anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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experimentX
  • experimentX
rational operations??
anonymous
  • anonymous
summ , multiplication division..
experimentX
  • experimentX
with complex number or real number?

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anonymous
  • anonymous
complex
anonymous
  • anonymous
www.dms.uaf.edu/~rybkin/MathPhysicsLectureNotes.pdf follow this
anonymous
  • anonymous
maybe prooving by induction for each operation separatly...
experimentX
  • experimentX
i was thinking the same
experimentX
  • experimentX
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??
anonymous
  • anonymous
@Rohangrr in the link you posted says just check that that's true, but its not prooved
anonymous
  • anonymous
@myko give up a medal then
experimentX
  • experimentX
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)) ???
anonymous
  • anonymous
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
anonymous
  • anonymous
@experimentX ??
experimentX
  • experimentX
yeah ... let's try it!!!
experimentX
  • experimentX
but damn ... what did i do above???
anonymous
  • anonymous
idk
experimentX
  • experimentX
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??????
experimentX
  • experimentX
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
anonymous
  • anonymous
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....
anonymous
  • anonymous
ok, got it too
experimentX
  • experimentX
damn ... man!!! it's what i said as before ...
anonymous
  • anonymous
??
anonymous
  • anonymous
i don't need a simple check, i need proof
experimentX
  • experimentX
conj(a+b) = conj(a) + conj(b) --- we have to operate on conjugate with conjugate
anonymous
  • anonymous
induction works fine
anonymous
  • anonymous
thx for help
experimentX
  • experimentX
prove this for n=2 prove this for n=3 assume this for n=n prove this for n=n+1
anonymous
  • anonymous
thats what i did

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