Non-Real Cube RootsDate: 01/28/2001 at 19:28:22 From: Nicole Subject: Cube Roots The real cube root of -8 is -2. -8 also has 2 NON-real cube roots. Find those. (Use (a+bi)^3 and solve for a and b. i= imaginary number.) I tried solving, but as soon as I started, the i's totally confused me. I need to see it step by step. Date: 01/29/2001 at 20:55:39 From: Doctor Fenton Subject: Re: Cube Roots Hi Nicole, Thanks for writing to Dr. Math. Don't be afraid of the i's - just keep the real and imaginary parts of the complex number separate. You probably recall from algebra that (x+y)^3 = x^3 + 3*x^2*y + 3*x*y^2 + y^3 ; apply this to (a+bi)^3 to get (a+b*i)^3 = a^3 + 3*a^2*(b*i) + 3*a*(b*i)^2 + (b*i)^3 = a^3 + (3*a^2*b)*i + 3*a*(b^2*i^2) + b^3*i^3 . Now simplify, using i^2 = -1 and i^3 = -i , to get (a+bi)^3 = a^3 + (3*a^2*b)*i - 3*a*b^2 - b^3*i = (a^3 - 3*a*b^2) + (3*a^2*b - b^3)*i = -2 (which is -2 + 0*i ). When complex numbers are equal, then their real and imaginary parts must also be equal, so a^3 - 3*a*b^2 = -2 and 3*a^2*b - b^3 = 0 . Since you are looking for a non-real solution, b isn't 0, so divide the second equation by b: 3*a^2 - b^2 = 0 or 3*a^2 = b^2 . Replace b^2 in a^3 - 3*a*b^2 = -2 by 3*a^2, and you have a^3 - 3*a*(3*a^2) = -2 . This determines a, and you can determine two values of b from 3*a^2 = b^2 . I think you will be able to finish the problem from these hints. If you have further questions, please write us again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/