Line and contour integrals.
(Lecture 13 of Mathematical Methods II.)
We have seen last lecture that the complex integral of a function $f(z)=u(x,y)+iv(x,y)$ leads to
\begin{equation} \tag{1} \oint_\mathcal{C}f(z)\,dz=\int_\mathcal{C}(u\,dx-v\,dy)+i\int_\mathcal{C} (v\,dx+u\,dy)\,, \end{equation}
and on a close contour $\mathcal{C}$, the right hand side can be evaluated through Green's theorem $\oint_\mathcal{C}P\,dx+Q\,dy=\iint_\mathcal{S}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy$ to yield a complex version of it. We rearrange Eq.~(1) as:
\begin{align} \tag{2} \oint_\mathcal{C}f(z)\,dz&=-\iint_\mathcal{S}(\partial_xv+\partial_yu)dx\,dy+i\iint_\mathcal{S}(\partial_xu-\partial_yv)dx\,dy\\ &=2i\iint_\mathcal{S}\left(\frac{1}{2}(\partial_x+i\partial_y)(u+iv)\right)dx\,dy\\ \end{align}
and express it in term of $\partial z^*$ the operator defined—along with $\partial z$—through so-called Wirtinger derivates as:
\begin{equation} \tag{3} \frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\quad\text{and}\quad\frac{\partial}{\partial z^*}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) \end{equation}
This comes from $$\partial_z=\frac{\partial x}{\partial z}\frac{\partial}{\partial x}+\frac{\partial y}{\partial z}\frac{\partial}{\partial y}$$ and computing. Coming back to Eq. (2), this yields the complex Green theorem:
\begin{equation} \oint_\mathcal{C}f(z)\,dz=2i\iint_\mathcal{S}\frac{\partial f}{\partial z^*}dx\,dy\,. \end{equation}
We need to characterize a bit more the operators $\partial_z$ and $\partial_{z^*}$ to make sense of this result. They are obtained by passing from:
\begin{equation} \tag{4} df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy \end{equation}
to
\begin{equation} \tag{5} df=\frac{\partial f}{\partial z}dz+\frac{\partial f}{\partial z^*}dz^* \end{equation}
by substituting in Eq.~(4) $dx=(dz+dz^*)/2$ and $dy=(dz-dz^*)/(2i)$ (this itself following from differentiating $z=x+iy$ and $z^*=x-iy$), rearranging and identifying. In writing Eq.~(5), we have assumed that $z$ and $z^*$ are independent variables. This results, in particular, in $\partial z/\partial z^*=0$ and $\partial z^*/\partial z=0$. From such algebraic properties, we arrive at the following elegant expressions of holomorphic functions:
\begin{equation} \tag{6} \partial_zf(z_0)=f'(z_0)\quad\text{and}\quad\partial_{z^*}f(z_0)=0\,. \end{equation}
The latter equation is the Cauchy-Riemann condition (as is checked straightforwardly). The former comes from $\partial_x f=f'$ from the independence of the derivative from the path taken for differentiation and (with slightly more details given) $$(\partial_y f)(z_0)=i\lim_{\Delta y\rightarrow0}[f(x+iy+i\Delta y)-f(x+iy)]/(i\Delta y)=if'(z_0)$$ so that $f'(z_0)=-i\partial_yf$ which, summing with $\partial_xf$ provides $2f'$.
Applying such rules to the complex form of the Green theorem, we find as a particular case, calling $A$ the area of the domain enclosed by a contour $\mathcal{C}$:
\begin{equation} \tag{7} A=\frac{1}{2i}\oint_\mathcal{C}z^*dz\,. \end{equation}
This illustrates vividly how a complex integral in the plane depends, in general, on the contour of integration.
The complex Green theorem also brings us to a more important result, that relates the case where $f(z,z^*)$ is holomorphic, that is, $z^*$ independent, in which case:
Cauchy theorem—If $f$ is holomorphic (or equivalently, analytic) in a simply connected region $\mathcal{R}$ and also on the contour $\mathcal{C}$ enclosing it, then:
\begin{equation} \tag{8} \oint_\mathcal{C}f(z)\,dz=0\,. \end{equation}
As a consequence of the Cauchy theorem, if $z_0$ and~$z_1$ are two complex points in a simply connected region where $f$ is holomorphic, then $\int_{z_0}^{z_1}f(z)\,dz$ is independent of the path (that remains within the said region) of integration, which allows such a notation.
The theory of integration and differentiation are linked through the following identity:
\begin{equation} \tag{9} \int_{z_0}^{z_1}f(z)\,dz=F(z_1)-F(z_0) \end{equation}
with $F'(z)=f(z)$. As an example, $\int_0^i z\,dz=(1/2)z^2\Big|_0^i=-1/2$, as we have seen in the previous lecture on the particular case of integrating along the $y$ axis. The path is in fact completely arbitrary.
It is essential that the region be simply connected, that is, that there are no holes, for the theorem to hold. In fact, let us consider the case of the contour integral circling the divergence of the inverse function, i.e., $\oint_\mathcal{C}\frac{dz}{z-a}$ over a contour~$\mathcal{C}$ that surrounds~$a$. We can deform the contour into a circle~$C$ centered on~$a$, i.e., such that $|z-a|=r$, and the integral reads:
\begin{equation} \tag{10} \oint_C\frac{dz}{z-a}=\int_{0}^{2\pi}\frac{ire^{i\theta}d\theta}{re^{i\theta}} =i\int_{0}^{2\pi}d\theta=2i\pi \end{equation}
since $z=a+re^{i\theta}$ implying $dz=ire^{i\theta}d\theta$. If $\mathcal{C}$ does not surround~$a$, the integral is zero.
In the same way, we can also compute for any positive integer~$n\neq1$:
\begin{equation} \oint_C\frac{dz}{(z-a)^n}=\int_{0}^{2\pi}\frac{ire^{i\theta}d\theta}{r^ne^{in\theta}} =\frac{i}{r^{n-1}}\int_0^{2\pi}e^{i(1-n)\theta}\,d\theta =\frac{i}{r^{n-1}}\frac{1}{i(1-n)}e^{i(1-n)\theta}\Bigg|_0^{2\pi}=0 \end{equation}
so although there is a pole in $a$, the contour integral still gives zero. The important result to keep constantly in mind is, for $\mathcal{C}$ surrounding $a$ (always zero otherwise):
\begin{equation} \oint_\mathcal{C}\frac{dz}{(z-a)^n}= \begin{cases} 2i\pi & \text{if }n=1\,, \\ 0 & \text{otherwise}\,. \end{cases} \end{equation}
