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Homesecond form of Cauchy integral theorem

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# second form of Cauchy integral theorem

###### Theorem.

Let the complex function $f$ be analytic in a simply connected open domain $U$ of the complex plane, and let $a$ and $b$ be any two points of $U$. Then the contour integral

$\displaystyle\int_{\gamma}f(z)\,dz$ | (1) |

is independent on the path $\gamma$ which in $U$ goes from $a$ to $b$.

Example. Let’s consider the integral (1) of the real part function defined by

$f(z):=\mbox{Re}(z)$ |

with the path $\gamma$ going from the point $O=(0,\,0)$ to the point $Q=(1,\,1)$. If $\gamma$ is the line segment $OQ$, we may use the substitution

$z:=(1\!+\!i)t,\quad dz=(1\!+\!i)\,dt,\quad 0\leqq t\leqq 1,$ |

and (1) equals

$\int_{0}^{1}t\!\cdot\!(1\!+\!i)\,dt=\frac{1}{2}\!+\!\frac{1}{2}i.$ |

Secondly, we choose for $\gamma$ the broken line $OPQ$ where $P=(1,\,0)$. Now (1) is the sum

$\int_{{OP}}\mbox{Re}(z)\,dz+\int_{{PQ}}\mbox{Re}(z)\,dz=\int_{0}^{1}x\,dx+\int% _{0}^{1}i\,dy=\frac{1}{2}\!+\!i.$ |

Thus, the integral (1) of the function depends on the path between the two points. This is explained by the fact that the real part function $f$ is not analytic — its real part $x$ and imaginary part 0 do not satisfy the Cauchy-Riemann equations.

## Mathematics Subject Classification

30E20*no label found*

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