Question

I don't know where to begin with this question:
http://dl.dropbox.com/u/6717478/7.png

I think it might have something to do with a proof by induction, but I am having trouble setting that up.

Answer 1

Good luck with that

Answer 2

http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem#Simple_proof

Answer 3

Zarkon....so resourcefull

Answer 4

I can't follow the last 2 lines of the proof
could I please have a little explanation?

Answer 5

If I remember correctly, this was the reasoning I was presented with some time ago.

Since all the coefficients are real, if you have one complex root, you need to be able to multiply it by another complex root to balance it out. The conjugate is what does this for you.

I learned this a while ago, and it may be incorrect, but it seems to make sense.

Answer 6

what are you balancing out?

Answer 7

if there was only one complex root, you would be forced to have a complex coefficient. Also, if you multiply all the complex roots together, they must be real.

Answer 8

but it says that c is a root of f(x) in the proposition so it doesn't need to be multiplied by the conjugate right?

Answer 9

but c is in the complex. So to ensure that all the coefficients are real, we have to multiply c by its conjugate.

Answer 10

from what I can tell in the proof they are just taking the conjugate of the complex term
then they say that it ends up being the conjugate of 0, which is 0

Answer 11

ok how about this
if c is a complex root of f(x)
then the sum of all the terms of the polynomial = 0 (3rd line of that proof)
take the conjugate of both sides
by the properties of the conjugate you know that the conjugate of the sum will be the sum of the individual conjugates.
and the conjugate of 0 is 0
so then the sum of all the terms of the polynomial = 0 also when using the conjugate
which tells us that if c is a complex root of f(x), then so is the conjugate of c

Answer 12

From what I can tell, in the second to last line of that proof, they first say that that conjugate of (z+w) is the conjugate of z plus the conjugate of w, then that the conjugate of zw is the conjugate of z times the conjugate of w, then that the conjugate of z^n is the
(conjugate of z)^n. Finally, that's just the original equation with the conjugate instead of the root. Since they both equal 0, they must both be roots.

Answer 13

basically what you just said.

Answer 14

see page 13 http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF

Answer 15

alright I think this is making sense now. It's simpler then I thought it would be actually. But, maybe I'm just doing something wrong.

Answer 16

I don't even know what they are trying to prove there

Answer 17

number 14, top of page 13 they have a short proof for the CJRT that's very similar to what you described.

Answer 18

oh yeah that is actually really helpful, thank you