Cauchy-Goursat Theorem

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This proof is about Cauchy's Theorem on the value of integrals in complex analysis. For other uses, see Cauchy's Theorem.


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:

$\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = 0$


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:

\(\displaystyle \int_{C_1} f \left({z}\right) \ \mathrm d z\) \(=\) \(\displaystyle \int_{a_1}^{b_1} \dfrac{\mathrm d {C_1} } {\ \mathrm d t} f \left({C_1 \left({t}\right)}\right) \ \mathrm d t\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_1 \left({a_1}\right)}^{C_1 \left({b_1}\right)} f \left({C_1}\right) \ \mathrm d {C_1}\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_2 \left({a_2}\right)}^{C_2 \left({b_2}\right)} f \left({C_2}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{a_2}^{b_2} \frac{\mathrm d {C_2} } {\mathrm d t} f \left({C_2 \left({t}\right)}\right) \ \mathrm d t\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_2} f \left({z}\right)\ \mathrm d z\)                    



$\Box$


Step 2

\(\displaystyle \oint_\gamma f \left({z}\right) \, \mathrm d z\) \(=\) \(\displaystyle \oint_{: = C_1 + \left({- C_2}\right)} f \left({z}\right) \, \mathrm d z\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_1} f \left({z}\right) \, \mathrm d z - \int_{C_2} f \left({z}\right) \, \mathrm d z\)                    
\(\displaystyle \) \(=\) \(\displaystyle \int_{C_1} f \left({z}\right) \, \mathrm d z - \int_{C_1} f \left({z}\right) \, \mathrm d z\)                    
\(\displaystyle \) \(=\) \(\displaystyle 0\)                    

$\blacksquare$


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,

\(\displaystyle \oint_\gamma f \left({z}\right) \, \mathrm d z\) \(=\) \(\displaystyle \int_0^{2 \pi} i e^{i t} e^{2 i t} \, \mathrm d t\)                    
\(\displaystyle \) \(=\) \(\displaystyle i \int_0^{2 \pi} e^{3it} \, \mathrm d t\)                    
\(\displaystyle \) \(=\) \(\displaystyle \frac{i}{3i} \left({ e^{6 i \pi} - e^{i 0} }\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 3 \left({\cos 6 \pi + i \sin 6 \pi - \cos 0 - i \sin 0}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 3 \left({1 - 1}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle 0\)                    

$\blacksquare$


Source of Name

This entry was named for Augustin Louis Cauchy and Édouard Jean-Baptiste Goursat.


Also known as

This result is also known as Cauchy's Integral Theorem or the Cauchy Integral Theorem.