Theorem 6.6.1. Morera’s theorem.
Let \(f\) be a continuous function in a simply connected domain \(D\text{.}\) If \(\int_Cf(z)\,dz=0\) for every closed contour \(C\) in \(D\text{,}\) then \(f\) is analytic in \(D\text{.}\)
Section 6.6 The Theorems of Morera and Liouville
In this section we investigate some of the qualitative properties of analytic and harmonic functions. Our first result shows that the existence of an antiderivative for a continuous function is equivalent to the statement that the integral of \(f\) is independent of the path of integration. This result is stated in a form that will serve as a converse of the Cauchy-Goursat theorem.
Proof.
We select a point \(z_0\) in \(D\) and define \(F(z)\) by
\begin{equation*}
F(z) = \int_{z_0}^{z}f(\xi)\,d\xi\text{,}
\end{equation*}
where the notation indicates the integral is taken on any contour that begins at \(z_0\) and ends at \(z\text{.}\) The function \(F(z)\) is well defined because, if \(C_1\) and \(C_2\) are two contours in \(D\)—both with initial point \(z_0\) and terminal point \(z\text{,}\) then \(C=C_1-C_2\) is a closed contour in \(D\text{,}\) and by hypothesis,
\begin{equation*}
0 = \int_C f(\xi)\,d\xi =\int_{C_1}f(\xi)\,d\xi - \int_{C_2} f(\xi)\,d\xi
\end{equation*}
Since \(f\) is continuous, we know that for any \(\varepsilon>0\) there exists \(\delta>0\) such that \(|f(\xi)-f(z)|\lt \varepsilon\) whenever \(|\xi -z| \lt \delta\text{.}\) Now we can use the same steps as those in the proof of Theorem 6.4.1 to show that \(F\,'(z) = f(z)\text{.}\) Hence \(F(z)\) is analytic on \(D\text{,}\) and Corollary 6.5.10 implies that \(F\,'(z)\) and \(F\,''(z)\) are also analytic. Therefore \(f\,'(z) = F\,''(z)\) exists for all \(z\) in \(D\text{,}\) proving that \(f(z)\) is analytic in \(D\text{.}\)
Cauchy ’s integral formula shows how the value \(f(z_0)\) can be represented by a contour integral. If we choose the contour of integration \(C\) to be a circle with center \(z_0\text{,}\) then we can show that the value \(f(z_0)\) is the integral average of the values of \(f(z)\) at points \(z\) on the circle \(C\text{,}\) a fact that the following theorem elucidates.
Theorem 6.6.2. Gauss’s mean value theorem.
If \(f\) is analytic in a simply connected domain \(D\) that contains the circle \(C_R(z_0)=\{z:|z-z_0|=R\}\text{,}\) then
\begin{equation*}
f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f(z_0+Re^{i\theta})\,d\theta
\end{equation*}
Proof.
We parametrize the circle \(C_R(z_0)\) by
\begin{equation*}
C_R(z_0) : z(\theta ) = z_0+Re^{i\theta}, \text{ and } dz = iRe^{i\theta}\,d\theta, \text{ for } \quad0 \le \theta \le 2\pi\text{,}
\end{equation*}
and use this parametrization along with Cauchy ’s integral formula to obtain
\begin{equation*}
f(z_0) = \frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+Re^{i\theta})iRe^{i\theta}}{Re^{i\theta}}\,d\theta = \frac{1}{2\pi}\int_0^{2\pi}f(z_0+Re^{i\theta})\,d\theta\text{.}
\end{equation*}
We now prove an important result concerning the modulus of an analytic function.
Theorem 6.6.3. Maximum modulus principle.
Let \(f\) be analytic and nonconstant in the domain \(D\text{.}\) Then \(|f(z)|\) does not attain a maximum value at any point \(z_0\) in \(D\text{.}\)
Proof.
