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In the text "Function Theory of One Complex Variable" by Robert E.Greene and Steven G.Krantz I'm having trouble obtaining a proof of $\text{Proposition (1)}$ via Cauchy-Goursat may I have hints on finishing the proof including any alternative approaches to the problem ? The attempted proof can be followed from $\text{Lemma (1)}$

$\text{Proposition (1)}$

Let $f$ be holomorphic on $\mathbb{C} \setminus \{ 0 \}.$ Let $s_1, s_2 > 0$. Prove that if $\Psi_{1}$ and $\Psi_{2}$ are counterclockwise oriented squares of center $0$ and side length $s_1$ and $s_2$, respectively, then in $(1.1)$

$$ \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta = \frac{1}{2 \pi i} \oint_{\Psi_2} f(\zeta) \, d\zeta \tag{1.1} $$

$\text{Remark}$

We are not assuming that the sides of these squares are parallel to the coordinate axis, An important hint is to note that the function $f$ has an holomorphic antiderative on $\, \{ z = x+iy : y > 0 \}.$ and $\, \{ z = x+iy : y > 0 \}.$ Use these to evaluate the integrals over the top and bottom halves of the given curves.

$\text{Lemma (1)}$

Recall from $\text{Proposition (1)}$ that are integrals are defined over a counterclockwise oriented square for the sake of our proof one should consider this contour with it's vertices being defined as $\phi$, $\omega$, $\Gamma$, $\gamma$ and finally $0$.To fully realize the ovulation of our integral we will need the Cauchy-Goursat Theorem which is fully developed in $(1.2)$

$\text{Theorem 1.2 (Cauchy Goursat)}$

Let $U$ be an open subset of C which is simply connected, let $f : U → C$ be a holomorphic function, and let ${\displaystyle \!\,\Gamma } \!\,$ be a rectifiable path in $U$ whose start point is equal to its end point. Then in $(1.2)$

$$\oint_\Gamma f(z)\,dz = 0. \tag{1.2}$$

In view of $(1.2)$ one trivially achieves the following developments for the LHS side of $(1.1)$ in $(1.3)$

\begin{align} & \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta \\[10pt] = {} & \frac{1}{2 \pi i} \bigg( \oint_\omega^\phi f(z)i \, dy + \oint_\omega^\Gamma f(z) \, dx + \oint_\Gamma^\gamma f(z) \, dx + \oint_\phi^\gamma f(x)i \, dy \bigg ) = 0 \tag{1.3} \end{align}

In view of the approach in $(1.3)$ we obtain similar conclusions in $(1.4)$

\begin{align} & \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta) \, d\zeta \\[10pt] = {} & \frac{1}{2 \pi i} \bigg( \oint_\omega^\phi f(z)i\,dy + \oint_\omega^\Gamma f(z) \, dx + \oint_\Gamma^\gamma f(z) \, dx + \oint_\phi^\gamma f(x)i\,dy \bigg ) = 0 \tag{1.4} \end{align}

Hence, in conclusion, from $(1.4)$ and $(1.3)$ one has

$$ \bigg( \frac{1}{2 \pi i} \oint_{\Psi_1} f(\zeta)\,d\zeta = 0 \bigg) = \bigg( \frac{1}{2 \pi i} \oint_{\Psi_2} f(\zeta) \, d\zeta = 0 \bigg) $$

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  • $\begingroup$ It seems like the Residue Theorem can be applied to $(1.1)$ $\endgroup$
    – Zophikel
    Commented Jul 7, 2018 at 17:24

1 Answer 1

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I don't understand your notation

$$\int_{\text{path one}}^{\text{path two}}$$

You meant to split the square path into its edges, right? In this case you cannot apply Goursat, the region over which the function $f$ is holomorphic is not simply connected (it has one hole at $0$), hence the two integrals can be non-zero.

I don't have the book at hand, but I think it is saying the following.

Call $R_1, R_2$ the two squares (paths). Assume that they their corresponding edges parallel, and assume that $R_1$ is inside $R_2$. Split the in-between region into four trapezoids (or eight triangles if you wish), and call their contours $T_1, \ldots, T_4$. Note that $$\int_{R_2} = \int_{R_1} + \int_{T_1} + \int_{T_2} + \int_{T_3} + \int_{T_4}.$$ You can check that this is true since some paths on the right hand side cancel out (one time you integrate one direction, one time the opposite). Probably a drawing would explain better. Then by Goursat the integrals along the paths $T_i$ are zero, since they are contained in a region over which $f$ is holomorphic.

From this two facts you get proposition (1)

$$\int_{R_2} f \, dz= \int_{R_1} f \, dz.$$

For an example, try to compute the integral of $f(z) = 1/z$ along a square path of arbitrary edge length. You should get $2 \pi i$, which is the same result you get if you integrate it along a circle or arbitrary radious with the same orientation (in fact, if two paths are homotopic, or two closed paths are free homotopic, then the integral along a path is equal to the integral along the other).

Edit. I just saw your comment to the OP. Yes, you could apply the residue theorem, in which case the two integrals are just the residue at the origin, but I belive your book prove the residue theorem after this proposition.

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  • $\begingroup$ The notation $ϕ, ω, Γ, γ$ denotes the vertices of the square, since the $f$ is holomorphic one would evaluate $f$ over the top and bottom halfs of our given Contour. Hence why I used Cauchy-Goursat for this particular problem. Also since one can split the integrals and get some arbitrary constant couldn't one use the notation that the two integrals are equivalent to the same constant to achieve a proof ? $\endgroup$
    – Zophikel
    Commented Jul 7, 2018 at 18:52
  • $\begingroup$ To elaborate more and to clarify since $(1.2)$ does not hold it seems one would have to parametrize the integral along the given segments and evaluate each of the integrals individually. $\endgroup$
    – Zophikel
    Commented Jul 7, 2018 at 19:07

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