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Derivative of sinc(z) in the complex plane

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]z=x+iy;

    f(z)=sin(z)/z[/itex]
    find f'(z) and the maximal region in which f(z) is analytic.

    2. Relevant equations

    The sinc function is analytic everywhere.

    3. The attempt at a solution
    Writing f(z) as [itex](z^{-1})sin(z)[/itex] and differentiating with respect to z using the chain rule I get...
    [itex](-z^{-2})sin(z)+cos(z)/z[/itex]

    However, this seems to simple since the context of the chapter of the book this problem comes from is cauchy-riemann. I would suspect I need to put f(z) in the form Re{f(z)}=u(x,y) and Im{f(z)}=v(x,y). Then [itex]df/dz[/itex] would be [itex]du/dx+idv/dx[/itex]. If that is the case then I'm in trouble because the I can't separate the imaginary part from the real part of [itex]sinc(z)[/itex].
     
  2. jcsd
  3. Sep 10, 2013 #2

    UltrafastPED

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    You can always remove the complex value from the denominator:

    1. Assume an expression like F(z)/G(z)

    2. Multiply it by unity of the form G*(z)/G*(x) where * is the complex conjugation operator.

    3. Your expression now looks like F(z)G*(z)/[G(z)G*(z)] = F(z)G*(z)/|G(z)|^2.
     
  4. Sep 10, 2013 #3
    OK thanks. Now I'm getting somewhere.
    So now I have the following:

    [itex]sin(z)/z=\frac {1}{|z|^2}[z^*sin(z)] [/itex]
    [itex]=\frac{1}{(|z|^2)}[(x-iy)\frac{(e^z-e^{-z})}{2i}][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[(x-iy)(e^z-e^{-z})][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{(x+iy)}-xe^{-(x+iy)}-iye^{(x+iy)}+iye^{-(x+iy)}][/itex]

    I'm not really sure what to do now. I can't group the e^(iy)'s. I'm guessing I'm missing some sort of trick or something.

    Thanks!
     
  5. Sep 10, 2013 #4

    UltrafastPED

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    Now go back to the definitions of sin(z); definitions are your friend. :-)
     
  6. Sep 10, 2013 #5
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{(x+iy)}-xe^{-(x+iy)}-iye^{(x+iy)}+iye^{-(x+iy)}][/itex]
    [itex]=\frac{1}{|z|^2}[x*sin(z)-y*cos(z)][/itex]

    I'm not sure how this helps me though because I still have imaginary parts in both terms because of the z. Or, is this when I use cos(z)=Re{e^{iz}} ? That seems like a stretch but it's the only connection I'm making atm.
     
  7. Sep 10, 2013 #6

    UltrafastPED

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    In your reconstruction the final line is incorrect. That cos(z) is wrong.

    In any case you want to find the derivative ... if your goal is to find the real and imaginary parts you can find them either before or after the derivative. But you must carry out the Cauchy-Riemann tests as well ...

    We've been on a side issue: how to remove the complex values from the denominator. You probably don't want to do that at the start -
     
  8. Sep 10, 2013 #7
    OK, if we ignore my attempt at a solution, how would you suggest I attack the original question?
     
  9. Sep 10, 2013 #8

    vanhees71

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    Just take the Laurent series expansion of [itex]\sin z/z[/itex] and show that its convergence radius is infinity and that it has no singularities anywhere.
     
  10. Sep 10, 2013 #9
    Since you are saying I only need to find the maximal region of convergence and show that there are no singularities in order to solve; can I assume you think my original solution to finding the derivative was correct? Specifically
    (−z^-2)sin(z)+cos(z)/z
     
  11. Sep 10, 2013 #10
    If you muscle through it you can. You mean you can't separate the real and imaginary part of the expression:

    [tex]\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i(x+iy)}[/tex]

    Bet you can. You know, convert the exponents to sines and consines, multiply top and bottom by conjugate of the denominator, bingo bango.
     
