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When is the expression real

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    I have the expression ##a_2\overline{β} + a_3β## involving complex variables, where ##|β|≤1##. I was wondering, is it possible to determine for what ##a_2## and ##a_3## the expression is always real, for every ##β## satisfying ##|β|≤1##; or is there insufficient information?

    2. Relevant equations


    3. The attempt at a solution

    I was under the impression an inquiry constituted an attempt. At any rate, I am not asking anyone to solve the problem; I am merely asking if solving such a problem is possible given the information. This doesn't come from a textbook, although I am sure it is considered a standard, textbook question.

    How is this for an attempt?

    ##a_2 \overline{\beta} + a_3 \beta = r##, where ##r## is some real number.

    ##a_2 = \frac{r - a_3 \beta}{\overline{\beta}} = e^{i \theta} - a_3 e^{2 i \theta}##
     
  2. jcsd
  3. May 8, 2015 #2

    PeroK

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    Not very impressive. Maybe the answer is pretty obvious.

    Edit: when is the sum of two complex numbers real?
     
    Last edited: May 8, 2015
  4. May 8, 2015 #3

    vela

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    Try writing ##\beta = x + iy## and then require the imaginary part of the expression you wrote to equal 0.
     
  5. May 8, 2015 #4
    Wouldn't I have to do the same for ##a_2## and ##a_3##, seeing as I am trying to solve for these?
     
  6. May 8, 2015 #5
    As PeroK noted, my situation is a particular case of a more general result--namely, that ##z_1 + z_2 = 0 \implies Im~z_2 = -Im~z_1##. But this doesn't seem particularly illuminating as ##Im~(a_2 \overline{\beta}) = - Im~(a_3 \beta)## produces a somewhat intricate equation.
     
  7. May 8, 2015 #6

    PeroK

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    I would say that ##Im(z_1) = -Im(z_2)## is very illuminating. Can you think of when that is the case?
     
  8. May 8, 2015 #7
    When what is the case? When ##Im~(z_1) = -Im~(z_2)## is true? Isn't it true when ##y_2 = -y_1##?
     
  9. May 8, 2015 #8

    PeroK

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    Complex conjugates have that property!
     
  10. May 8, 2015 #9
    I don't think we can claim that ##z_1## and ##z_2## are complex conjugates as they could have different real parts.
     
  11. May 8, 2015 #10

    PeroK

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    No, but IF they are complex conjugates, then they have that property.
     
  12. May 8, 2015 #11
    Sure, I would agree with that, but I can't conclude from that that ##a_2 \overline{\beta}## and ##a_3 \beta## are complex conjugates, if that's what you are aiming at.
     
  13. May 8, 2015 #12

    PeroK

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    Not directly, no. But if ##a_2## and ##a_3## are complex conjugates?
     
  14. May 8, 2015 #13
    Yes, I agree. But I am trying to deduce what ##a_2## and ##a_3## are, not assign a property to them. My problem is, how do I not know there are more cases?
     
  15. May 8, 2015 #14

    PeroK

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    That's the next question. You know that if ##a_2## and ##a_3## are conjugates, then ##a_2 \bar{\beta} + a_3 \beta## is real for all ##\beta##.

    Now you have to show that they must be conjugates. Hint: consider ##\beta = i##.
     
  16. May 8, 2015 #15

    jbunniii

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    ##\alpha_2## and ##\alpha_3## need not be conjugates. A complex number is real if and only if it equals its own conjugate, so the requirement that ##\alpha_2 \overline{\beta} + \alpha_3 \beta## is real is equivalent to
    $$\alpha_2 \overline{\beta} + \alpha_3 \beta = \overline{\alpha_2} \beta + \overline{\alpha_3} \overline{\beta}$$
    Rearranging, this is the same as
    $$\beta(\alpha_3 - \overline{\alpha_2}) = \overline{\beta}(\overline{\alpha_3} - \alpha_2)$$
    Writing ##\alpha_3 = a + bi## and ##\alpha_2 = c + di##, the above becomes
    $$\beta(a - c + i(b - d)) = \overline{\beta}(a - c - i(b + d))$$
    This is one linear equation with four unknowns, so there are three degrees of freedom in specifying ##a,b,c,d##: you can set any three of them freely, and solve for the fourth. If it was necessary for ##\alpha_2## and ##\alpha_3## to be conjugates, you would only have two degrees of freedom available.
     
  17. May 8, 2015 #16

    PeroK

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    Well, except this was supposed to be the case for all ##\beta##. If the expression is real for all ##\beta## then ##a_2## and ##a_3## must be conjugates.
     
  18. May 8, 2015 #17

    jbunniii

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    Oops, sorry, I missed that! In that case, I agree: consider ##\beta = i##.
     
  19. May 8, 2015 #18

    Ray Vickson

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    If we write ##v = a_3 - \bar{a_2}## your condition is that ##\beta v = \overline{\beta v}##, so ##\beta v = ## real for all ##|\beta| \leq 1##. Write ##\beta = r e^{i t}## and ##v = R e^{i w}##. If ##rR \neq 0## this implies ##e^{i(t+w)} =## real for all ##w \in [0, 2 \pi)##, so ##\sin(t+w) = 0## for all ##t##. Do you think that is possible?
     
    Last edited: May 8, 2015
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