Finding roots of z^4 + iz^3 - z^2 - iz + 1 using z^5 = i

[Clef]

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

 

[HallsofIvy]

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?

 

[Clef]

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

??

 

[kuruman]

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 ?

 

[terminator88]

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