Stuck on Complex Equations: Can You Help?

[cotufa]

Homework Statement

Solve

$$z^3 - 3z^2 + 6z - 4 = 0<br />$$

The Attempt at a Solution

I tried factoring a z and quadratic equation but went nowhere

Input apreciated

 

[gabbagabbahey]

Homework Statement

Solve

$$z^3 - 3z^2 + 6z - 4 = 0<br />$$

The Attempt at a Solution

I tried factoring a z and quadratic equation but went nowhere

Input apreciated

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.

 

[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...

 

[cotufa]

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.

Ok so 1 works for z0

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

 

[Useful nucleus]

Use any polynomial dividing technique

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

So your eqn is:

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

The rest simple algebra!

 

[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).

 

[cotufa]

THanks for the help