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### Complex Polynomial

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

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