First post, so I'll start on a neat topic. There is a distinct difference between real and complex analysis, the difference is not only the existence of the imaginary unit $i,$ but the failure of common inverse functions. The most important function in analysis is the function $\exp$ defined for every complex number $z$ by the equation

\[\exp(z) = \sum_{n=0}^{\infty}\frac{z^n}{n!}.\]

There are various ways to define this function, but the above equation is the most concise. Common notation for this function is $\exp(z) = e^z.$ A natural question then is whether or not there exists a function $f$ such that for every complex number $z$ we have

\[e^{f(z)}=z.\]

The answer is no for the general case of the entire complex plane. However, if $z$ happens to be a positive real number, then the answer is yes. In this case the function $f$ is unique, and we denote it by $\ln.$ For every positive number $x$ we have

\[\ln(x)= \int_1^x{{\frac{1}{t}dt}.\]

Something goes wrong in the complex case which prevents the existence of a universal inverse of $\exp.$ For one thing, the notion of integration of complex valued functions is much different than that of real valued functions. For functions like $\exp$ we have for any closed path $\gamma$ that

\[\int_{\gamma}\exp(z)dz = 0 \]

(For definitions of these concepts, see Rudin's Real and Complex Analysis page 200-201.) The notion of a path integral is not so difficult, as it is simply the same thing as one might have encountered in a vector calculus course. Functions which satisfy the above equation for any closed path $\gamma$ in a region $\Omega$ are said to be holomorphic in $\Omega.$ This fact can be used to show that the path of integration of holomorphic functions does not matter, and one can simply pick the easiest path to evaluate the integral.

This is where the problem occurs, there is a function $\ln$ in complex analysis, but it fails to be holomorphic in any region which includes a portion of the negative real axis. The question then is what exactly happens if we have a given path $\gamma$ and we perform the integration

\[\int_{\gamma}{{\frac{1}{z}dz}?\]

The answer is (intuitively): It depends on how many times the path "loops" around the origin. For every loop the path makes, the integral picks up another $2\pi i.$ This actually has some useful consequences, as it makes the Cauchy integral formula a possibility, which leads to power series representation of functions holomorphic on a given region. There is a sophisticated theorem which says that for any closed path $\gamma$ the above integral, when divided by $2\pi i$ is always an integer. The proof of this is quite subtle, but I can present the special case of a circle. Furthermore, I will consider the full "winding number," which is actually defined to be

\[Ind_{\gamma}(\alpha)\int_{\gamma}{{\frac{1}{z-\alpha}dz}.\]

for every complex number $\alpha.$ The main purpose of this is that it translates the origin to another location, letting us consider how many times the path loops around any given point.

A circle is the set of points of distance $r>0$ from a given center point $\alpha$. Complex variables gives a very simple way to express this , for every $t\in[0,2\pi]$ we define $\gamma(t)= \alpha + re^{it}$ This path $\gamma$ traces out the path of a circle, it begins and ends on the point $\alpha + r$, and loops exactly once around the center $\alpha$. We will let it loop $n$ times around $\alpha$ by taking instead $t\in[0,2n\pi]$

Suppose that $ | z - \alpha | < r $, then

$

\begin{aligned}

\frac{1}{2\pi i}\int_{\gamma}{\frac{1}{w-z}dw}= \frac{1}{2\pi i}\int_{\gamma}{\frac{1}{(w-\alpha)-(z-\alpha)}dw}\\

\\= \frac{ir}{2\pi i}\int_0^{2n\pi}{\frac{e^{it}}{re^{it}-(z-\alpha)}}\\

\\= \frac{1}{2\pi}\int_0^{2n\pi}{\frac{1}{1- \frac{z-\alpha}{re^{it}}}dt}\\

\\= \frac{1}{2\pi}{\sum_{m=0}^{\infty}\int_0^{2n\pi}\left(\frac{z-\alpha}{re^{it}}}\right)^mdt}\\

\\= n + 0 + 0 + \ldots\\

\end{aligned}

$

Although a simple fact, it has some nice consequences, from this and path independence one can prove power series representation of holomorphic functions. This gives us eventually Liouville's theorem, which can be used to prove some very deep theorems in the theory of Banach algebras.

2 hours ago

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