# Morera's theorem

(Redirected from Morera's Theorem)

Here we follow standard texts, such as Spiegel (1964) or Levinson and Redheffer (1970). 

If for every closed contour $C$ within a region ${\mathcal {R}}$ of the complex and

$\oint _{C}f(z)\;dz=0$ then $f(z)$ is analytic everywhere in ${\mathcal {R}}$ .

## Proof:

Suppose that the hypothesis is false, and that in some portion of ${\mathcal {R}}$ , there exists some contour $\gamma$ such that

$\oint _{\gamma }f(z)\;dz=A$ where $A$ is nonzero.

We assume that the real and imaginary parts of $f(z)$ are continuously differentiable in ${\mathcal {R}}$ , (otherwise $df(z)/dz$ would not exist at some points in ${\mathcal {R}}$ and $f(z)$ would not be analytic by the definition of analyticity.)

Rewriting the integral by subsituting real and imaginary parts $f(z)=u(x,y)+iv(x,y)$ and $dz=dx+idy$ we obtain

$\oint _{\gamma }f(z)\;dz=\int _{\gamma }(u+iv)(dx+idy)=\oint _{\gamma }(udx-vdy)+i\oint _{\gamma }(vdx+udy)=A$ .

Because the partial derivatives of $u(x,y)$ and $v(x,y)$ exist, we may invoke the 2D version of Green's theorem for the real and imaginary parts

$\int _{C}f(z)\;dz=-\left(\int _{\mathcal {R}}\int \left({\frac {\partial v(x,y)}{\partial x}}+{\frac {\partial u(x,y)}{\partial y}}\right)\;dx\;dy\right)+i\left(\int _{\mathcal {R}}\int \left({\frac {\partial u(x,y)}{\partial x}}-{\frac {\partial v(x,y)}{\partial y}}\right)\;dx\;dy\right)=A.$ Because $A=a+ib\neq 0$ at least one of the Cauchy-Riemann equations is not satisfied, and the theorem is proven. However, this proof is less satisfactory, because we had to assume that the real and imaginary parts of $f(z)$ were differentiable.

# Alternate proof:

In general, Morera's theorem is a statement that if $f(z)$ is continuous, then it has an anti-derivative $F(z)$ , which is an analytic function for all $z$ in the region ${\mathcal {R}}$ . Furthermore, this forces $f(z)$ to be analytic, as well.

Suppose that we define a function $F(z)=\int _{a}^{z}f(u)\;du$ and form the derivative of $F(z)$ by the formal definition of a differentiation

$\lim _{\Delta z\rightarrow 0}\left|{\frac {F(z+\Delta z)-F(z)}{\Delta z}}-f(z)\right|=\lim _{\Delta z\rightarrow 0}\left|{\frac {\int _{a}^{z+\Delta z}f(u)\;du-\int _{a}^{z}f(u)\;du-f(z)\;\Delta z}{\Delta z}}\right|$ .

This expression should vanish in the limit, if $F(z)$ is the anti-derivative of $f(z)$ .

We can rewrite this limit as

$\lim _{\Delta z\rightarrow 0}\left|{\frac {\int _{a}^{z+\Delta z}f(u)\;du-\int _{a}^{z}f(u)\;du-f(z)\;\Delta z}{\Delta z}}\right|=\lim _{\Delta z\rightarrow 0}\left|{\frac {\int _{z}^{z+\Delta z}\left[f(u)-f(z)\right]\;du}{\Delta z}}\right|$ .

Because $f(z)$ is assumed to be continuous, by the formal definition of continuity, whenever $|f(u)-f(z)|<\epsilon$ there exists, for all $u$ a $\delta$ such that $|u-z|<\delta$ .

Given $\delta >|\Delta z|$ this is true, allowing us to estimate the bound on the integrand with $\epsilon$ $\lim _{\Delta z\rightarrow 0}\left|{\frac {F(z+\Delta z)-F(z)}{\Delta z}}-f(z)\right|\leq \lim _{\Delta z\rightarrow 0}{\frac {1}{\Delta z}}\left|\int _{z}^{z+\Delta z}\left[f(u)-f(z)\right]\;du\right|<\epsilon$ .

Because $\epsilon$ is arbitrary $F^{\prime }(z)=f(z)$ . $F(z)$ is analytic by the definition of analyticity as the existence of the derivative. Furthermore, because $F(z)$ is defined by an integral, this result holds for any integration path in ${\mathcal {R}}$ we have

$\oint _{C}f(u)\;du=\int _{a}^{z}f(u)\;du+\int _{z}^{a}f(u)\;du=0$ for arbitrary $a$ and $z$ in ${\mathcal {R}}$ , which in turn shows that $f(z)$ must be analytic for all $z$ in ${\mathcal {R}}$ , proving Morera's theorem.

## Further ramifications of this second proof

It is possible to show from the Cauchy-Riemann equations that the derivative of an analytic function is, itself, analytic, and is infinitely differentiable in its region of analyticity. It is furthermore possible to construct a new function $F_{0}(z)$ that is the anti-derivative of $F(z)$ . This process may also be repeated ad infinitum.