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Complex Analysis - Residues

  1. Feb 4, 2007 #1
    Hello!

    I am studing for my Complex Analysis exam and solving the exercises for Residues given by the professor.

    The problem is that for some exercises I get to a solution different from the one of the professor :bugeye:, and I am not sure that the mistake is in my calculations.

    I would greatly appreciate it, if somebody could solve it and tell me what a solution he/she came up with.

    Here is the exercise:
    Calculate the residue of the complex-valued function [tex]f(z)[/tex] at [tex]z=-\imath[/tex], as:

    [tex]f(z) = \frac{\sin(z)}{(z^2 + 1)^2}[/tex] ​

    My answer:
    [tex]Res(f(z),\imath) = -\frac{1}{4e}[/tex]​
    The professor's answer:
    [tex]Res(f(z),\imath) = \frac{\imath}{2}\cosh(1)[/tex]​

    Thanks a lot!
     
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 4, 2007 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, what did you get? Post your attempt, and we'll be able to see if you went wrong.
     
  4. Feb 4, 2007 #3
    Thank you for getting involved :smile:, here is what I did, I used the formula:

    [tex] Res(f(z), z_0) = \lim_{z \to z_0} \left( \frac{1}{(m-1)!}\, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right)
    \right) [/tex]​

    where [tex]m[/tex] is the order of the pole, and [tex]z_0[/tex] is the pole.

    In the particular case of this exercise,

    [tex]
    \begin{align*}
    m &= 2 \\
    z_0 &= -i
    \end{align*}
    [/tex]



    [tex]
    \begin{align*}
    Res(f(z), -i) &= \lim_{z \to z_0} \left( \frac{1}{(m-1)!}\, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right)
    \right) \\
    &= \lim_{z \to -i}
    \left(
    \frac{1}{1!} \,
    \dfrac{d}{dz} \left( (z+i)^2 f(z) \right)
    \right)
    \end{align*}
    [/tex]
     
    Last edited: Feb 4, 2007
  5. Feb 4, 2007 #4
    My Calculations

    Here are the calculations:


    [tex]
    \begin{align*}
    Res(f(z), -i) &= \lim_{z \to z_0}
    \left(
    \frac{1}{(m-1)!} \, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right)
    \right) \\
    &= \lim_{z \to -i}
    \left(
    \frac{1}{1!} \, \dfrac{d}{dz} \left( (z+i)^2 f(z) \right)
    \right) \\
    &= \lim_{z \to -i}
    \left(
    \dfrac{d}{dz} \left( (z+i)^2 \frac{\sin(z)}{(z+i)^2 (z-i)^2} \right)
    \right) \\
    &= \lim_{z \to -i}
    \left(
    \dfrac{d}{dz} \left(\frac{\sin(z)}{(z-i)^2} \right)
    \right) \\
    &= \lim_{z \to -i}
    \left(
    \dfrac{\cos(z) \cdot (z-i)^2 - \sin(z) \cdot 2(z-i)(z-i)'}{(z-i)^4}
    \right) \\
    &= \dfrac{\cos(-i) \cdot (-i-i)^2 - \sin(-i) \cdot 2(-i-i)}{(-i-i)^4} = \dfrac{\cos(-i) \cdot 4(-1) +4i \sin(-i)}{(-2i)^4} \\
    &= \dfrac{-4\cos(-i) +4i \sin(-i)}{16} = \dfrac{-\cos(-i) +i \sin(-i)}{4} = \dfrac{-\cos(i) -i \sin(i)}{4}\\
    &= -\dfrac{\cos(i) + i \sin(i)}{4} = -\dfrac{e^{i \cdot i}}{4}= -\dfrac{e^{-1}}{4} = -\dfrac{1}{4e}\\

    \end{align*}
    [/tex]
     
    Last edited: Feb 4, 2007
  6. Feb 4, 2007 #5
    The Professor's Solution

    Here is the solution of the professor. This is a screenshot of the page of his lecture notes on which he solves the exercise. In his calculations:
    [tex] j=\sqrt{-1}[/tex]​
     

    Attached Files:

    Last edited: Feb 4, 2007
  7. Feb 5, 2007 #6

    benorin

    User Avatar
    Homework Helper

    iflare, the fact of the matter is that your prof goofed the derivative and your work is correct.
     
  8. Feb 5, 2007 #7
    That is just great :smile:, thank you very much for the help!
     
  9. May 28, 2007 #8
    i don't think it's so great that the prof goofed here!
     
  10. May 28, 2007 #9

    mathwonk

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    Science Advisor
    Homework Helper

    i havent taught this in a long time, so this may be wrong, but i think the idea is to separate the pole from the holomorphic part at -i.

    so we have sin(z)/(z-i)^2 as the holomorphic part and 1/(z+i)^2 as the polar part.

    now we want to multiply these together and pick off the coefficient of
    1/(z+i)

    Now to get the coefficient of 1/(z+i) it seems we just need the derivative of the holo part at -i.

    by the quotient rule that should be [cos(z)(z-i) - 2sin(z)]/(z-i)^3, all evaluated at z= -i. yipes!

    i.e. -[cos(i)+i sin(i)]/4 = -1/4e,

    using the fact that cos(z) = (1/2)[e^z + e^(-z)], etc...

    so its much easier than it looks above, but still tedious for me.
     
    Last edited: May 28, 2007
  11. May 29, 2007 #10

    mathwonk

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    Science Advisor
    Homework Helper

    did you understand my post? the point was that if you multiply a holomorphic power series by 1/(z-a)^2, the residue at a will be given by the coefficient of the linear term of the powers eries. i.e.all the exponents of (z-a) go down by 2, so the linear etrm becomes the -1 order term.
     
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