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Complex Analysis: Contour Integration Question

  1. Apr 8, 2016 #1
    1. The problem statement, all variables and given/known data
    State, with justification, if the Fundamental Theorem of Contour Integration can be applied to the following integrals. Evaluate both integrals.

    \begin{eqnarray*}
    (i) \hspace{0.2cm} \int_\gamma \frac{1}{z} dz \\
    (ii) \hspace{0.2cm} \int_\gamma \overline{z} dz \\
    \end{eqnarray*}

    where

    \begin{eqnarray*}
    \gamma(t) = \exp(i2\pi t), -\frac{1}{4} \leq t \leq \frac{1}{3}
    \end{eqnarray*}

    2. Relevant equations

    Fundamental Theorem of Contour Integration:

    \begin{eqnarray*}
    \text{Suppose }f : S \rightarrow C\text{ is continuous, }\\
    \gamma : \left[a:b\right] \rightarrow S\text{ is a smooth path.}\\
    \text{Suppose }F : S \rightarrow C\text{ such that }F'(z) = f(z) \text{ }\forall z \in S.\\
    \text{Then } \int_\gamma f dz = F(\gamma(b)) - F(\gamma(a))
    \end{eqnarray*}


    Complex Line Integral Formula:

    \begin{eqnarray*}
    \text{Let }U \subset C \text{ be an open path-connected set. }\\
    \text{Let }f : U \rightarrow C \text{ be a continuous function.} \\
    \text{Let }\gamma : \left[a:b\right] \rightarrow S \text{ be a smooth path in U.}\\
    \text{Then the complex line integral of f along }\gamma \text{ is:} \\
    \int_\gamma f dz = \int_{a}^{b} f(\gamma(t))\gamma'(t)dt, \hspace{0.2cm} \text{where }\gamma'(t) = x'(t) + iy'(t)
    \end{eqnarray*}



    3. The attempt at a solution

    Hi everyone,

    Here's my attempt so far:


    \begin{eqnarray*}
    \text{(i) }\hspace{0.2cm} \int_\gamma \frac{1}{z} dz
    \end{eqnarray*}

    Can't use Fundamental Theorem as 1/z has no antiderivative. Even though we can differentiate Log(z) to get 1/z, this is only defined locally, not generally, as Log(z) is itself based on the complex exponential, which is a periodic function.

    Therefore, use the complex line integral formula:

    \begin{eqnarray*}
    \gamma(t) &=& \exp(i2\pi t) \\
    \gamma'(t) &=& i2\pi \exp(i2\pi t) \\
    f'(\gamma(t)) &=& \frac{1}{\exp(i2\pi t)} \\
    &=& \exp(-i2\pi t)\\
    \hspace{0.2cm} \int_\gamma \frac{1}{z} dz &=& \int_{-\frac{1}{4}}^{\frac{1}{3}}\exp(-i2\pi t)i2\pi\exp(i2\pi t)dt \\
    &=& \int_{-\frac{1}{4}}^{\frac{1}{3}}i2\pi dt \\
    &=&\left[i2\pi t\right]_\frac{-1}{4}^\frac{1}{3} \\
    &=& i\frac{7\pi}{6}
    \end{eqnarray*}


    \begin{eqnarray*}
    \text{(ii) }\hspace{0.2cm} \int_\gamma \overline{z} dz
    \end{eqnarray*}

    Again, can't use Fundamental Theorem as z conjugate has no antiderivative. Therefore, use the complex line integral formula:

    \begin{eqnarray*}
    \gamma(t) &=& \exp(i2\pi t) \\
    \gamma'(t) &=& i2\pi \exp(i2\pi t) \\
    f(\gamma(t)) &=& \overline{\exp(i2\pi t)} \\
    &=& \exp(-i2\pi t)\\
    \int_\gamma \overline{z} dz &=& \int_{-\frac{1}{4}}^{\frac{1}{3}}\exp(-i2\pi t)i2\pi\exp(i2\pi t)dt \\
    &=& i\frac{7\pi}{6} \\
    \end{eqnarray*}

    as before.


    It seems strange that I get the same answer for both parts. I am reasonably happy that my answer for (ii) is correct. I think I may be wrong about the first one, particularly in my reasoning that it has no antiderivative. Also, seeing as the questions are from a chapter on contour integration, it seems strange that neither one can use the fundamental theorem.

    I'd really appreciate if anyone could help point out where I may be going wrong!

    Thanks






     
  2. jcsd
  3. Apr 8, 2016 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't think that's a concern, because the contour contains only numbers with unit modulus, and for such numbers the reciprocal is the same as the conjugate, ie ##\tfrac{1}{z}=\bar z##.

    I think you can use the Fundamental Theorem at least for the ##1/z## case, because all that's necessary is for the antiderivative to exist on an open set containing the contour ##D\equiv Image\ \gamma##. It doesn't have to exist everywhere on ##\mathbb{C}##. So we can define our open set ##S## as all points no more than 0.001 away from a point in ##D## (an open 'ribbon' containing ##D##) then you can choose as antiderivative of ##1/z## a branch of the complex logarithm function that doesn't have a cut in ##S##, which is easy because ##S## doesn't entirely encircle the origin.

    I'm not sure whether the conjugate has an antiderivative on ##S##. I can't think of one. I suspect it doesn't as, given the way the question is worded, it seems likely that the Fund Theorem will be applicable to one but not the other of the two cases. Unfortunately, my Complex Analysis is too rusty to produce a proof that the conjugate has no antiderivative on ##S##.
     
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