(By contraposition): Assume the contrary and suppose that there exists a point \(z_0\) in \(D\) such that
\begin{equation}
|f(z)| \le |f(z_0)|\tag{6.6.1}
\end{equation}
holds for all \(z\) in \(D\text{.}\) If \(C_R(z_0)\) is any circle contained in \(D\text{,}\) Theorems 6.6.2 and Theorem 6.2.19 imply that
\begin{equation}
|f(z_0)| = \left|\frac{1}{2\pi} \int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta\right| \le \frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})|\,d\theta\text{,}\tag{6.6.2}
\end{equation}
for \(0 \le r \le R\text{.}\) We now treat \(|f(z)| = |f(z_0+re^{i\theta})|\) as a real-valued function of the real variable \(\theta\) and use Inequality (6.6.1) to get
\begin{equation}
\frac{1}{2\pi}\int_0^{2\pi} |f(z_0+re^{i\theta})|\,d\theta \le \frac{1}{2\pi}\int_0^{2\pi}|f(z_0)|\,d\theta = |f(z_0)|\text{,}\tag{6.6.3}
\end{equation}
\begin{equation*}
|f(z_0)| = \frac{1}{2\pi} \int_0^{2\pi}\big|f(z_0+re^{i\theta})\big|\,d\theta\text{,}
\end{equation*}
which we rewrite as
\begin{equation}
\frac{1}{2\pi i}\int_0^{2\pi}\Big(\big|f(z_0)\big| - \big|f(z_0+re^{i\theta})\big|\Big)\,d\theta =0, \text{ for } 0 \le r \le R\text{.}\tag{6.6.4}
\end{equation}
A theorem from calculus states that if the integral of a nonnegative continuous function taken over an interval is zero, then that function must be identically zero. Since Inequality (6.6.1) implies that the integrand in Equation (6.6.4) is a nonnegative real-valued function, we conclude that it is identically zero; that is,
\begin{equation}
\big|f(z_0)\big| = \big|f(z_0+re^{i\theta})\big|, \text{ for } 0 \le r \le R \text{ and } 0 \le \theta \le 2\pi\text{.}\tag{6.6.5}
\end{equation}
If the modulus of an analytic function is constant in a closed disk, then the function is constant in that closed disk by Theorem 3.2.12. Therefore we conclude from Identity (6.6.5) that
\begin{equation}
f(z)=f(z_0), \text{ for all } z \text{ in the closed disk } \overline{D}_R(z_0)\text{,}\tag{6.6.6}
\end{equation}
where \(\overline{D}_R(z_0) = \{z:|z-z_0| \le R\}\text{.}\) Now we let \(\zeta\) denote an arbitrary point in \(D, \, C\) be a contour in the original domain \(D\) that joins \(z_0\) to \(\zeta\text{,}\) and 2\(d\) denote the minimum distance from \(C\) to the boundary of \(D\text{.}\) We can find consecutive points \(z_0,\,z_1,\,z_2, \ldots, \,z_n=\zeta\) along \(C\text{,}\) with \(|z_{k+1}-z_k| \le d\text{,}\) such that the disks \(D_k=\{z:|z-z_k| \le d\}\text{,}\) for \(k=0,\,1,\ldots ,\,n\text{,}\) are contained in \(D\) and cover \(C\) as illustrated in Figure 6.6.4.
Each disk \(D_k\) contains the center \(z_{k+1}\) of the next disk \(D_{k+1}\text{,}\) so it follows that \(z_1\) lies in \(D_0\) and, from Equation (6.6.6) \(|f(z)|\) also reaches its maximum value at \(z_1\text{.}\) An identical argument to the one given above will show that
\begin{equation}
f(z)=f(z_1)=f(z_0, \text{ for all } z \in D_1\text{.}\tag{6.6.7}
\end{equation}
We proceed inductively to get
\begin{equation*}
f(z) = f(z_{k+1}) = f(z_k), \text{ for all } z \in D_{k+1}, \; 0 \le k\lt n-1\text{,}
\end{equation*}
from which it follows that \(f(\zeta )=f(z_0)\text{.}\) Therefore \(f\) is constant in \(D\text{,}\) which contradicts the assumption of our theorem. With this contraposition, the proof is complete.
