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Simple proof

  1. Dec 5, 2004 #1
    Can anyone help me with this:

    Let w be a complex number with the property [tex]w \leq 2[/tex].

    Prove that w can be written as a sum of to complex numbers on the unit circle.
    That is; prove that w can be written as [tex]w = z_1 + z_2[/tex], where [tex]|z_1| = 1[/tex] and [tex]|z_2| = 1[/tex].

    I really can't come up with a consistent proof, although it is pretty obvious :/
     
  2. jcsd
  3. Dec 5, 2004 #2
    What does w<=2 mean for a complex number ?
     
  4. Dec 5, 2004 #3

    dextercioby

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    He misspelled it.Obviously it should have been a modulus.
    The proof is trivial.
    Let [itex] w [/itex] be a complex number such as [itex] |w|\leq 2 =|z_{1}|+
    |z_{2}|[/itex],where [itex] |z_{1}|=1;|z_{2}|=1 [/itex].Remember that the only constraint u have on "w" is the fact that its modulus is less or equal to 2.You state:
    Proposition:[itex] w=z_{1}+z_{2} [/itex].
    Proof:If [itex] |z_{1}|=1;|z_{2}|=1 [/itex],then [itex] z_{1}=\exp(i\phi_{1});z_{2}= \exp(i\phi_{2}) [/itex].It's simply applying Euler's formula and the fact that the trigonometric function cosine (for real arguments) is always less or equal to one,that u prove the following [itex] |z_{1}+z_{2}|\leq 2 [/itex],with the zeds having the form of complex exponentials of unity modulus.
    Once u've proven that,u can chose [itex] w=z_{1}+z_{2} [/itex],as u've just proven that the RHS in modulus is less or equal to 2,which was the only constraint upon "w"(actually upon his modulus).
    Endproof.

    Daniel.
     
  5. Dec 5, 2004 #4
    Just how does |z_1 + z_2| <= 2 imply w = z_1 + z_2?
     
  6. Dec 5, 2004 #5

    dextercioby

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    From the text of the problem,one deduces that "w" is a complex number of arbitrary form as long as his modulus is less/equal to 2.So its trivial to chose him of that particular form,as long as that particular form complies with the constraint.

    I've just thought of a geometrical approach,based on the possibility of representing complex numbers in terms of vectors in the complex Gauss plane.Basically u have to prove that any vector of modulus less/equal to 2 can be seen as the diagonal of the paralelelogram built on 2 unit vectors.It can be shown,i guess.
     
    Last edited: Dec 5, 2004
  7. Dec 5, 2004 #6
    I'm not really convinced of your proof. w is /given/. You seem to be saying that for /any/ z_1, z_2 on the unit circle, w = z_1 + z_2?
     
  8. Dec 5, 2004 #7

    dextercioby

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    When u say that "w" is given,well actually it's "given" only a constraint upon its modulus.Its phase is arbitrary,just like in the case of z_1 and z_2.Let's say we fix the phase and let its modulus take any value between zero and 2.The modulus of "w"in terms of phases of z_1 &z_2 is given by:
    [tex] |w|=\sqrt{2+2\cos(\phi_{1}-\phi_{2})} [/tex].And u see,[itex] \phi_{1} [/itex] and [itex] \phi_{2} [/itex] can run freely just as long as they obey the condition for a fixed phase of "w",call it "phi":
    [tex]\tan\phi=\frac{\sin\phi_{1}+\sin\phi_{2}}{\cos\phi_{1}+\cos\phi_{2}}
    [/tex].
    But,by fixing both the phase and the modulus of "w",then u can find explicitely the phases of z_1 and z_2 and your decomposition for fixed "w" is unique.For example,if "a"is the modulus of "w" and "phi"s its phase,one finds the phase of z_2:[itex] \phi_{2}=\arccos(\frac{a}{2})-\phi [/itex] and the phase of z_1 has a messy expression.
     
  9. Dec 5, 2004 #8

    HallsofIvy

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    dextercioby:"When u say that "w" is given,well actually it's "given" only a constraint upon its modulus.Its phase is arbitrary"

    No. From the original question, w is a specific complex number such that |w|< 2.
     
  10. Dec 5, 2004 #9

    dextercioby

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    If the number is fully specified,then the last paragraph of my prior post applies and the decomposition is unique,meaning u can find the phases of z_1 and z_2 in terms of the modulus and the phase of "w".
     
  11. Dec 5, 2004 #10

    shmoe

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    A simplification- assume w is a non-negative real number, so 0<=w<=2. It should be easier to find a complex z1, z2 that fit your conditions in this case (draw a triangle).

    The general case for w complex is just a rotation away.
     
  12. Dec 5, 2004 #11
    Well, hmm... I really don't like the geometrical "construction proofs", but I don't fully understand dextercioby's approach...

    And thank you all for your time, btw
     
  13. Dec 5, 2004 #12

    shmoe

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    The sugestion to draw a triangle was to help you work out z1 and z2. From a picture it should be clear how to pick the real components of z1 and z2 (remember the sides of the triangle reperesenting z1 and z2 will both have length 1). Then show how you can select the imaginary components to satisfy the rest of your conditions. Do this for say w=1 to try a specific example.

    There's no reason to be afraid of a picture-addition and multiplication of complex numbers is a very geometric thing. Using geometric intuition to lead to an analytic answer is nothing to shy away from. You can then hide the geometry if you like, and show off a solution that looks like divinely inspired algebra.
     
  14. Dec 5, 2004 #13
    Thank you! This is a _very_ good approach. Once again Physicsforums saved my day :)
     
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