Cauchy-Goursat Theorem
This proof is about Cauchy's Theorem on the value of integrals in complex analysis. For other uses, see Cauchy's Theorem.
Contents
Theorem
Let $U$ be a simply connected open subset of the complex plane $\C$.
Let $\gamma : \left[{a \,.\,.\, b}\right] \to U$ be a closed contour in $U$.
Let $f: U \to \C$ be holomorphic in $U$.
Then:
- $\displaystyle \oint_\gamma f \left({z}\right) \rd z = 0$
Proof
Step 1
Let $C_1$ and $C_2$ be two contours such that:
- $\gamma := C_1 + \left({- C_2}\right)$
- $C_1$ has domain $\left[{a_1 \,.\,.\, b_1}\right]$
and:
- $C_2$ has domain $\left[{a_2 \,.\,.\, b_2}\right]$
Then:
- $C_1 \left({a_1}\right) = C_2 \left({a_2}\right)$
and
- $C_1 \left({b_1}\right) = C_2 \left({b_2}\right)$
Thus:
| \(\displaystyle \int_{C_1} f \left({z}\right) \rd z\) | \(=\) | \(\displaystyle \int_{a_1}^{b_1} \dfrac{\rd {C_1} } {\ \mathrm d t} f \left({C_1 \left({t}\right)}\right) \rd t\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{C_1 \left({a_1}\right)}^{C_1 \left({b_1}\right)} f \left({C_1}\right) \rd {C_1}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{C_2 \left({a_2}\right)}^{C_2 \left({b_2}\right)} f \left({C_2}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{a_2}^{b_2} \frac{\mathrm d {C_2} } {\rd t} f \left({C_2 \left({t}\right)}\right) \rd t\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{C_2} f \left({z}\right) \rd z\) | $\quad$ | $\quad$ |
The accuracy of the above is suspect -- for a start there is a missing $\rd$ in the third line -- and in any case it needs to be linked to results.
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$\Box$
Step 2
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| \(\displaystyle \oint_\gamma f \left({z}\right) \rd z\) | \(=\) | \(\displaystyle \oint_{: = C_1 + \left({- C_2}\right)} f \left({z}\right) \rd z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{C_1} f \left({z}\right) \rd z - \int_{C_2} f \left({z}\right) \rd z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{C_1} f \left({z}\right) \rd z - \int_{C_1} f \left({z}\right) \rd z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |
$\blacksquare$
Example
Let $\gamma \left({t}\right) = e^{i t}$.
Give $\gamma$ the domain $\left[{0, 2 \pi}\right)$.
Now, let $f \left({z}\right) = z^2$.
Then:
| \(\displaystyle \oint_\gamma f \left({z}\right) \, \mathrm d z\) | \(=\) | \(\displaystyle \int_0^{2 \pi} i e^{i t} e^{2 i t} \, \mathrm d t\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle i \int_0^{2 \pi} e^{3 i t} \, \mathrm d t\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac i {3 i} \left({e^{6 i \pi} - e^{i 0} }\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 3 \left({\cos 6 \pi + i \sin 6 \pi - \cos 0 - i \sin 0}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 3 \left({1 - 1}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |
$\blacksquare$
Also known as
This result is also known as Cauchy's Integral Theorem or the Cauchy Integral Theorem.
Source of Name
This entry was named for Augustin Louis Cauchy and Édouard Jean-Baptiste Goursat.
Historical Note
The Cauchy-Goursat Theorem was actually first investigated and proved by Carl Friedrich Gauss, but it was just one of the things that he failed to get round to publishing.
He stated it in a letter of $1811$ to his friend Friedrich Wilhelm Bessel, where he says:
- This is a very beautiful theorem whose fairly simple proof I will give on a suitable occasion. It is connected with other beautiful truths which are concerned with series expansions.
Augustin Louis Cauchy was, however, the first to publish a proof.
Karl Weierstrass independently discovered this theorem during his exercise to rebuild the theory of complex analysis from first principles.