We sometimes state the maximum modulus principle in the following form.
Theorem 6.6.5. Maximum modulus principle.
Let \(f\) be analytic and nonconstant in the bounded domain \(D\text{.}\) If \(f\) is continuous on the closed region \(R\) that consists of \(D\) and all its boundary points \(B\text{,}\) then \(|f(z)|\) assumes its maximum value, and does so only at point(s )\(z_0\) on the boundary \(B\text{.}\)
Example 6.6.6.
Let \(f(z)=az+b\text{.}\) If we set our domain \(D\) to be \(D_1(0)\text{,}\) then \(f\) is continuous on the closed region \(\overline{D}_1(0) = \{z:|z| \le 1\}\text{.}\) Prove that
\begin{equation*}
\max_{|z| \le 1}|f(z)| = |a| + |b|\text{,}
\end{equation*}
and that this value is assumed by \(f\) at a point \(z_0 = e^{i\theta _0}\) on the boundary of \(D_1(0)\text{.}\)
Solution.
From the triangle inequality and the fact that \(|z| \le 1\) in \(\overline{D}_1(0)\text{,}\) it follows that
\begin{equation}
|f(z)| = |az+b| \le |az|+|b| \le |a|+|b|\text{.}\tag{6.6.8}
\end{equation}
If we choose \(z_0=e^{i\theta _0}\text{,}\) where \(\theta _0\in \arg b-\arg a\text{,}\) then
\begin{align*}
\arg (az_0) \amp = \arg a+\arg z_0\\
\amp = \arg a+(\arg b-\arg a)\\
\amp = \arg b\text{,}
\end{align*}
so the vectors \(az_0\) and \(b\) lie on the same ray through the origin. This is the requirement for the Inequality (6.6.8) to be an equality (see Exercise 1.3.19, Section 1.3). Hence \(|az_0+b| = |az_0|+|b| = |a|+|b|\text{,}\) and the result is established.
Theorem 6.6.7. Cauchy ’s inequalities.
Let \(f\) be analytic in the simply connected domain \(D\) that contains the circle \(C_R(z_0)=\{z:|z-z_0|=R\}\text{.}\) If \(|f(z)| \le M\) holds for all points \(z\in C_R(z_0)\text{,}\) then
\begin{equation*}
\big|f^{(n)}(z_0)\big| \le \frac{n!M}{R^n}, \text{ for } n=1,\,2,\ldots\text{.}
\end{equation*}
Proof.
Let \(C_R(z_0)\) have the parametrization
\begin{equation*}
C_R(z_0) : z(\theta) = z_0+Re^{i\theta} \text{ and } dz = iRe^{i\theta}\,d\theta, \text{ for } 0 \le \theta \le 2\pi\text{.}
\end{equation*}
We use Cauchy ’s integral formula and write
\begin{equation*}
f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_{C_R(z_0)}\frac{f(z)}{(z-z_0)^{n+1}}\,dz = \frac{n!}{2\pi i}\int_0^{2\pi}\frac{f(z_0+Re^{i\theta}) iRe^{i\theta}}{R^{n+1}e^{i(n+1)\theta}}\,d\theta\text{.}
\end{equation*}
Combining this result with the ML inequality (Theorem 6.2.19), we obtain
\begin{align*}
\left| f\,^{(n)}(z_0) \right| \amp = \left|\frac{n!}{2\pi i} \int_0^{2\pi}\frac{f(z_0+Re^{i\theta}) iRe^{i\theta}}{R^{n+1}e^{i(n+1)\theta}}\,d\theta\right|\\
\amp \le \frac{n!}{2\pi}\int_0^{2\pi}\Big|f(z_0+Re^{i\theta})\Big| \left|\frac{iRe^{i\theta}}{R^{n+1}e^{i(n+1)\theta}}\right|\,d\theta\\
\amp \le \frac{n!}{2\pi}\int_0^{2\pi}M\frac{1}{R^n}\,d\theta\\
\amp = \frac{n!}{2\pi R^n}M2\pi\\
\amp = \frac{n!M}{R^n}\text{.}
\end{align*}
Theorem 6.6.8 shows that a nonconstant entire function cannot be a bounded function.
