Complex PolynomialDate: 11/22/96 at 08:54:56 From: Peter Johansson Subject: Complex Polynomial Hi, I have been sitting trying to solve a complex equation for some hours but getting nowhere. It looks like this: z^4 + 3iz^3 - (4 + i)z^2 - 3iz + 3 + i = 0. I would really be happy if you could send the whole solution. Peter Date: 04/14/97 at 14:44:13 From: Doctor Greg Subject: Re: Complex Polynomial Dear Peter, This is a nice problem! The first thing that I notice is that this is a polynomial equation of degree 4: 4 3 2 z + 3iz - (4+i)z - 3iz + (3+i) = 0 Now, there are formulas for the roots of a fourth degree equation in terms of the coefficients, so if worst comes to worst, one can always use those formulas. However, when you're given a homework problem, it's often faster to try to guess some roots. Okay, let's try z = 0... nope, we get (3+i). Let's try z=1... we get 1 + 3i - (4+i) - 3i + 3+i... =0. Hey, it works! So, this means that (z-1) is a factor of (z^4 + ...). So, now that we have a factor, we can divide it into the original fourth degree polynomial and continue by tring to factor the cubic. We get: 4 3 2 3 2 z + 3iz - (4+i)z - 3iz+(3+i)=(z-1)(z + (1+3i)z + (-3+2i)z - 3-i) So, we play the same game with the cubic! (Again, there are also formulas for the roots of a cubic in terms of the coefficients, but if we can quickly guess a root, we can avoid having to invoke the formula.) Try z=0... oh, we don't have to try this since we already did it. Try z=1... nope (although we did already try z=1, it is possible that 1 is a multiple root.) Try z=i... nope Try z=-1... success! So, a second factor is (z+1). From here, you can factor out z+1 from the cubic and then use the quadratic formula to find the roots of the resulting quadratic equation. Hope this helps... and let us know if you're still having trouble! -Doctors Greg and Ceeks, The Math Forum Check out our web site! 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/