Matematik

Binom ligning z^6=-729

20. september 2021 af MajaXm - Niveau: Universitet/Videregående

Hej, hvordan løses den binome ligning z^6=-729?

Hvilken metode skal benyttes?

Brugbart svar (0)

Svar #1
20. september 2021 af peter lind

πskriv -729 som 729^eiπ+2pπi de mulige løsninger bliver 729e^{(iπ+2pπi)/6}

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Svar #2
20. september 2021 af mathon

$\small \begin{array}{lllllll}&& z^6=&-729 =729\cdot e^{\textit{\textbf{i}}\cdot \left (\pi+p\cdot 2\pi \right )}\qquad p\in\left \{ 0,1,2,3,4,5 \right \}\\\\&&z=& 729^{\frac{1}{6}}\cdot e^{\textit{\textbf{i}}\cdot \left (\pi+p\cdot 2\pi \right )^{\frac{1}{6}}}\\\\&& z=&3\cdot e^{{\textit{\textbf{i}}\cdot \left (\frac{\pi}{6}+p\cdot \frac{\pi}{3} \right )}}\qquad p\in\left \{ 0,1,2,3,4,5 \right \}\\\\\\&& \end{array}$

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Svar #3
20. september 2021 af mathon

$\small \begin{array}{llllll} \textup{hvoraf:}\\& z=&\left\{\begin{array}{lll} 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6} \right )}&=&\frac{3\sqrt{3}}{2}+\textit{\textbf{i}}\cdot\frac{3}{2}\\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{\pi}{3} \right )}&=&\textit{\textbf{i}}\cdot\cdot 3 \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{2\pi}{3} \right )}&=&-\frac{3\sqrt{3}}{2}+\textit{\textbf{i}}\cdot\frac{3}{2} \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{3\pi}{3} \right )}&=&-\frac{3\sqrt{3}}{2}-\textit{\textbf{i}}\cdot\frac{3}{2} \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{4\pi}{3} \right )}&=&\textit{\textbf{i}}\cdot\left ( -3 \right ) \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{5\pi}{3} \right )}&=&\frac{3\sqrt{3}}{2}-\textit{\textbf{i}}\cdot\left (\frac{3}{2} \right ) \end{array}\right. \end{array}$

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Svar #4
20. september 2021 af janhaa

#3
$\small \begin{array}{llllll} \textup{hvoraf:}\\& z=&\left\{\begin{array}{lll} 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6} \right )}&=&\frac{3\sqrt{3}}{2}+\textit{\textbf{i}}\cdot\frac{3}{2}\\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{\pi}{3} \right )}&=&\textit{\textbf{i}}\cdot\cdot 3 \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{2\pi}{3} \right )}&=&-\frac{3\sqrt{3}}{2}+\textit{\textbf{i}}\cdot\frac{3}{2} \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{3\pi}{3} \right )}&=&-\frac{3\sqrt{3}}{2}-\textit{\textbf{i}}\cdot\frac{3}{2} \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{4\pi}{3} \right )}&=&\textit{\textbf{i}}\cdot\left ( -3 \right ) \\ 3\cdot e^{\textit{\textbf{i}}\cdot\left ( \frac{\pi}{6}+\frac{5\pi}{3} \right )}&=&\frac{3\sqrt{3}}{2}-\textit{\textbf{i}}\cdot\left (\frac{3}{2} \right ) \end{array}\right. \end{array}$

hvilken layout er dette?

Latex?

Svar #5
21. september 2021 af MajaXm

#2
$\small \begin{array}{lllllll}&& z^6=&-729 =729\cdot e^{\textit{\textbf{i}}\cdot \left (\pi+p\cdot 2\pi \right )}\qquad p\in\left \{ 0,1,2,3,4,5 \right \}\\\\&&z=& 729^{\frac{1}{6}}\cdot e^{\textit{\textbf{i}}\cdot \left (\pi+p\cdot 2\pi \right )^{\frac{1}{6}}}\\\\&& z=&3\cdot e^{{\textit{\textbf{i}}\cdot \left (\frac{\pi}{6}+p\cdot \frac{\pi}{3} \right )}}\qquad p\in\left \{ 0,1,2,3,4,5 \right \}\\\\\\&& \end{array}$

Forstår ikke helt hvad der sker i sidste led i den øverste linje. Hvorfor kan man gøre det?

ellers mange tak for hjælpen:)

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