} \\\\ && \boxed{ f(0)=-154\\ x_1 = 1+2\sqrt{3}\\ x_2 = 1-2\sqrt{3}\\ x_3 = 3-\sqrt{2}\\ x_4 = 3+\sqrt{2} } \\\\ a &=& \dfrac{-154}{(1+2\sqrt{3})(1-2\sqrt{3})(3-\sqrt{2})(3+\sqrt{2})} \\\\ &=& \dfrac{-154}{(1-4\cdot 3)(9-2)} \\\\ &=& \dfrac{-154}{(-11)\cdot 7} \\\\ &=& \dfrac{ 154}{77} \\\\ \mathbf{a} &=& \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{f(1)=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ a(x-x_1)(x-x_2)(x-x_3)(x-x_4) } \\\\ f(1) &=& 2\left(1-(1+2\sqrt{3})\right) \left(1-(1-2\sqrt{3})\right) \left(1-(3-\sqrt{2})\right) \left(1-(3+\sqrt{2})\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) (-2+\sqrt{2}) (-2-\sqrt{2})) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \left((-2)^2-2\right) \\ &=& 2(2\sqrt{3}) (-2\sqrt{3}) \cdot 2\\ &=& -16\cdot 3 \\ \mathbf{f(1)} &=& \mathbf{-48} \\ \hline \end{array}\)

Let \(z = 2 e^{(10 \pi i)/21}\)  and \(w = e^{(\pi i)/7}\).

Then what is \(|(z+w)^6|\),

the magnitude of \((z+w)^6\)?

\(\begin{array}{|rcll|} \hline z = 2 e^{i\left(\frac{10}{21} \pi \right) } && w = e^{i\left(\frac{\pi}{7} \right) } \\ \boxed{\frac{10}{21} \pi = \frac{7}{21}\pi + \frac{3}{21}\pi \\ = \frac{1}{3}\pi + \frac{1}{7}\pi } \\ z = 2 e^{i\left(\frac{1}{3}\pi + \frac{1}{7}\pi \right) } \\ z = 2 e^{i\left(\frac{\pi}{3} \right) }e^{i\left(\frac{\pi}{7} \right) } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z+w &=& 2 e^{i\left(\frac{\pi}{3} \right) }e^{i\left(\frac{\pi}{7} \right) } + e^{i\left(\frac{\pi}{7} \right) } \\ &=& e^{i\left(\frac{\pi}{7} \right)} \left( 2 e^{i\left(\frac{\pi}{3} \right)}+1 \right) \\ && \boxed{ e^{i\left(\frac{\pi}{3} \right)} = \cos\left(\frac{\pi}{3} \right) + i\sin\left(\frac{\pi}{3} \right) \\ = \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \\ 2e^{i\left(\frac{\pi}{3} \right)} =1+i\sqrt{3} } \\\\ &=& e^{i\left(\frac{\pi}{7} \right)} \left(1+i\sqrt{3} +1 \right) \\ &=& e^{i\left(\frac{\pi}{7} \right)} \left(2+i\sqrt{3} \right) \\\\ && \boxed{2+i\sqrt{3} = \sqrt{2^2+(\sqrt{3})^2} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } \\ = \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } } \\\\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } e^{i\left(\frac{\pi}{7} \right)} \\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) +i\left(\frac{\pi}{7} \right)} \\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} + \frac{\pi}{7} ) } \\ \left(z+w \right)^6 &=& \left(\sqrt{7}\right)^6 \cdot e^{i\cdot 6\arctan(\frac{\sqrt{3}}{2} + \frac{\pi}{7} ) } \\ |\left(z+w \right)^6| &=& \left(\sqrt{7}\right)^6 \\ &=& 7^3 \\ \mathbf{|\left(z+w \right)^6|} &=& \mathbf{343} \\ \hline \end{array}\)

...Thank you, Alan...

At what point does the graph of the parametric equations

x = 3-2t, y = -2 + 5t, intersect the graph of the parametric equations

x = -1+3s,y = 5-9s? The answer is the point of intersection (x,y).

\(\begin{array}{|lrcll|} \hline 1) & 3-2t &=& -1+3s \\ & 3s+2t &=& 4 \quad | \quad \cdot 3 \\ &\mathbf{ 9s+6t} &=& \mathbf{12} \qquad (1) \\\\ 2) & -2+5t &=& 5-9s \\ & 9s+5t &=& 7 \\ &\mathbf{ 9s+5t} &=& \mathbf{7} \qquad (2) \\ \hline (1)-(2): & 9s+6t -9s-5t &=& 12-7 \\ &\mathbf{t} &=& \mathbf{5} \\ \hline \text{The point of intersection } (x,y)\\ & x &=& 3-2t \\ & x &=& 3-2\cdot 5 \\ & \mathbf{x} &=& \mathbf{-7} \\\\ & y &=& -2+5t \\ & y &=& -2+5\cdot 5 \\ & \mathbf{y} &=& \mathbf{23} \\ \hline & \mathbf{(x,y)} &=& \mathbf{(-7,\ 23)} \\ \hline \end{array}\)