complex roots occuring in conjugate pairs (1 Viewer)

[stag_j]

what types of questions would actually require us to prove that complex roots of polymonials occur in conjugate pairs?
i understand that usually it is sufficient just to state this rule and then say that the conjugate will also be a root.
but sometimes when i see the answers to these questions they also have the proof of the rule.
do you think its worth including just to be safe?

 

[Affinity]

erh.. depends.. I would say 1-2 marks .. no proof required

3+ prove

 

[Harimau]

Proving is quite a simple task once you practice it several times, i suggest you do this. Its a popular proof...

 

[Affinity]

the proof is easy.. but takes 1.5 mins to write down

 

[underthesun]

1.5 minutes for 3 marks = profit!

 

[Rahul]

do we prove it by using the factor theorem?

 

[felix_js]

is this the "polynomial has real coefficients, therefore has complex conjugate roots" ??

 

[tonberry_kun]

Originally posted by stag_j
what types of questions would actually require us to prove that complex roots of polymonials occur in conjugate pairs?

dont u also have to say "with REAL coefficients"?

 

[stag_j]

yes the polynomial has to have real coefficients for this rule to apply.
for the proof ill just give the example of one of the questions i was doing.
i dont know of an easy way to to the symbol for conjugate so i'm going to use |x| to mean conjugate. so dont go getting mixed up and think im talking about modulus...

2z^3-3z^2+18z+10=0
|2z^3-3z^2+18z+10|=|0|
|2z^3|-|3z^2|+|18z|+|10|=0
|2|.|z|^3-|3|.|z|^2+|18|.|z|+|10|=0
2.|z|^3-3.|z|^2+18|z|+10=0
therefore if x+iy is a root of the equation, |x+iy| = x-iy will also be a root of the equation.

1-3i is given as a root, so 1+3i is also a root.
using sum of roots or dividing the polymonial by (z-1-3i)(z-1+3i) gives last root as -1/2

 

[tonberry_kun]

gets confusing cuz i keep thinking that's modulus

 

[stag_j]

do you have a better idea? if so then ill change it.
that was just the best i could think of at the time.

 

[Harimau]

Originally posted by stag_j
do you have a better idea? if so then ill change it.
that was just the best i could think of at the time.

That is the quickest way to do it...

 

[Affinity]

use Z' for conjugate of Z.

 

[Grizzly]

Two "proofs", one from cambridge, pg 124
the other from excel, pg 126.

Cambridge interpration:

1. Suppose P(x) has real coefficients, and a non real zero a (alpha).

2. Consider D(x) = ( x - a ) . ( x - a' )

2. D(x) = x^2 - ( a - a' )x + a.a'
= x^2 - 2Re(a)x + |a|^2

3. Therefore, d(x) also has real coefficients.

4. It is known, P(x)=D(x)Q(x) + R(x) [ Degree of R < D ]

5. Then, P(x) = ( x - a ).(a - a').Q(x) + ( cx + d) [ c, d real]

6. P(a) = ca + d

7. P(a') = ca' + d
= (ca+d)'
= [P(a)]'

8. Therefore, P(a) = 0 and P(a')



Quoting from Excel :

If x = a+ib it is a complex zero of the polynomial P(x) of degree >=2, with real coeffiecents, then x = a - ib is also a zero of P(x). It follows that if x = a+ib is a root, of the real poly. equation P(x) = 0, then x = a - ib is also a root


Not really a "proof" by excel

Hoped that helped

 

[Constip8edSkunk]

Originally posted by Grizzly
5. Then, P(x) = ( x - a ).(a - a').Q(x) + ( cx + d) [ c, d real]

this should be P(x) = ( x - a ).(x - a').Q(x) + ( cx + d)

and the excel one isnt a proof, its basically just restating the theory

 

[Affinity]

complicated.. just take conjugates of both sides like stag_j did

 

[Grizzly]

this should be P(x) = ( x - a ).(x - a').Q(x) + ( cx + d)

ops, yup, sorry, a typo
thanks