Theorem 6.6.8. Liouville’s theorem.
If \(f\) is an entire function and is bounded for all values of \(z\) in the complex plane, then \(f\) is constant.
Proof.
Suppose that \(|f(z)| \le M\) holds for all values of \(z\text{,}\) and let \(z_0\) denote an arbitrary point. Using the circle \(C_R(z_0)=\{z:|z-z_0| =R\}\) and Cauchy ’s Inequality with \(n=1\) yields
\begin{equation*}
\big|f\,'(z_0)\big| \le \frac{M}{R}\text{.}
\end{equation*}
Because \(R\) can be arbitrarily large we must have \(f\,'(z_0)=0\text{.}\) But \(z_0\) was arbitrary, so \(f\,'(z)=0\) for all \(z\text{.}\) If the derivative of an analytic function is zero for all \(z\text{,}\) then by Theorem 3.2.13 the function must be constant. Therefore, \(f\) is constant.
Example 6.6.9.
Show that the function \(f(z)=\sin z\) is not a bounded function.
Solution.
We established this characteristic with a somewhat tedious argument in Section 5.4. All we need do now is observe that \(f\) is not constant, and hence it is not bounded.
We can use Liouville’s theorem to establish an important theorem of elementary algebra.
Theorem 6.6.10. The fundamental theorem of algebra.
If \(P\) is a polynomial of degree \(n \ge 1\text{,}\) then \(P\) has at least one zero.
Proof.
(By contraposition): We will show that if \(P(z) \ne 0\) for all \(z\text{,}\) then the degree of \(P\) must be zero. Suppose that \(P(z) \ne 0\) for all \(z\text{.}\) This supposition implies that the function \(f(z) = \frac{1}{P(z)}\) is an entire function. Our strategy for the rest of the proof is as follows: We will show that \(f\) is bounded. Then Liouville’s theorem will imply that \(f\) is constant, and since \(f=\frac{1}{P}\text{,}\) this conclusion will imply that the polynomial \(P\) is constant, which will mean that its degree must be zero.
First we write \(P(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0\) and consider the equation
\begin{equation}
|f(z)| = \frac{1}{|P(z)|} = \frac{1}{|z|^n}\frac{1}{\left|a_n+\frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2}+\cdots +\frac{a_1}{z^{n-1}}+\frac{a_0}{z^n}\right|}\text{.}\tag{6.6.9}
\end{equation}
For \(k=1,\ldots n\text{,}\) \(\frac{|a_{n-k}|}{|z^k|} \to 0\) as \(|z| \to \infty\text{,}\) so
\begin{equation*}
a_n+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}+\cdots +\frac{a_0}{z^n} \to a_n, \text{ as } |z| \to \infty\text{.}
\end{equation*}
Combining this result with Equation (6.6.9) gives
\begin{equation*}
|f(z)| \to 0, \text{ as } |z| \to \infty\text{.}
\end{equation*}
In particular, we can find a value of \(R\) such that
\begin{equation}
|f(z)| \le 1 \text{ for all } |z| \ge R\text{.}\tag{6.6.10}
\end{equation}
If \(f(z)=u(x,y) +iv(x,y)\text{,}\) we have
\begin{equation*}
|f(z)| = \Big([u(x, y)]^2+[v(x,y)]^2\Big)^\frac{1}{2}\text{.}
\end{equation*}
which is a continuous function of the two real variables \(x\) and \(y\text{.}\) A result from calculus regarding real functions says that a continuous function on a closed and bounded set is bounded. Hence \(|f(z)|\) is a bounded function on the closed disk \(\overline{D}_R(0)\text{.}\) Thus there exists a positive real number \(K\) such that
\begin{equation*}
|f(z)| \le K, \text{ for all } |z| \le R\text{.}
\end{equation*}
Combining this with Inequality (6.6.10) gives
\begin{equation*}
|f(z)| \le M \text{ for all } z\text{,}
\end{equation*}
where \(M=\max \{K,1\}\text{.}\) By Liouville’s theorem, \(f\) is constant, so that the degree of \(f\) is zero. This observation completes the argument.
