Question

Suppose that a complex number u is obtained as a result of a finite number of rational operations applied to the complex numbers z1,z2,z3,..zn. Prove that the same operations applied to the complex conjugates of z1,z2,z3..zn leads to the number ü, conjugate of u.

Answer 1

rational operations??

Answer 2

summ , multiplication division..

Answer 3

with complex number or real number?

Answer 4

complex

Answer 5

www.dms.uaf.edu/~rybkin/MathPhysicsLectureNotes.pdf follow this

Answer 6

maybe prooving by induction for each operation separatly...

Answer 7

i was thinking the same

Answer 8

let a+ib be a complex number, so a-ib is it's conjugate, try adding c+id to both of them,

(a+c)+(b+d)j for our number
(a+c)+(d-b)j or conjugate ... Oo whats this??

Answer 9

@Rohangrr
in the link you posted says just check that that's true, but its not prooved

Answer 10

@myko give up a medal then

Answer 11

let's move on to multiplication:
(a+ib) x (c+id) = (ac-bd)+i(ad+bd))
for conjugate
(a+ib) x (c-id) = (ac+bd)+i(ad-bd)) ???

Answer 12

I think it might be easyer by induction:
Summ:
supose true for n=2... Conj(z1+z2) = Conj(z1)+Conj(z2)
for n=3 ...Conj(z1)Conj(z2)Conj(z3)=Conj(z1+z2) + Conj(z3) = Conj(z1+z2+z3)
Similarly for multiplication

Answer 13

@experimentX
??

Answer 14

yeah ... let's try it!!!

Answer 15

but damn ... what did i do above???

Answer 16

idk

Answer 17

let's prove the basic first ..
(a+ib) x (c+id) = (ac-bd)+i(ad+bd))
for conjugate
(a-ib) x (c+id) = (ac+bd)+i(ad-bd)) ???

let's see for conjugate
(a-ib) x (c-id) = (ac-bd)-i(ad+bd)) ???

maybe you meant ... operation on conjugate bu conjugate??????

Answer 18

same for
(a+c)+(b+d)j for our number
(a+c)-(d+b)j or conjugate ... for addition.

@myko i think you operation on conjugate by conjugate

Answer 19

I have to prove Conj(z1,z2,z3...zn) = Conj(z1)Conj(z2)...Conj(zn)
and its true for every operation separatly
I am only stragling with division by induction....

Answer 20

ok, got it too

Answer 21

damn ... man!!! it's what i said as before ...

Answer 22

??

Answer 23

i don't need a simple check, i need proof

Answer 24

conj(a+b) = conj(a) + conj(b) --- we have to operate on conjugate with conjugate

Answer 25

induction works fine

Answer 26

thx for help

Answer 27

prove this for n=2
prove this for n=3

assume this for n=n
prove this for n=n+1

Answer 28

thats what i did