# Morera's Theorem

1. Jul 8, 2013

### Tsunoyukami

I'm having difficulty understanding this question - and if I am interpreting it correctly, how to go about doing so.

First, we will need the following theorem:

"Morera's Theorem If $f$ is a continuous function on a domain $D$ and if
$$\int_{\gamma} f(z) dz = 0$$
for every triangle $\gamma$ that lies, together with its interior, in $D$, then $f$ is analytic on $D$." (Complex Variables, 2nd. edition; Stephen D. Fisher, pg. 129)

"24. Use Morera's Theorem and an interchange of the order of integration to show that each of the following functions is analytic on the indicated domain; find a power-series expansion for each function by using the known power series for the integrand and interchanging the summation and integration.

a) $\int_{0}^{1} \frac{dt}{1 -tz}$ on $|z| < 1$" (Complex Variables, 2nd. edition; Stephen D. Fisher, pg. 134)

I have not attempted this problem because I am unsure of what exactly it is asking me to do. Here is my interpretation:

1) Write the integral with respect to z (be sure to change the bounds of the integral as well) as opposed to z.
2) Apply Morera's Theorem to show that the function is analytic.
3) Use the "known power series for the integrand" to write out a power series...as in basically copy out the known power series?

I'm just not really sure what to do or how to go about doing so. If my above interpretation is correct how do I complete the first step? Do I simply write:

$\int_{0}^{1} \frac{dt}{1 -tz}$ on $|z| < 1$
$\int_{0}^{2\pi} \frac{dz}{1-tz}$ on $0 \leq t \ leq 1$

I feel like that is almost blatantly incorrect...

Any guidance is very much appreciated - thanks!

2. Jul 8, 2013

### HallsofIvy

Staff Emeritus
Well, first, you are right that this is "blatantly incorrect". t going from 0 to 1 has nothing to do with z going from 0 to $2\pi$. They are independent variables.

The problem says "find a power-series expansion for each function by using the known power series for the integrand and interchanging the summation and integration". Have you written out the "known power series for the integrand"? It is assuming that you know the sum for a geometric series:
$$\sum r^n= \frac{1}{1- r}$$.

3. Jul 8, 2013

### Ray Vickson

Presumably, the question wants you to show that
$$\int_{\gamma} \int_{t=0}^1 \frac{dt}{1-zt} \:dz = \int_{t=0}^1 \int_{\gamma} \frac{dz}{1-tz} \; dt$$
for any triangle $\gamma$ in $|z| < 1$, and to use Morers's Theorem to derive some analyticity properties of the original integral.

4. Jul 8, 2013

### Tsunoyukami

So I can set my function such that r = zt? Then:

$$\frac{1}{1 - r} = \sum r^{n}$$
$$\frac{1}{1 - zt} = \sum (zt)^{n}$$

Not too sure where to go from here...or if what I've done is even valid. Should I then write

$$\sum (zt)^{n} = \sum t^{n}z^{n} = \sum t^{n} (z - 0)^{n}$$

Which means I have a power series centered at 0 and the coefficient of the nth term is $a_{n} = t^{n}$?

But then how does the integral from 0 to 1 come into play? Is this important because this series is convergent only in a disc of radius 1? In that case I would end of with an integral over $\gamma$ of a power series - or am I missing something?

Sorry, I just feel rather confused by this problem.

Additionally, I'm not sure I understand Morera's Theorem. Is it essentially the "opposite" of Cauchy's Theorem? Cauchy's Theorem states that for any analytic function on D the path integral of a piecewise smooth simple closed curve in D whose inside lies in D is equal to 0. Morera's Theorem states that if the path integral for every triangle (and its interior) in D is equal to 0, then the function is analytic.

I guess what's bugging me is...why is this true only if the closed curve is a triangle? I mean, it holds true for every triangle in D - so why not take a union of a bunch of triangles contained within D to form other non-triangle shapes? I would have expected Morera's Theorem to apply to any piecewise smooth simple closed curve...

5. Jul 10, 2013

### Tsunoyukami

I asked the professor about the question and this is the correct interpretation. However, I'm still not really sure how to go about doing this.

In order to show that the function is analytic in |z| < 1 I need to use Morera's Theorem, which states that I must show $\int_{\gamma} f(z) = 0$ for every triangle in |z| < 1.

My next question is what is $f(z)$? Does $f(z) = \frac{1}{1 - tz}$ or does $f(z) = \int_{0}^{1} \frac{1}{1 - tz} dt$?

I don't really see how the integral with respect to $\gamma$ is easier...the only way I can think of to make it easier is to use properties of analytic functions (which I can't do because I'm trying to prove that the function is analytic!)

6. Jul 12, 2013

### Tsunoyukami

I apologize for triple posting now (really...I'm sorry), but for the sake of completeness, I will continue with this problem as I believe I have made some progress.

First, let $f(z) = \int_{0}^{1} \frac{dt}{1 - tz}$ on $|z| < 1$. We want to compute the line integral over $f(z)$ and we can do this by writing the following:

$$\int_{\gamma} \left[ \int_{0}^{1} \frac{dt}{1 - tz} \right] dz = \int_{0}^{1} \left[ \int_{\gamma} \frac{dz}{1 - tz} \right] dt$$

In this formulation we integrate over $\gamma$ first with respect to $z$ while considering $t$ constant. We call this 'inner' function $g(z) = \frac{1}{1 -tz}$ and note that this function $g(z)$ is analytic on the disc $|z| < 1$. Therefore, for any closed curve $\gamma$, $\int_{\gamma} g(z) dz = 0$ by Cauchy's Theorem. Then:

$$\int_{\gamma} \left[ \int_{0}^{1} \frac{dt}{1 - tz} \right] dz = \int_{0}^{1} \left[ \int_{\gamma} \frac{dz}{1 - tz} \right] dt = \int_{0}^{1} 0 dt = c$$

A constant function is analytic (and in fact, is entire).

That being said, I'm not sure if there is more I should do before proceeding to the second portion of the question or if this is sufficient. Any thoughts?