Solving a polynomial with complex properties

In summary, a polynomial with complex properties is a mathematical expression involving complex numbers and can be solved using various techniques. The degree of a polynomial with complex properties is determined by its highest exponent, and the steps for solving it involve simplifying, factoring, and using methods such as the quadratic formula or synthetic division. These polynomials can have multiple solutions, with the solutions being real or complex depending on the roots of the polynomial and the discriminant.
  • #1
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Homework Statement


If z = cosθ + isinθ, show that zn + z-n = 2cosnθ. Hence solve the equation 3z4 - z3 + 4z2 - z + 3 = 0 for complex z.
(Hint: Reduce the given equation to an equation with the unknown cosθ)


Homework Equations


z = cosθ + isinθ
zn + z-n = 2cos(nθ)
3z4 - z3 + 4z2 - z + 3 = 0
cos2θ = 2cos2θ - 1

The Attempt at a Solution


From de Moivre's Theorem
zn = cos(nθ) + isin(nθ) AND z-n = cos(-nθ) + isin(-nθ) = cos(nθ) - isin(nθ)
Hence zn + z-n
= cos(nθ) + isin(nθ) + cos(nθ) - isin(nθ)
= 2cos(nθ)

3z4 - z3 + 4z2 - z + 3 = 0
Divide both sides by z2
3z2 - z + 4 - z-1 + 3z-2 = 0

Rearranging to
3z2 + 3z-2 - z - z-1 + 4 = 0
3(z2 + z-2) - (z + z-1) + 4 = 0

Using zn + z-n = 2cos(nθ)
3(2cos2θ) - (2cosθ) + 4 = 0

Substituting cos2θ = 2cos2θ - 1
3(2[2cos2θ - 1]) - 2cosθ + 4 = 0

Simplifies down to
6cos2θ - cosθ - 1 = 0

Solving for cosθ
(2cosθ - 1)(3cosθ + 1) = 0
Therefore cosθ = 1/2 OR cosθ = -1/3
cosθ = 1/2, θ = π/3 or 5π/3
cosθ = -1/3, θ = cos-1(-1/3)
(Note: π = funny looking pi)

Hence (using complex conjugate is also a root of polynomial)
z1 = 1/2 + sqrt(3)/2 i
z2 = 1/2 - sqrt(3)/2 i
z3 = -1/3 + sin[cos-1(-1/3)] i
z4 = -1/3 - sin[cos-1(-1/3)] i

I'm pretty confident that I've done it right until the cosθ = -1/3 and getting sin[cos-1(-1/3)] in the roots z3 and z4. So my question is have I the whole question correctly? If not can you show me how it should be done. Thanks in advance!
 
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  • #2
You did it correctly, though you should simplify sin[arccos(-1/3)]. You can always check your answers by plugging them back into the original equation.
 
  • #3
Ok thankyou (:

umm another question unrelated
but now I have another concern

for a function to be STRICTLY increasing, is the point 0 actually included? Because me and my friends all have different opinions. So the gradient at x = 0 is 0 so would the function still be considered as strictly increasing if it included x=0?
 
  • #4
Strictly increasing means f'(x)>0 for all x. If you have f'(x)≥0, the function is monotonically increasing.
 
  • #5
Just to add one point to your original post, when I did questions like these in Further Math, it was always best to show that z=0 was not a root before dividing by any power of z.
 

1. What is a polynomial with complex properties?

A polynomial with complex properties is a mathematical expression consisting of variables, coefficients, and exponents that involve complex numbers. Complex numbers are numbers in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1). Polynomials with complex properties can be solved using techniques such as factoring, synthetic division, and the quadratic formula.

2. How do I determine the degree of a polynomial with complex properties?

The degree of a polynomial with complex properties is the highest exponent of its variable. For example, the polynomial 3x^2 + 5x + 2 has a degree of 2. In general, the degree of a polynomial is equal to the number of terms when the polynomial is written in its standard form.

3. What are the steps for solving a polynomial with complex properties?

The steps for solving a polynomial with complex properties may vary depending on the specific polynomial and the method used. However, a general approach is to first simplify the polynomial by combining like terms and then factor it if possible. If factoring is not possible, the polynomial can be solved using the quadratic formula or synthetic division.

4. Can a polynomial with complex properties have multiple solutions?

Yes, a polynomial with complex properties can have multiple solutions. This is because complex numbers have both a real and imaginary component, allowing for multiple solutions to exist. For example, the polynomial x^2 + 1 has two complex solutions, x = i and x = -i.

5. How do I know if the solutions to a polynomial with complex properties are real or complex?

The solutions to a polynomial with complex properties can be determined by solving for the roots of the polynomial. If the roots are complex numbers, then the solutions will also be complex. However, if the roots are real numbers, then the solutions will also be real. Additionally, the discriminant of a quadratic polynomial, b^2 - 4ac, can be used to determine if the solutions will be real or complex.

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