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Complex analysis: residue integration question

  1. Nov 9, 2014 #1
    I'm asked to evaluate the following integral: [itex]\int_{c} \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz[/itex] where c is the unit circle. This function has a simple pole at [itex]z=\frac{1}{3}[/itex] and a second order pole at [itex]z=\frac{1}{2}[/itex], both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): [itex]Res f(z) = \frac{1}{(1-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[2z-1]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}[/itex] for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.
     
    Last edited: Nov 9, 2014
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  3. Nov 9, 2014 #2

    lurflurf

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    The formula you should have used is $$
    Res f(z) = \frac{1}{(2-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[z-\dfrac{1}{2}]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}
    $$
    That is why you got 2=4(1/2) instead of 1/2 and 6=3*2 instead of 2
     
  4. Nov 9, 2014 #3
    Oh, wow that makes a lot of sense. I got in the habit of multiplying [itex] f(z_0)[/itex] by whichever term was creating the singularity in the denominator because a lot of the first problems we did would have something simple like [itex](z-1)[/itex] there. So then, [itex] (z-z_0) [/itex] would equal exactly what was there. Anyway, thank you so much!
     
  5. Nov 9, 2014 #4
    Ah, and yes that factorial term definitely should have been [itex] (2-1)![/itex], although of course it doesn't matter in this case.
     
  6. Nov 9, 2014 #5
    One more question for you actually. So, is this the most general residue formula? This should apply to nearly any case?
     
  7. Nov 10, 2014 #6

    mathwonk

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    I am not very good at computations, but I am getting residues opposite to what you seem to say, i.e. res = 1/2 at z=1/2, and res = 2 at z= 1/3.

    My reasoning is as follows: the residue at a, is the coefficient of 1/(z-a) in the Laurent expansion in powers of (z-a). So at z= 1/3 we should have 1/(z-1/3) . g(z), where g(z) = (30z^2 -23z + 5)/3(2z-1)^2. Then when we expand g in powers of (z-1/3), the constant term will be the residue, i.e. the coefficient of 1/(z-1/3) after we multiply by the 1/(z-1/3) out front. so this is just g(1/3) = (30/9 - 23/3 + 5)/3(1/3)^2 = 2.

    Then for the other one, at z= 1/2, the function can be written as 1/(z-1/2)^2 h(z), where h(z) = (30z^2-23z+5)/4(3z-1). then the residue after multiplying by the factor 1/(z-1/2)^2, will be the linear coefficient of the expansion of h(z), i.e. should be the derivative of h(z). By the quotient rule, this is (1/4)(7/2 - 3)/(1/4) = (1/8)/(1/4) = 1/2.


    Am I missing something? My point in doing it this way is to emphasize not memorizing formulas. but going back to the meaning of the concept. But of course one wants to get it right, and I am not super confident after apparently getting a different answer. Or maybe I just misunderstood what was said.
     
  8. Nov 10, 2014 #7
    Those are the correct residues. I made a mistake in finding them the first time.
     
  9. Nov 10, 2014 #8

    mathwonk

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    Thank you. My recommended point of view, as I said, is to write the function as a product of form 1/(z-a)^m . g(z), where g is holomorphic at a.

    Then we want the coefficient of (z-a)^(m-1) in the Taylor expansion of g. this is of course just g^(m-1)(a) /(m-1)!, i.e. the (m-1) st derivative of g at a, divided by (m-1)!


    I suggest this because your error was apparently in recalling the residue formula incorrectly. I hope this method makes that less likely, since all you have to recall this way is that you want the coefficient of 1/(z-a). In the product 1/(z-a)^m . g(z), that will be the coefficient of (z-a)^(m-1) in the expansion of g.
     
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