Definition 9.3.1.
A function \(f\) is meromorphic in the region \(G\) if \(f\) is holomorphic in \(G\) except for poles.
Section 9.3 Argument Principle and Rouché’s Theorem
In the previous section we saw how to compute integrals via residues, but in many applications we actually do not have an explicit expression for a function that we need to integrate (or this expression is very complicated). However, it may still be possible to compute the value of a function at any given point. In this situation we cannot immediately apply the Residue Theorem because we don’t know where the singularities are. Of course, we could use numerical integration to compute integrals over any path, but computationally this task could be very resource intensive. But if we do know the singularities, we can compute the residues numerically by computing a finite number of the integrals over small circles around these singularities. And after that we can apply the residue theorem to compute the integral over any closed path very effectively: we just sum up the residues inside this path. The argument principle that we study below, in particular, addresses this question. We start by introducing the logarithmic derivative.
Suppose we have a differentiable function \(f\text{.}\) Differentiating \(\fLog f\) (where \(\fLog\) is some branch of the logarithm) gives \(\frac{f'}{f}\text{,}\) which is one good reason to call this quotient the logarithmic derivative of \(f\text{.}\) It has some remarkable properties, one of which we would like to discuss here.
Now let’s say we have two functions \(f\) and \(g\) holomorphic in some region. Then the logarithmic derivative of their product behaves very nicely:
\begin{equation*}
\frac{ (fg)' }{ fg } \ = \ \frac{ f' g + f g' }{ fg } \ = \ \frac{ f'
}{ f } + \frac{ g' }{ g } \, \text{.}
\end{equation*}
We can apply this fact to the following situation: Suppose that \(f\) is holomorphic in a region \(G\) and \(f\) has (finitely many) zeros \(z_1, \dots,
z_j\) of multiplicities \(n_1, \dots, n_j\text{,}\) respectively. By Theorem 8.2.1, we can express \(f\) as
\begin{equation*}
f(z) \ = \ ( z - z_1 )^{n_1} \cdots ( z - z_j )^{n_j} g(z) \,\text{,}
\end{equation*}
where \(g\) is also holomorphic in \(G\) and never zero. Let’s compute the logarithmic derivative of \(f\) and play the same remarkable cancellation game as above:
\begin{align}
\frac{ f'(z) }{ f(z) } \amp \ = \ \frac{n_1 (z - z_1)^{n_1-1}
\cdots (z - z_j)^{n_j} \, g(z) + \cdots }{ ( z - z_1 )^{n_1}
\cdots ( z - z_j )^{n_j} \, g(z) } \nonumber\notag\\
\amp \ = \ \frac{ n_1 }{ z
- z_1 } + \frac{ n_2 }{ z - z_2 } + \cdots + \frac{ n_j }{ z -
z_j } + \frac{ g'(z) }{ g(z) } \, \text{.}\tag{9.1}
\end{align}
Something similar happens if \(f\) has finitely many poles in \(G\text{.}\) In Exercise 9.4.18, we invite you to prove that, if \(p_1, \dots, p_k\) are all the poles of \(f\) in \(G\) with order \(m_1, \dots, m_k\text{,}\) respectively, then the logarithmic derivative of \(f\) can be expressed as
\begin{equation}
\frac{f'(z)}{f(z)} \ = \ - \frac{ m_1 }{ z - p_1 } -
\frac{m_2}{z - p_2} - \dots - \frac{ m_k }{ z - p_k } + \frac{
g'(z) }{ g(z) } \, \text{,}\tag{9.2}
\end{equation}
where \(g\) is a function without poles in \(G\text{.}\) Naturally, we can combine the expressions for zeros and poles, as we will do in a moment.
Theorem 9.3.2. Argument Principle 1
The name Argument Principle stems from interpreting the integral \(\int_\gamma \frac{f'}{f}\) as the change in the argument of \(f(z)\) as \(z\) traverses \(\gg\text{,}\) since \(\fLog(f(z))' =
\frac{ f'(z) }{ f(z) }\text{.}\)
.
