Question

Find the other complex roots of 4(z^8)+17(z^4)+4=0 if we know 1+i and hence 1-i and 1/(1+i) and 1/(1-i) are solutions.

Answer 1

\[4z^8+17z^4+4=0\]
you wanna find the complex roots as in:
\[z^n=r^n(\cos(n\theta)+isin(n\theta))\]
?

Answer 2

Where is your complex number ? you need to find the roots of ?

Answer 3

Yeah there should be 8 roots. What do you mean where is my complex number?

Answer 4

You're given 4 solutions and has to find the 4 remaining.
Normally im used to solve for the complex roots when the complex number is given like: a+ib.

Answer 5

Yes any form is fine

Answer 6

Im too tired, to help you, im sorry mate : /

Answer 7

its fine :(

Answer 8

All right, then, if you know 4 solutions, you can then use polynomial division to extract the last equivalent expression. Try doing that.

By this, I mean, for your polynomial \(f(x)\), there is some other polynomial \(g(x)\) such that:
\[
f(x)=(x-(1+i))(x-(1-i))(x-(1+i)^{-1})(x-(1-i)^{-1})\;g(x)
\]

Answer 9

Yes/no? Would you like me to expand on this?

Answer 10

Actually, never mind, I'm blind.

What kind of function is your polynomial, check that and you'll have your answer without having to solve anything out.

Answer 11

so i get \[f(z)=(z^4-3z^3+4.5z^2-3z+1)g(z)\]

Answer 12

Disregard my previous to last comment.

Think of the property of even exponents. A sum of even exponents yields what kind of function?

Answer 13

Say, we have a function:
\[
f(x)=\sum_{n} a_nx^{2n}
\]What does that say about:
\[
f(-x)?
\]

Answer 14

This is a really pretty problem, and it's a lot easier than it seems; there's very little work involved.

Answer 15

I have a question, please answer me @any mathematicians
if I let z^4 = u, then I have 4u^2 +17u+4=0, from then I have u = -0.25 and u = -4
consider u = -0.25 it means z^4 = -0.25 so \(z^2 = \pm0.5\)
and continue with the rest
wonder whether I am on the right track? appreciate any respond

Answer 16

does that mean the negative of the roots are also a solutions?

Answer 17

sorry, z^2 =\(\pm0.5i\)

Answer 18

Yessir. If we have a solution \(f(a)\), then, immediately, \(f(-a)\) is a solution.

Answer 19

oooh thank you :)

Answer 20

It's pretty, huh?

And, sure thing.

Answer 21

Here's a little bit to expand:

Since \(f(x)\) is an even function, we have:
\[
f(x)=f(-x)
\]So, for some root \(a\), we have:
\[
f(a)=f(-a)=0
\]So, for any root, its negation is immediately a root, itself.

Note that this property also applies to odd functions.