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How do you find three cube roots of -64? - Answers
Not calculus, but -64 has three cube roots, two of which are complex numbers. Note, sqrt(-1) = i. 4*e^(pi*i/3) is a root because [4*e^(pi*i/3)]^3 = 64*e^(pi*i) = 64*-1= -64 4*e^(-pi*i/3) is a root because [4*e^(-pi*i/3)]^3 = 64*e^(-pi*i) = 64*-1= -64 and of course -4 = 4*e^(pi*i) is a root because (-4)^3 = -64. To find these, set up the equation -64 = [A*e^(pi*i*B)]^3 and solve for A and B. If you don't like using complex exponentials, any complex number can also be written as a + b*i so you could solve -64 = (a + bi)^3. If you use the complex exponential form, you should get the three answers above. If you use the a + b*i form you should get the equivalent {2 +/- 3.464i, -4}. Remember either equations has three solutions because it is cubic.
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How do you find three cube roots of -64? - Answers
Not calculus, but -64 has three cube roots, two of which are complex numbers. Note, sqrt(-1) = i. 4*e^(pi*i/3) is a root because [4*e^(pi*i/3)]^3 = 64*e^(pi*i) = 64*-1= -64 4*e^(-pi*i/3) is a root because [4*e^(-pi*i/3)]^3 = 64*e^(-pi*i) = 64*-1= -64 and of course -4 = 4*e^(pi*i) is a root because (-4)^3 = -64. To find these, set up the equation -64 = [A*e^(pi*i*B)]^3 and solve for A and B. If you don't like using complex exponentials, any complex number can also be written as a + b*i so you could solve -64 = (a + bi)^3. If you use the complex exponential form, you should get the three answers above. If you use the a + b*i form you should get the equivalent {2 +/- 3.464i, -4}. Remember either equations has three solutions because it is cubic.
DuckDuckGo
How do you find three cube roots of -64? - Answers
Not calculus, but -64 has three cube roots, two of which are complex numbers. Note, sqrt(-1) = i. 4*e^(pi*i/3) is a root because [4*e^(pi*i/3)]^3 = 64*e^(pi*i) = 64*-1= -64 4*e^(-pi*i/3) is a root because [4*e^(-pi*i/3)]^3 = 64*e^(-pi*i) = 64*-1= -64 and of course -4 = 4*e^(pi*i) is a root because (-4)^3 = -64. To find these, set up the equation -64 = [A*e^(pi*i*B)]^3 and solve for A and B. If you don't like using complex exponentials, any complex number can also be written as a + b*i so you could solve -64 = (a + bi)^3. If you use the complex exponential form, you should get the three answers above. If you use the a + b*i form you should get the equivalent {2 +/- 3.464i, -4}. Remember either equations has three solutions because it is cubic.
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- og:descriptionNot calculus, but -64 has three cube roots, two of which are complex numbers. Note, sqrt(-1) = i. 4*e^(pi*i/3) is a root because [4*e^(pi*i/3)]^3 = 64*e^(pi*i) = 64*-1= -64 4*e^(-pi*i/3) is a root because [4*e^(-pi*i/3)]^3 = 64*e^(-pi*i) = 64*-1= -64 and of course -4 = 4*e^(pi*i) is a root because (-4)^3 = -64. To find these, set up the equation -64 = [A*e^(pi*i*B)]^3 and solve for A and B. If you don't like using complex exponentials, any complex number can also be written as a + b*i so you could solve -64 = (a + bi)^3. If you use the complex exponential form, you should get the three answers above. If you use the a + b*i form you should get the equivalent {2 +/- 3.464i, -4}. Remember either equations has three solutions because it is cubic.
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