1
The name Argument Principle stems from interpreting the integral \(\int_\gamma \frac{f'}{f}\) as the change in the argument of \(f(z)\) as \(z\) traverses \(\gg\text{,}\) since \(\fLog(f(z))' =
\frac{ f'(z) }{ f(z) }\text{.}\)
Suppose \(f\) is meromorphic in a region \(G\) and \(\gg\) is a positively oriented, simple, closed, piecewise smooth path that does not pass through any zero or pole of \(f\text{,}\) and \(\gg \sim_G 0\text{.}\) Denote by \(Z(f,\gg)\) the number of zeros of \(f\) inside \(\gg\) counted according to multiplicity and by \(P(f,\gg)\) the number of poles of \(f\) inside \(\gg\) counted according to order. Then
\begin{equation*}
\frac{1}{2 \pi i} \i \frac{f'}{f} \ = \ Z(f,\gg) - P(f,\gg) \,\text{.}
\end{equation*}
Proof.
Suppose the zeros of \(f\) inside \(\gg\) are \(z_1, \dots,z_j\) of multiplicities \(n_1, \dots, n_j\text{,}\) respectively, and the poles inside \(\gg\) are \(p_1, \dots, p_k\) with order \(m_1, \dots, m_k\text{,}\) respectively. (You may meditate about the fact why there can be only finitely many zeros and poles inside \(\gg\text{.}\)) In fact, we may shrink \(G\text{,}\) if necessary, so that these are the only zeros and poles in \(G\text{.}\) By (9.1) and (9.2),
\begin{equation*}
\frac{ f'(z) }{ f(z) } \ = \ \frac{ n_1 }{ z - z_1 } + \dots +
\frac{ n_j }{ z - z_j } - \frac{ m_1 }{ z - p_1 } - \dots -
\frac{ m_k }{ z - p_k } + \frac{ g'(z) }{ g(z) } \, \text{,}
\end{equation*}
where \(g\) is a function that is holomorphic in \(G\) (in particular, without poles) and never zero. Thanks to Cauchy’s Theorem 4.3.4 and Exercise 4.5.4, the integral is easy:
\begin{align*}
\i \frac{ f' }{ f } \amp \ = \ n_1 \i \frac{ \diff{z}
}{ z - z_1 } \, + \, \cdots \, + \, n_j \i \frac{ \diff{z}
}{ z - z_j }\\
\amp \qquad - \, m_1 \i \frac{ \diff{z} }{ z - p_1 }
\, - \, \cdots \, - \, m_k \i \frac{ \diff{z} }{ z - p_k }
+ \i \frac{ g' }{ g }\\
\amp \ = \ 2 \pi i \left( n_1 + \dots + n_j - m_1 -
\dots - m_k \right) + \i \frac{ g' }{ g } \, \text{.}
\end{align*}
Finally, \(\frac{g'}{g}\) is holomorphic in \(G\) (because \(g\) is never zero in \(G\)), so that Corollary 4.3.8 gives
\begin{equation*}
\i \frac{g'} g \ = \ 0 \, \text{.}
\end{equation*}
As mentioned above, this beautiful theorem helps to locate poles and zeroes of a function \(f\text{.}\) The idea is simple: we can first numerically integrate \(\frac{f'}{f}\) over a big circle \(\gamma\) that includes all possible paths over which we potentially will be integrating \(f\text{.}\) Then the numerical value of \(\frac{ 1 }{ 2 \pi i } \int_{\gamma} \frac{ f' }{ f }\) will be close to an integer that, according to the Argument Principle, equals \(Z(f,\gamma)-P(f,\gamma)\text{.}\) Then we can integrate \(\frac{f'}{f}\) over a smaller closed path \(\gamma_1\) that encompasses half of the interior of \(\gamma\) and find \(Z(f,\gamma_1)-P(f,\gamma_1)\text{.}\) Continuing this process for smaller and smaller regions will (after certain verification) produce small regions where \(f\) has exactly one zero or exactly one pole. Integrating \(f\) over the boundaries of those small regions that contain poles and dividing by \(2\pi i\) gives all residues of \(f\text{.}\)
Another nice related application of the Argument Principle is a famous theorem due to Eugene Rouché (1832–1910).
