finding roots
- Thread starter tsh44
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mad_mathematician
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- Nov 9, 2004
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z^3 + 8 = 0
z^3 = -8
z= -2
thats the real root, and im not sure how you'd find the complex roots.
z^3 = -8
z= -2
thats the real root, and im not sure how you'd find the complex roots.
pka
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- Jan 29, 2005
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Because \(\displaystyle z^3 = - 8\) z is one of the three cube roots of –8.
Because \(\displaystyle - 8 = 2^3 (\cos (\pi ) + i\sin (\pi )) = 2^3 e^{i\pi }\), the three roots are:
\(\displaystyle \begin{array}{l}
2(\cos (\pi /3) + i\sin (\pi /3)) = 2e^{i\pi /3} \\
2(\cos (\pi ) + i\sin (\pi )) = 2e^{i\pi } \\
2(\cos ( - \pi /3) + i\sin ( - \pi /3)) = 2e^{ - i\pi /3} \\
\end{array}\).
Because \(\displaystyle - 8 = 2^3 (\cos (\pi ) + i\sin (\pi )) = 2^3 e^{i\pi }\), the three roots are:
\(\displaystyle \begin{array}{l}
2(\cos (\pi /3) + i\sin (\pi /3)) = 2e^{i\pi /3} \\
2(\cos (\pi ) + i\sin (\pi )) = 2e^{i\pi } \\
2(\cos ( - \pi /3) + i\sin ( - \pi /3)) = 2e^{ - i\pi /3} \\
\end{array}\).
Just solve the quadratic in the factorization to find the two complex roots.
You can use the quadratic formula.
\(\displaystyle \L\\z^{2}-2z+4=0\)
\(\displaystyle \L\\\frac{-(-2)+\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1+\sqrt{3}i\)
\(\displaystyle \L\\\frac{-(-2)-\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1-\sqrt{3}i\)
You can use the quadratic formula.
\(\displaystyle \L\\z^{2}-2z+4=0\)
\(\displaystyle \L\\\frac{-(-2)+\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1+\sqrt{3}i\)
\(\displaystyle \L\\\frac{-(-2)-\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1-\sqrt{3}i\)