Corollary 6.6.11.
Let \(P\) be a polynomial of degree \(n \ge 1\text{.}\) Then \(P\) can be expressed as the product of linear factors. That is,
\begin{equation*}
P(z)=A(z-z_1) (z-z_2) \cdots (z-z_n)
\end{equation*}
where \(z_1,\,z_2, \ldots, \,z_n\) are the zeros of \(P\text{,}\) counted according to multiplicity, and \(A\) is a constant.
Exercises Exercises
1.
Factor each polynomial as a product of linear factors.
(a)
\(P(z)=z^4+4\text{.}\)
Solution.
\((z+1+i)(z+1-i)(z-1+i)(z-1-i)\text{.}\)
(b)
\(P(z)=z^2+(1+i)z+5i\text{.}\)
(c)
\(P(z)=z^4-4z^3+6z^2-4z+5\text{.}\)
Solution.
\((z+i)(z-i)(z-2+i)(z-2-i)\text{.}\)
(d)
\(P(z)=z^3-(3+3i) z^2+(-1+6i)z+3-i\text{.}\) \hint{Show that \(P(i)=0\text{.}\)}
2.
Let \(f(z)=az^n+b\text{,}\) where the region is the disk \(R=\{z:|z| \le 1\}\text{.}\) Show that \(\max\limits_{|z| \le 1}|f(z)| = |a|+|b|\text{.}\)
3.
Show that \(\cos z\) is not a bounded function.
Solution.
We know that the complex cosine is an entire function that is not a constant. By Liouville’s Theorem, it is not bounded.
4.
Let \(f(z)=z^2\text{.}\) Evaluate the following, where \(R\) represents the rectangular region defined by the set
\begin{equation}
R=\{z=x+iy:2 \le x \le 3 \text{ and } 1 \le y \le 3\}\text{.}\tag{6.6.11}
\end{equation}
(a)
\(\max\limits_{z\in R} |f(z)|\text{.}\)
(b)
\(\min\limits_{z\in R} |f(z)|\text{.}\)
(c)
\(\max\limits_{z\in R}\mathrm{Re}[f(z)]\text{.}\)
(d)
\(\min\limits_{z\in R}\mathrm{Im}[f(z)]\text{.}\)
5.
Let \(f\) be analytic in the disk \(D_5(0)\) and suppose that \(|f(z)| \le 10\) for \(z\in C_3(1)\text{.}\)
(a)
Find a bound for \(|f^{\,(4)}(1)|\text{.}\)
Solution.
\(|f\,^{(4)}(1)| \le \frac{4!(10)}{2\pi 3^5}6\pi = \frac{80}{3^3}\text{.}\) (Explain.)
(b)
Find a bound for \(|f^{\,(4)}(0)|\text{.}\)
\hint{\(\overline{D}_2(0) \subseteq \overline{D}_3(1)\text{.}\) Use Theorems 6.6.5 and Theorem 6.6.7.}
Solution.
\(|f\,^{(4)}(0)| \le \frac{4!(10)}{2\pi 2^5}4\pi = 15\text{.}\) (Explain.)
