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Solving a polynomial with complex properties!

  • Thread starter moocav
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  • #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!
 

Answers and Replies

  • #2
vela
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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
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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
vela
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Strictly increasing means f'(x)>0 for all x. If you have f'(x)≥0, the function is monotonically increasing.
 
  • #5
rock.freak667
Homework Helper
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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.
 

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