### Author Topic: What is wrong with using the FTOC II in complex variable ?  (Read 591 times)

#### Ende Jin

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##### What is wrong with using the FTOC II in complex variable ?
« on: November 09, 2018, 06:48:49 PM »
In a tutorial, the TA showed us why FTOC II can be used in the complex context, the proof is simply just making the derivative of F into the imaginary part $v'$ and real part $u'$. Because that is a line integral, thus ultimately,  replace the $\Gamma$ with the parameterization of $\gamma : [a, b] \rightarrow \mathbb{C}$, we are integrating the derivative of $u \circ \gamma: [a,b] \rightarrow \mathbb{R}$ which is just a integral in the real line. Now after using the real-function version of FTOC II on the derivative of a real-valued function, we get $u,v$ back and end up with $F$ again.

This is the outline of the proof. However, I don't see that F needs to be analytic on a simply-connected domain in this proof. That F can be analytic on only an annulus (i.e. differentiable on the range of parameterization), even F is undefined outside the annulus, we can still get that the integration is zero when integrating on a closed curve. But it is absolutely wrong because it contradicts the chapter of singularities.

Which part is wrong?
« Last Edit: November 09, 2018, 06:51:07 PM by Ende Jin »

#### Victor Ivrii

Assume that you need to calculate $I=\int_\gamma (P\,dx +Q\,dy)$ and $P=U_x$, $Q=U_y$. Then $P\,dx +Q\,dy= dU$ and $I= U(x_1,y_1)-U(x_0,y_0)$ where $\gamma$ goes from $(x_0,y_0)$ to $(x_1,y_1)$. So far there is no analytic functions or simple connectivity of the domain.
However $P=U_x$, $Q=U_y$ implies $P_y=Q_x$. If we integrate $f\,dz$ then $P=f$, $Q=if$ and this is equivalent to $f_y=if_x$ which is a Cauchy-Riemann condition (plug $f=u+iv$).
Further, if domain is not simply connected, then even "$f$ is analytic" does not imply that there exists an analytic single-valued function $F$ such that $F'=f$ (for non-analytic functions the "derivative": is not defined). F.e. in $\mathbb{C}\setminus 0$ let $f=z^{-1}$; then $F=\log (z)$ is a multivalued function and $I$ is an increment of $F$ along $\gamma$, depending on $\gamma$ (more precisely, it depends on the equivalency class $[\gamma]$ of $\gamma$: $\gamma_1 \sim \gamma_2$ if one could  be continuously morphed into another without leaving domain. Only in simple connected domains all curves with the same start and end points are equivalent.