Theorem 9.3.3. Rouché’s Theorem.
Suppose \(f\) and \(g\) are holomorphic in a region \(G\) and \(\gg\) is a positively oriented, simple, closed, piecewise smooth path, such that \(\gg \sim_G 0\) and \(|f(z)|>|g(z)|\) for all \(z \in \gg\text{.}\) Then
\begin{equation*}
Z(f+g,\gg) \ = \ Z(f,\gg) \,\text{.}
\end{equation*}
This theorem is of surprising practicality. It allows us to locate the zeros of a function fairly precisely. Here is an illustration.
Example 9.3.4.
All the roots of the polynomial \(p(z) = z^5 + z^4 + z^3 + z^2 + z + 1\) have modulus less than two. To see this, let \(f(z) = z^5\) and \(g(z) = z^4 + z^3 + z^2 + z + 1\text{.}\) Then for \(z \in C[0,2]\)
2
The Fundamental Theorem of Algebra (Theorem 5.3.2) asserts that \(p\) has five roots in \(\C\text{.}\) What’s special about the statement of Example 9.3.4 is that they all have modulus \(\lt 2\text{.}\) Note also that there is no general formula for computing roots of a polynomial of degree 5. (Although for this \(p\) it’s not hard to find one root—and therefore all of them.)
\begin{align*}
|g(z)| \ \amp \le \ |z|^4 + |z|^3 + |z|^2 + |z| + 1 \ = \ 16 + 8 + 4
+ 2 + 1 \ = \ 31 \\
\amp \lt \ 32 \ = \ |z|^5 \ = \ |f(z)| \, \text{.}
\end{align*}
So \(g\) and \(f\) satisfy the condition of the Theorem 9.3.3. But \(f\) has just one root, of multiplicity 5 at the origin, whence
\begin{equation*}
Z(p, \, C[0,2]) \ = \ Z(f+g, \, C[0,2]) \ = \ Z(f, \, C[0,2]) \ = \ 5 \,\text{.}
\end{equation*}
Proof.
By (9.1) and the Argument Principle (Theorem 9.3.2)
\begin{align*}
Z(f+g,\gg) \amp \ = \ \frac 1 {2 \pi i} \i \frac{ (f+g)' }{
f+g }\\
\amp \ = \ \frac 1 {2 \pi i} \i \frac{ \left( f \left( 1 + \frac
g f \right) \right)' }{ f \left( 1 + \frac g f \right) }\\
\amp \ = \
\frac 1 {2 \pi i} \i \left( \frac{ f' }{ f } + \frac{ \left( 1
+ \frac g f \right)'}{ 1 + \frac g f } \right)\\
\amp \ = \ Z(f,\gg) + \frac 1 {2 \pi i} \i \frac{\left( 1 +
\frac g f \right)' }{ 1 + \frac g f } \, \text{.}
\end{align*}
We are assuming that \(| \frac g f | \lt 1\) on \(\gg\text{,}\) which means that the function \(1 + \frac g f\) evaluated on \(\gg\) stays away from \(\R_{ \le 0 }\text{.}\) But then \(\Log ( 1 + \frac g f)\) is a well-defined holomorphic function on \(\gg\text{.}\) Its derivative is
\begin{equation*}
\dfrac{ \left( 1 + \frac g f \right)' }{ 1 + \frac g f }
\end{equation*}
which implies by Corollary 4.2.7 that
\begin{equation*}
\frac 1 {2 \pi i} \int_\gamma \frac{ \left( 1 + \frac g f
\right)' }{ 1 + \frac g f } \ = \ 0 \, \text{.}
\end{equation*}
