Complex Polynomial Solutions: z^4 + iz^3 - z^2 - iz + 1 = 0 | Polar Form

Clef
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Homework Statement



(b) Using your answer to part (a), write down all complex roots of the polynomial
z4 + iz3 - z2 - iz + 1:

Homework Equations



(a) Determine all of the complex solutions of
z5 - i = 0:
Write your answers in polar form with -pi < x < pi.

The Attempt at a Solution



I've worked out all of the complex solutions of part a to be :

0= 1e^(ipi/10)

1= 1e^(ipi/2)

2= 1e^(9pi/10)

3= 1e^ (-7pi/10)

4= 1e^(-3pi/10)


But I have no idea how to go about the second part, having only solved complex polynomials with Z not iZ. :S
 
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Use the fact that z^5- a^5= (z- a)(z^4+ az^3+ a^2z^2+ a^3z+ a^4)
With a= i, using the facts that i^2= -1, i^3= -i, i^4= 1, and i^5= i, that becomes z^5- i= (z- i)(z^4+ iz^3- z^2- iz^3+ 1.

Do you see the point?
 
okay, I understand how that is the expanded form.
So would the solutions for part b) just be the polar forms of part a) divided by (z-i)?

Which would in turn make it:

(z5-a5)/(z1-a3)

=

z4-a2

=

z4-1

??
 
The answer to part (b) would be easier to see if you convert the roots of the polynomial from complex exponentials to complex numbers in the form z = a + i b. Once you've done this, is there another way you can express the equation z5 - i = 0 ?
 
Clef said:

Homework Statement



(b) Using your answer to part (a), write down all complex roots of the polynomial
z4 + iz3 - z2 - iz + 1:

Homework Equations



(a) Determine all of the complex solutions of
z5 - i = 0:
Write your answers in polar form with -pi < x < pi.

The Attempt at a Solution



I've worked out all of the complex solutions of part a to be :

0= 1e^(ipi/10)

1= 1e^(ipi/2)

2= 1e^(9pi/10)

3= 1e^ (-7pi/10)

4= 1e^(-3pi/10)


But I have no idea how to go about the second part, having only solved complex polynomials with Z not iZ. :S

Are you by any chance taking calculus 2 at melbourne university? LoL
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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