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Residue of cot(z)

  1. Nov 28, 2013 #1
    1. The problem statement, all variables and given/known data
    So guys..the title says it! I need to find the residue of cot(z) at z=0.


    2. Relevant equations
    For this situation, since the pole order is 1

    [itex]Residue=\lim_{z \to z_{0}}(z-z_{0})f(z)[/itex]


    3. The attempt at a solution
    So here's what Im doing in steps:

    First, the singularity is at z=0. So [itex]z_{0}=0[/itex].

    Then I multiply both sides by [itex](z-z_{0})=z[/itex]...to get [itex](z-z_{0})f(z)=zcot(z)[/itex]

    Now taking the limit of this is as z = 0 is [itex]0 \times \frac{cos(0)}{sin(0)}=0[/itex]...but this is wrong, the residue is 1...

    I know its something stupid that im doing but what is it? even if i expand sin and cos I still end up with 0...
     
  2. jcsd
  3. Nov 29, 2013 #2
    Study this arguement carefully and see if that doesn't remind you of some elementary calculus:

     
  4. Nov 29, 2013 #3
    Yea I guess you're supposed to use L'Hopital's rule to find the behaviour of the function towards a limit....textbook didnt really say that
     
  5. Nov 29, 2013 #4
    For most problems you encounter, there won't be a textbook to tell you anything at all. Take a minute to understand the trick.
     
  6. Nov 29, 2013 #5
    thanks for the hint tho bro!!
     
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