# Cauchy Integral Theorem

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:

$\displaystyle \gamma : = C_1 + \left({- C_2}\right)$

, $C_1$ has domain $\left[{a_1 , b_1}\right]$, and $C_2$ has domain $\left[{a_2 , b_2}\right]$.

Then:

$\displaystyle C_1 \left({a_1}\right) = C_2 \left({a_2}\right)$

and

$\displaystyle C_1 \left({b_1}\right) = C_2 \left({b_2}\right)$

Thus:

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

$\Box$

### Step 2

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

$\blacksquare$

## 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,

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

## Source of Name

This entry was named for Augustin Louis Cauchy.