  12. Sep 10, 2013 #11

    vanhees71

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    I don't know, what you are allowed to assume. For me the most simple solution is to look for the Laurent expansion around 0 and showing that the series is convergent on the entire complex plane, showing that the continuation of the function to the entire complex plane
    [tex]f(z)=\begin{cases}
    \frac{\sin z}{z} & \text{for} \quad z \neq 0,\\
    1 & \text{for} \quad z=0
    \end{cases}
    [/tex]
    in fact is an entire function, i.e., holomorphic on [itex]\mathbb{C}[/itex].

    Your derivative was correct for [itex]z \neq 0[/itex].
     
  13. Sep 10, 2013 #12
    Since my derivative was correct, I'm wondering if I can always simply pretend my complex function is a real function and differentiate as usual then go back and consider any singularities, branch cuts, etc....
     
  14. Sep 11, 2013 #13
    [itex]sin(z)/z=\frac {1}{|z|^2}[z^*sin(z)] [/itex]
    [itex]=\frac{1}{(|z|^2)}[(x-iy)\frac{(e^{iz}-e^{-iz})}{2i}][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[(x-iy)(e^{iz}-e^{-iz})][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{i(x+iy)}-xe^{-i(x+iy)}-iye^{i(x+iy)}+iye^{-i(x+iy)}][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{ix-y}-xe^{-ix+y}-iye^{ix-y}+iye^{-ix+y}][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{-y}e^{ix}-xe^ye^{-ix}-iye^{-y}e^{ix}+iye^ye^{-ix}][/itex]
    [itex]=\frac{-i}{(2*|z|^2)}[xe^{-y}(cosx+isinx)-xe^y(cosx-isinx)-iye^{-y}(cosx+isinx)+iye^y(cosx-isinx)][/itex]

    [itex]=\frac{1}{(2*|z|^2)}[-ixe^{-y}(cosx+isinx)+ixe^y(cosx-isinx)-ye^{-y}(cosx+isinx)+ye^y(cosx-isinx)][/itex]

    [itex]=\frac{1}{(2*|z|^2)}[
    -ixe^{-y}cosx
    +xe^{-y}sinx
    +ixe^ycosx
    +xe^{y}sinx
    +ye^{-y}cosx
    -iye^{-y}sinx
    +ye^ycosx
    -iye^ysinx][/itex]

    [itex]=\frac{1}{(2*|z|^2)}[
    +xe^{-y}sinx
    +ye^{-y}cosx
    +xe^{y}sinx
    +ye^{y}cosx

    -ixe^{-y}cosx
    -iye^{-y}sinx
    +ixe^ycosx
    -iye^ysinx

    ][/itex]

    [itex]u(x,y)=e^{-y}(xsinx+ycosx)+e^y(xsinx+ycosx)[/itex]
    [itex]v(x,y)=-e^{-y}(xcosx+ysinx)+e^y(xcosx-ysinx)[/itex]

    I was missing an 'i' in the exponent at the very beggining. I'm almost there I think.
     
    Last edited: Sep 11, 2013
  15. Sep 11, 2013 #14

    vanhees71

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    What is this good for?:confused:
     
  16. Sep 11, 2013 #15
    In theory, putting f(z) in the form u(x,y)+iv(x,y) allows you to use cauchy-riemann to determine if f(z) is analytic. However, in practice, attempting this has driven me to drinking...so it is good for something!
     
  17. Sep 11, 2013 #16
    That's the best response possible.

    while this type of question is rather pedantic it is important for the student to become familiar with the Cauchy-Riemann equations in order to appreciate that they are a very strong condition for analycity (which suggests that only a small class of functions are analytic) and these topics are often taught prior to being introduced to Laurent series or power series representations of complex functions.

    Furthermore, by asking students to solve problems in this manner before they learn the Laurent series the student develops an appreciation for the power of the Laurent series and the fact that they are infinitely differentiable.


    EDIT: Furthermore, the Cauchy-Riemann equations prove useful in proving several important theorems and corollaries that will follow.
     
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