# Complex number equation

1. Sep 22, 2008

### cotufa

1. The problem statement, all variables and given/known data

Solve

$$z^3 - 3z^2 + 6z - 4 = 0$$

3. The attempt at a solution

I tried factoring a z and quadratic equation but went nowhere

Input apreciated

2. Sep 22, 2008

### gabbagabbahey

Well there is a constant term in the equation, so z is clearly not a factor.

The first thing to take note of is that this is a 3rd degree polynomial and so there must be 3 roots. Moreover, at least one of those roots must be real. So try to find a real root by inspection: plug in z=0, z=1, z=-1, z=2...etc. until you find a root z_0 and then factor out a (z-z_0) to obtain a quadratic equation you can then solve to find the other two roots.

3. Sep 22, 2008

### chaoseverlasting

One thing that you could try out is assume $$a+ib$$ is a solution to the equation, plug it in and equate the real and imaginary parts individually equal to zero. That should give you two degree 3 equations which you can individually try to solve...

4. Sep 22, 2008

### cotufa

Ok so 1 works for z0

If I factor a z-1 how would the equation look?

5. Sep 22, 2008

### Useful nucleus

Use any polynomial dividing technique

(Z^3- 3Z^2+ 6 Z-4)/(z-1)= z^2 -2z+4

(z^2 -2z+4) (z-1)=0

The rest simple algebra!

6. Sep 22, 2008

### Defennder

Simply divide the f(z) expression by z-1 using polynomial long division. The result is how it would look (if you did it correctly).

7. Sep 22, 2008

### cotufa

THanks for the help