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Proof that the number is Real

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    z1, z2 are complex numbers.
    If z1z2 =/= -1
    and |z1| = |z2| = 1

    then number :
    z1 + z2
    ________
    1 + z1z2

    is real.
    2. Relevant equations


    3. The attempt at a solution
    z1 = (a+bi), z2 = (c+di)

    Should i use this extended form or is there a shorter path ? because it's pretty long.
     
    Last edited: May 12, 2016
  2. jcsd
  3. May 12, 2016 #2

    BvU

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    Hi Dank,

    Pity there are no relevant equations available to you. If you want a shorter path you might need something to work with. One thing I can think of is that for any complex number ##\bf\alpha## the form ##\bf\alpha^*\alpha## is real number. But in this case the given ##|{\bf z}|## seems to point in the direction of another kind of notation than the one you propose

    (he said, mysterically, because he didn't want to give it all away so easily:rolleyes:)
     
  4. May 12, 2016 #3

    BvU

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    Does that signify you picked up the hint and found a fairly short path indeed ?
     
  5. May 12, 2016 #4
    Ohh
    Sorry it was z1 + z2 and not z1 + z1, and also i marked * as multiplication in mistake and edited.

    So if i take the absolute value of the fraction i get : |z1+z2| over |1+z1z2|
    , can't get much further here using the data
     
  6. May 12, 2016 #5
    nope, since i had a typo and now it's fixed. so i bet the hints will change ;D
     
  7. May 12, 2016 #6

    BvU

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    That was clear. And the * in the denominator is indeed multiplication. The hints stay as they were.
    But going for absolute values isn't productive: basically you don't care about them if you want to show something has imaginary part zero.

    So I'll repeat the nudge: |z| = 1 should inspire a different notation : ## {\bf z} = \alpha + i\beta## has two unknowns. The notation I mean has only one :smile:
     
  8. May 12, 2016 #7
     
  9. May 12, 2016 #8
    Ok i think ill try somthing and post here in few min
     
    Last edited: May 12, 2016
  10. May 12, 2016 #9
    Taking the conjugate of all the fraction might yield something meaningful? but if i then multiply it with the fraction ill get a real number, but that doesn't show me why z is real still...
     
  11. May 12, 2016 #10
    I'm afraid i need another hint,
     
    Last edited: May 12, 2016
  12. May 12, 2016 #11

    BvU

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    You loose a few brownie points, but what the heck: what is the locus of points ##z_1## and ##z_2## in the complex plane ?
     
  13. May 12, 2016 #12
    Locus? maybe there is a different name for it, i'm studying linear algebra 1. hmm vectors you mean? or circle unit?
     
  14. May 12, 2016 #13

    BvU

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    Unit circle is the name. And yes, locus is the collection of positions ##z_i## can occupy in the complex plane.
    One way to pinpoint a complex number is by Re() and Im(), real and imaginary parts (your ##a## and ##b## ).
    Another way is with Abs() and Arg(), absolute value and angle . Since in this exercise ##Abs(z_i) = 1## that's what you want to adopt !
     
  15. May 12, 2016 #14
    Can write z1 = 1(cosx+isinx), and z2 = 1(cosy+isiny), where r = 1?
    and then plug it in the fraction?
     
  16. May 12, 2016 #15

    BvU

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    You can indeed. A short form for ##\cos\phi + i\sin\phi## is also available (saves writing effort :smile:).
     
  17. May 12, 2016 #16
    cosx + isinx + cosy + isiny over 1 +cos(x+y) + isin(x+y) +1
     
  18. May 12, 2016 #17

    BvU

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    Hehe, you reluctant to write $$
    e^{i\phi_1} + e^{i\phi_2}\over 1+e^{i\left ( \phi_1 +\phi_2\right )} $$ or something ?

    Now: what's the road ahead if you want to show this fraction is a real number ?
     
  19. May 12, 2016 #18
    Haven't learned using e at all yet, my first course in linear algebra 1.
     
  20. May 12, 2016 #19
    well, i would want the imaginary part to be gone from the fraction

    but sinx+siny =/= sin(x+y)i ;(
     
  21. May 12, 2016 #20

    BvU

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    Oh, sorry. Well, in that case my respect that you got this far already !
    We proceed with sin and cos. What's the way to ensure/make the denominator is real (remember post # 2)
     
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