6.
Let \(f\) be an entire function such that \(|f(z)| \le M|z|\) for all \(z\text{.}\)
(a)
Show that, for \(n \ge 2\text{,}\) \(f\,^{(n)}(z)=0\) for all \(z\text{.}\)
(b)
Use part (a) to show that \(f(z)=az+b\text{.}\)
7.
Establish the following minimum modulus principle.
(a)
Let \(f\) be analytic and nonconstant in the domain \(D\text{,}\) and continuous on the closed region \(R\) that consists of \(D\) and all its boundary points \(B\text{.}\) Show that, if \(f(z) \ne 0\) throughout \(R\text{,}\) then \(|f(z)|\) assumes its minimum value, but does so only at point(s) \(z_0\) on the boundary \(B\text{.}\)
Solution.
If \(|f(z) |\ge m\) for all \(z\) in \(D\text{,}\) where \(m>0\text{,}\) then the function \(\frac{1}{f}\) is analytic in \(D\text{.}\) Apply the maximum modulus Theorem to the function \(\frac{1}{f}\) to get your result.
(b)
Show that the requirement \(f(z) \ne 0\) in part (a) is necessary by finding a function for which the requirement fails, and whose minimum modulus is attained at some place other than the boundary.
8.
Let \(u(x,y)\) be harmonic for all \((x,y)\text{.}\) Show that
\begin{equation*}
u(x_0,y_0) = \frac{1}{2\pi}\int_0^{2\pi}u\big(x_0+R\cos \theta, \; y_0+R\sin \theta\big)\,d\theta\text{,}
\end{equation*}
where \(R>0\text{.}\) \hint{Let \(f(z)=u(x,y) +iv(x,y)\text{,}\) where \(v\) is a harmonic conjugate of \(u\text{.}\)}
9.
Establish the following maximum principle for harmonic functions: Let \(u(x,y)\) be harmonic and nonconstant in the simply connected domain \(D\text{.}\) Then \(u\) does not have a maximum value at any point \((x_0, \, y_0)\) in \(D\text{.}\)
Solution.
Let \(f(z) =u(z) +iv(z)\text{,}\) where \(v\) is a harmonic conjugate of \(u\text{,}\) so that \(f\) is analytic in \(D\text{.}\) The function \(F(z) =\exp\big(f(z)\big)\) is also analytic in \(D\text{,}\) so that \(|F|\) does not take on a maximum in \(D\) by the maximum modulus Theorem. But \(|F(z)|=\exp\big(u(z)\big)\) for all \(z\) (show why). This leads to the conclusion since \(u\) is a real valued function, and the real valued function \(\exp\) is an increasing function. Explain this last part in detail.
10.
Let \(f\) be an entire function with the property that \(|f(z)| \ge 1\) for all \(z\text{.}\) Show that \(f\) is constant.
11.
Let \(f\) be nonconstant and analytic in the closed disk \(\overline{D}_1(0)\text{.}\) Suppose that \(|f(z)|\) is constant for \(z \in C_1(0)\text{,}\) i.e., that there is some number \(K\) such that \(|f(z)| = K\) for all \(z \in C_1(0)\text{.}\) Show that \(f\) has a zero in \(\overline{D}_1(0)\text{,}\) i.e., that there exists some \(z_0 \in \overline{D}_1(0)\) such that \(f(z_0)=0\text{.}\) \hint{Use both the minimum modulus principle (see Exercise 6.6.7) and maximum modulus principle.}
Solution.
By contraposition) If \(f\) does not have a zero, then \(\frac{1}{f}\) is analytic in \(D_1(0)\text{,}\) so its maximum occurs on the boundary. Since \(f\) is constant on the boundary, we conclude that both the maximum and the minimum of \(f\) are the same, which means \(f\) is constant.
12.
Why is it important to study the fundamental theorem of algebra in a complex analysis course?
