Can You Solve This Complex Number Equation?

In summary, I tried factoring a z and quadratic equation but went nowhere. I then assumed a+ib was a solution and tried to solve for the other two roots, but it was simple algebra once I found a z-1.
  • #1
cotufa
16
0

Homework Statement



Solve

[tex]
z^3 - 3z^2 + 6z - 4 = 0

[/tex]

The Attempt at a Solution



I tried factoring a z and quadratic equation but went nowhere

Input apreciated
 
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  • #2
cotufa said:

Homework Statement



Solve

[tex]
z^3 - 3z^2 + 6z - 4 = 0

[/tex]

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.
 
  • #3
One thing that you could try out is assume [tex]a+ib[/tex] 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
gabbagabbahey said:
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?
 
  • #5
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!
 
  • #6
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
THanks for the help
 

1. What is a complex number equation?

A complex number equation is an equation that includes both real and imaginary numbers. It is written in the form of a + bi, where a is the real number and bi is the imaginary number.

2. How do you solve a complex number equation?

To solve a complex number equation, you can use the properties of complex numbers, such as the distributive property, to simplify the equation. Then, you can use the quadratic formula or other methods to find the values of the variables.

3. What is the difference between a real number equation and a complex number equation?

A real number equation only includes real numbers, while a complex number equation includes both real and imaginary numbers. Real numbers can be represented on a number line, while complex numbers have a real part and an imaginary part.

4. Can a complex number equation have multiple solutions?

Yes, a complex number equation can have multiple solutions. This is because the equation can have multiple values for the variables that satisfy the equation. These solutions can be represented as points on a complex plane.

5. What is the significance of complex number equations in science?

Complex number equations are used in many fields of science, such as physics and engineering, to model and solve real-world problems. They are particularly useful in analyzing systems with oscillatory behavior, such as electrical circuits and wave phenomena.

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