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Non-Real Cube Roots


Date: 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/   
    
Associated Topics:
High School Imaginary/Complex Numbers

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