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1) Let $f$ have an antiderivative $F$ in a region $D$ containing a $\underline{closed \,\,\,contour}$ $C$. Then \begin{equation} \int_Cf(z)dz = 0 \end{equation}

2) Also, Cauchy's theorem says: If f is $\underline{holomorphic}$ inside and on a $\underline{closed \,\,\,contour}$ $C$ then \begin{equation} \int_Cf(z)dz = 0 \end{equation}

Then why is Cauchy's theorem so important and what is the difference between this one and the previous one? It looks to me that the previous one required less conditions than this.

3) Finally, another theorem is: If f is $\underline{holomorphic}$ on a region $D$ and $C_1$, $C_2$ are $\underline{closed \,\,\,contours}$ in $D$, which can be continuously deformed into each other within $D$, then: \begin{equation} \int_{C_1}f(z)dz = \int_{C_2}f(z)dz \end{equation} Is it now obvious? Of course they are, they are zero!

4) Deformation Theorem: Let $f$ be $\underline{holomorphic}$ in a simply connected region $D$. Then: \begin{equation} \int_Cf(z)dz = 0 \end{equation} for every closed contour $C$ in $D$.

Where simply connected means: that any closed contour in $D$ can be continously deformed within $D$ to a point.

Again, is this not saying the same thing as above?? I really don't get the difference between all these theorems.

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(2) Holomorphic means simply that $\exists f'$. Why $\exists F$ s.t. $F' = f$ is "less"? Even more, what condition is easier to check?

(3) In this case, interesting things can happen out of the region. Check what happens to $f(z) = 1/z$ in $\Bbb C\setminus\{0\}$. And the integrals can be $\ne 0$!

(4) is a consequence of (3).

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  • $\begingroup$ So between (1) and (2) the difference is that for (2) we can differentiate the function at every point, while for (1) we can integrate it at every point? And I thought it was less because you don't have to know anything about the derivative. To be honest differentiation is easier, so it makes sense $\endgroup$ Nov 24, 2016 at 22:23
  • $\begingroup$ For (3) there is the same example in my book but I don't understand. We say that it works IF it is holomorphic on D. Which means differentiable at all points in D. Clearly $f(z) = 1/z$ is not differentiable at $z=0$, so we shouldn't consider it. Or should we? And also, you say outside of the region. Why should we care? The contours we are integrating on, they are inside the region! $\endgroup$ Nov 24, 2016 at 22:26
  • $\begingroup$ @Euler_Salter, take a path around 0. $\endgroup$ Nov 24, 2016 at 23:05
  • $\begingroup$ oh okay! So basically (3) is saying that now we don't consider the function holomorphic inside and on the contour. We only consider it holomorphic in the region, however we substitute that condition with the ability of continuously deforming any two contours inside the region? Does it mean that being able to continuously deforming two contours in a region, means the region itself has no gaps/holes or points in which is not defined? $\endgroup$ Nov 24, 2016 at 23:15
  • $\begingroup$ @Euler_Salter, exactly. Simply connected. $\endgroup$ Nov 24, 2016 at 23:18

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