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Complex Conjugate: just replace i by -i even in denominator or inside argument?

  1. Jan 20, 2012 #1
    to get a Complex Conjugate of a #, is it ok to just replace i by -i even in denominator or inside argument?
     
  2. jcsd
  3. Jan 21, 2012 #2

    Mark44

    Staff: Mentor

    If your complex number is a + ib, then its complex conjugate is a - ib.

    However, you can't just go in an replace the coefficient of i willy-nilly. So for example, if you have 2/(1 + 3i), you can't just rewrite this as 2/(1 - 3i), as the two expressions aren't equal.

    What you can do, though, and you can always do this, is multiply by 1, in whatever form is convenient.

    So 2/(1 + 3i) = 2/(1 + 3i) * (1 - 3i)/(1 - 3i) = (2 - 6i)/(1 + 9) = (2 - 6i)/10 = (1 - 3i)/5 or 1/5 - (3/5)i.
     
  4. Jan 21, 2012 #3
    I always thought you could just replace all i with -i in any case. Can you give me an example where that is not true?
     
  5. Jan 21, 2012 #4

    Mark44

    Staff: Mentor

    1 + 3i [itex]\neq[/itex] 1 - 3i

    In general, a complex number is not equal to its conjugate.

    If you're asking how to get the complex conjugate, then yes, all you need to do is to replace the coefficient of i by its opposite. My point was that you get a different number.
     
  6. Jan 21, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why was that necessary? The original post just asked about getting the complex conjugate. I presume that the OP knows that only for real numbers is the complex conjugate the same as the number itself. That was never in question.
     
  7. Jan 21, 2012 #6

    Mute

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    Homework Helper

    If [itex]\phi(z)[/itex], for z complex, is holomorphic and is real for [itex]z = x \in \mathbb{R}[/itex], then [itex](\phi(z))^\ast = \phi(z^\ast)[/itex]. (Wikipedia).
     
  8. Jan 21, 2012 #7

    Mark44

    Staff: Mentor

    It probably wasn't. I misunderstood what the OP was asking.
     
  9. Jan 21, 2012 #8
    this is basically the answer to my question, but i do need to unravel its meaning:

    1. holomorphic - basically well behaved, has derivatives.

    2. phi(z) is real for z in reals, this is a hard condition, because often the form of the expression is not such, but i have found that the trick with i -> -i still works, as in the lorentzian:

    [itex]\frac{i \tau }{\sqrt{2 \pi } (i+\tau \omega )}[/itex]

    when i replace i with -i and multiply the two together i think i get the right magnitude. tell me if i am wrong, or when this would not work

    [itex]\frac{\tau ^2}{2 \pi +2 \pi \tau ^2 \omega ^2} [/itex]


    3. is the dirac function holomorhpic? eg. how would i find the abs value of:
    [itex]\sqrt{2 \pi } \text{DiracDelta}\left[\frac{i}{\tau }+\omega \right][/itex]

    sorry for all the questions, thanks :)
     
  10. Jan 21, 2012 #9
    Are you looking to see when the conjugate is equal to the original expression? Or just that you can replace i with -i to get the conjugate?

    If the latter, that works in any expression whatsoever, as long as only addition and multiplication are involved (and subtraction and division of course). That's because conjugation is an automorphism of the complex numbers: it preserves addition and multiplication. So if you have a/b for complex a and b, then (a/b)* = a*/b* where '*' is the complex conjugate. And since complex conjugation fixes the reals (that is, a = a* for all real a) you can just replace i with -i and vice versa to compute the conjugate of any expression involving only addition and multiplication.

    Asking when the conjugate of an expression is equal to the original expression is a separate question, which seems to be confusing this thread. Which question are you asking?
     
  11. Jan 21, 2012 #10
    see my post right above yours, i know that the complex conjugate is only equal to the complex # itself when the number is real,

    i am asking how to get the complex conjugate
     
  12. Jan 22, 2012 #11

    Char. Limit

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    Gold Member

    Well, if you have something simple, like

    [tex]\frac{2 + 3 i}{3 - 4 i}[/tex]

    Then the complex conjugate will simply replace every instance of i in the equation with -i. This property also holds true when you add many other functions to the mix, like exponentiation to a real number, and I believe the sine, cosine, and exponentiation function as well. It won't always work, though. For example, I believe it fails in the case of the log function, that is, the complex conjugate of log(3+4i) isn't necessarily log(3-4i).

    EDIT: I stand corrected, it works for the log function as well, at least up to branch cuts!
     
  13. Jan 22, 2012 #12
    this is very good, that is what i was asking! Thank you,

    so it works almost all the time. do you know when, in what cases i should be careful when using it?
    ie anytime i dont use fractions and logs, and exponetnials and trig funcs?
     
  14. Jan 23, 2012 #13
    so i think it works for any function with a taylor expansion (~analytic), because conjugation carries through addition and multiplication and so through powers and thus exponents

    1. (z1 +z2)* = z1*+z2*
    2. (z1 z2)* = z1* z2*
    3. (z1^n)* = z1*^n
    4. [itex]\left(\frac{1}{\text{z1}}\right)^* = \left(\frac{1}{a+i b}\frac{a-i b}{a-i b}\right)*
    \frac{a+i b}{a^2+ b^2}=\frac{a+i b}{a^2+ b^2}\frac{a-i b}{a-i b}
    =\frac{a^2+b^2}{\left(a^2+ b^2\right)(a-i b)}=\frac{1}{a-i b}=\frac{1}{\text{z1}^*}
    [/itex]
    5. any negative integer power we can derive from 3 & 4.

    Conclusion:
    the complex conjugate of any product or sum of polynomials of complex numbers z1..zn is the same polynomials but of z1*...zn*
     
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