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 #1
avatar+23086 
+3

Simplify
\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8\)( \(i\) is imaginary)

 

\(\text{Let $z = \dfrac{3 + i \sqrt{3}}{2}$} \\ \text{Let $\overline{z} = \dfrac{3 - i \sqrt{3}}{2}$} \)

 

\(\begin{array}{|rcll|} \hline z\cdot \overline{z} &=& \left(\dfrac{3}{2}\right)^2 +\left( \dfrac{\sqrt{3}}{2} \right)^2 \\ &=& \dfrac{9}{4} + \dfrac{3}{4} \\ \mathbf{z\cdot \overline{z}} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right) &=& \dfrac{3}{2} + \dfrac{i \sqrt{3}}{2}+ \dfrac{3}{2} - \dfrac{i \sqrt{3}}{2} \\ &=& 2\cdot \dfrac{3}{2} \\ \mathbf{\left(z+\overline{z} \right)} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right)^2 = 3^2 &=& z^2+2\cdot z\cdot \overline{z} + \overline{z}^2 \\ 3^2 &=& z^2 + \overline{z}^2 +2\cdot 3 \\ z^2 + \overline{z}^2 &=& 9-6 \\ \mathbf{z^2 + \overline{z}^2} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z^2+\overline{z}^2 \right)^2 = 3^2 &=& z^4+2\cdot z^2\cdot \overline{z}^2 + \overline{z}^4 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot \left(z\cdot \overline{z}\right)^2 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot 3^2 \\ z^4 + \overline{z}^4 &=& 9-18 \\ \mathbf{z^4 + \overline{z}^4} &=& \mathbf{-9} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z^4+\overline{z}^4 \right)^2 = (-9)^2 &=& z^8+2\cdot z^4\cdot \overline{z}^4 + \overline{z}^8 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot \left(z\cdot \overline{z}\right)^4 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot 3^4 \\ z^8 + \overline{z}^8 &=& 81-2\cdot 81 \\ \mathbf{z^8 + \overline{z}^8} &=& \mathbf{-81} \\ \hline \end{array} \)

 

\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8 = \mathbf{z^8 + \overline{z}^8} = \mathbf{-81} \)

 

laugh

Aug 18, 2019
Aug 17, 2019
 #2
avatar+23086 
+2

Calculate a normal vector n to the plane through the points
triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below,
such that if
\(\mathbf{n} = \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}\),
and \(n_1 + n_2 + n_3 = 7\).

\(\text{Let $\vec{v} = \vec{B} - \vec{C}$ } \\ \text{Let $\vec{w} = \vec{A} - \vec{C}$ }\)

 

\(\begin{array}{|rcll|} \hline \vec{v} &=& \begin{pmatrix}-3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{v}} &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \\\\ \vec{w} &=& \begin{pmatrix}1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{w}} &=& \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\ \hline \end{array}\)

 

1.

Vector cross product

\(\small{ \begin{array}{|lrcll|} \hline & \vec{n} =\begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}&=& \vec{v} \times \vec{w} \\\\ & &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \times \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\\\ & &=& \begin{vmatrix}1&1&1 \\ -4&0&2 \\ 0&-3&-1 \end{vmatrix} \\\\ & &=& \begin{pmatrix} 6 \\ -4 \\ 12 \end{pmatrix} \\ \hline n_1+n_2+n_3: & 6-4+12 &=& 14 \quad | \quad \cdot \dfrac{7}{14} \\\\ & 6\cdot\dfrac{7}{14}-4\cdot\dfrac{7}{14}+12\cdot\dfrac{7}{14} &=& 14\cdot\dfrac{7}{14} \\\\ &\mathbf{ 3-2+6 }&=& \mathbf{7} \\\\ & \vec{n} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array} }\)

 

2.

Vector dot product

\(\begin{array}{|lrcll|} \hline & \vec{n}\cdot \vec{v} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} &=& 0 \\ (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ \hline & \vec{n}\cdot \vec{w} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} &=& 0 \\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ \hline (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ \hline \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ &2n_3&=&4n_1 \quad | \quad :2 \\ & \mathbf{n_3} &=& \mathbf{2n_1} \\\\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ & \mathbf{n_3} &=& \mathbf{-3n_2} \\\\ & n_3= 2n_1 &=& -3n_2 \\ & 2n_1 &=& -3n_2 \quad | \quad :2 \\ & \mathbf{n_1} &=& \mathbf{-\dfrac{3}{2}n_2} \\\\ (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ & -\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\ & -\dfrac{3}{2}n_2 -2n_2 &=& 7 \quad | \quad \cdot (-1 )\\ & \dfrac{3}{2}n_2 +2n_2 &=& -7 \\ & \dfrac{7}{2}n_2 &=& -7 \quad | \quad \cdot \dfrac{2}{7} \\ & n_2 &=& -7\cdot \dfrac{2}{7} \\ & \mathbf{n_2} &=& \mathbf{-2} \\ \hline & n_3 &=& -3n_2 \\ & n_3 &=& -3\cdot (-2) \\ & \mathbf{n_2} &=& \mathbf{6} \\ \hline & n_1 &=& -\dfrac{3}{2}n_2 \\ & n_1 &=& \left(-\dfrac{3}{2}\right)\cdot(-2) \\ & \mathbf{n_2} &=& \mathbf{3} \\\\ & \mathbf{\vec{n}} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array}\)

 

laugh

Aug 17, 2019
 #6
avatar+1342 
+2
Aug 17, 2019

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