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Limit problem!

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Using [itex] C_R [/itex] as,

    P1eVX.png

    ([itex] C_R [/itex] doesn't include the bottom line between R and -R)

    Prove that,

    [itex] \lim_{R \to \infty} \int_{C_R} \frac{1}{(z+i)^2} dz = 0 [/itex]



    2. The attempt at a solution


    Using the estimation lemma and trying to find a bound on [itex] \frac{1}{(z+i)^2} [/itex]

    So if [itex] |z| = R [/itex] then,

    [itex] |(z + i)^2| = (|z+i|)^2 \geq (|z| - |i| )^2 = (|z| - 1)^2 = (R - 1)^2 [/itex]

    >>>(Not sure of the integrity of this step)

    Thus,

    [itex] \frac{1}{(z+i)^2} \leq \frac{1}{(R - 1)^2} [/itex]

    So using estimation lemma and the fact that arclength of [itex] C_R [/itex] is [itex] \pi R [/itex] then,


    [itex] \left| \int_{C_R} \frac{1}{(z+i)^2} dz \right| \leq \int_{C_R} \left|\frac{1}{(z+i)^2} \right| |dz| \leq \frac{\pi R}{(R-1)^2} \to 0 [/itex] as [itex] R \to \infty [/itex]
     
  2. jcsd
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