Argument Principle and Rouche's Theorem

Argument Principle

Lemma 1

If \(z_i\) is a zero of order \(n_i\) of \(f(z)\), then

\[\frac{f'(z)}{f(z)} = \frac{n_i}{z-z_i} + h_i(z)\]

near \(z_i\) for some function \(h_i(z)\) analytic at \(z_i\).

proof. Consider the Laurent expansion near \(z_i\), since \(z_i\) is a zero of order \(n_i\), we have

\[\begin{align*} f(z) &= \sum_{n=n_i}^{\infty}a_n(z-z_i)^n\\ &= (z-z_i)^{n_i} \sum_{n=0}^\infty a_{n+n_i}(z-z_i)^n\\ &:= (z-z_i)^{n_i} g_i(z) \end{align*}\]

Note that we have pulled out all the zeros, hence \(g_i(z)\neq 0\) near \(z_i\) and at \(z_i\). Thus, we can write

\[\frac{f'(z)}{f(z)} = \frac{n_i(z-z_i)^{n_i - 1} g_i(z) + (z-z_i)^{n_i}g_i'(z)}{(z-z_i)^{n_i}g_i(z)} = \frac{n_i}{z-z_i} + \frac{g'_i(z)}{g_i(z)}\]

Similarly, if \(z_i\) is a pole of order \(p_i\) near \(z_i\), then \(f(z) = \sum_{n=-n_i}^\infty c_n(z-z_n)^n\) so that

\[\frac{f'(z)}{f(z)} = \frac{-p_i}{z-z_i} + \frac{g'_i(z)}{g_i(z)}\]

If \(z_i\) is of order \(0\), then \(\frac{f'(z)}{f(z)}\) is already analytic using the same expansion.

Theorem 2 Argument Principle

Suppose \(f(z)\) is analytic on a Jordon contour \(C\) and meromorphic (has finitely many singularities, and each of which are poles) inside of \(C\). Then

\[N - P = \frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)}dz = [\arg(f(z))]_C=:w(f(C), 0)\]

where \(N\) is the number of \(0\)'s in \(C_{int}\) counted with multiplicity, and \(P\) is thue number of poles counted with multiplicity.

proof. For \(N - P = \frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)}dz\),

\[\begin{align*} \frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)}dz &= \sum_{k=1}^m Res(\frac{f'}{f}, z_k)\\ &= \sum_{zeros} (\lim_{z\rightarrow z_i} (z-z_i)(\frac{n_i}{z-z_i} + \frac{g'_i(z)}{g_i(z)})) + \sum_{poles} (\lim_{z\rightarrow z_i} (z-z_i)(\frac{-p_i}{z-z_i} + \frac{g'_i(z)}{g_i(z)})) \\ &= \sum n_i - \sum p_i\\ &= N- P \end{align*}\]

For \(\frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)}dz = w(f(C), 0)\),

Let \(c:[a,b]\rightarrow\mathbb C\) be some parameterization of \(C\), take \(\theta:[a,b]\rightarrow \mathbb R\) s.t. \(f(c(t)) = |f(c(t))|e^{i\theta(t)} = r(c(t))e^{i\theta(t)}\) then,

\[\begin{align*} \oint_C \frac{f'(z)}{f(z)} dz &= \int_a^b \frac{f'(c(t))}{f(c(t))}c'(t)dt\\ &= \int_a^b \frac{r'(c(t))c'(t) e^{i\theta(t)} + r(c(t))i\theta'(t)e^{i\theta(t)} }{r(c(t))e^{i\theta(t)} }dt\\ &= \int_a^b \frac{r'(c(t))c'(t)}{r(c(t))}dt + \int_a^b \theta'(t) dt \\ &= \int_{r(c(a))}^{r(c(b))} \frac{dr}{r} + i\int_{\theta(a)}^{\theta(b)} d\theta\\ &= \log(r(c(b))) - \log(r(c(a))) + i(\theta(b) - \theta(a))\\ &= i(\theta(b) - \theta(a)) \end{align*}\]

So that \(\frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)}dz = \frac{\theta(b) - \theta(a)}{2\pi} = w(f(C), 0)\)

Applications of Argument Principle

Example Determine the number of zeros of \(f(z) = z^3 + 1\) in the first quadrant.

Consider \(C = C_1\cup C_2 \cup C_3\) parameterized by \(c_1: [0, 1]\rightarrow \mathbb C:= Rt, c_2(t): [0, \pi/2]\rightarrow \mathbb C= Re^{it}, c_3: [0, 1]\rightarrow \mathbb C = iR(1-t)\).

Using argument princple, note that \(f\) does not have any singularities, so that number of zeros inside \(C\) equals \(w(f(C), 0)\). Then, compute

\[f(c_1(t)) = 1+R^3 t^3, f(c_2(t)) = 1+ R^3 e^{i3t} f(c_3(t)) = 1 - iR(1-3)^3\]

Thus, for \(R > 1\), \(w(C, 0) =1\) is the number of zeros.

Rouche's Theorem

Theorem Let \(f\) and \(g\) by analytic on \(\partial C \cup C_{int}\) for a Jordon contour \(C\). If \(|f(z)| > |g(z)|\) for all \(z\in C\), then \(f(z)\) and \(f(z) + g(z)\) have the same number of zeros in \(C_{int}\).

proof (informal). Since \(|f(z)| > |g(z)|\) on \(C\), notably have \(f(z)\neq 0\) on \(C\). Let \(w(z) = \frac{f(z)+g(z)}{f(z)}\).
Observe that \(|w(z) - 1| = |\frac{g(z)}{f(z)}| < 1\) on \(C\). i.e. \([\arg(w(z))]_C = 0\). Thus, \(w(z) \in B_1(1)\) has the same number of zeros and singularities. Furthermore, \(N=P=0\)

Example 1

Show that \(p(z) = z^4 + z^2 + 1\) has 4 roots inside \(C_2(0)\).

Let \(f(z) = z^4, g(z) = z^2 + 1\). \(\forall |z| = 2\), obviously \(|f(z)| = 16 > 5 = |f(z)|\) so that apply Rouche's Theorem, \(p = f+g\) has the same number of 0's as \(f\), which is 4.

Example 2

Show that \(h(z) = 3z^2 - \cos z\) has exactly \(2\) roots inside \(C_1(0)\).

Let \(f(z) = 3z^2, g(z) = -\cos(z)\). \(\forall |z| = 1. |f(z)| = 3\) and \(|g(z)| \leq |\cos z | \leq e\). so that by Rouche's Theorem, \(p = f+g\) has the 2 roots as of \(f\).

Example 3

Show that \(h(z) = z^5 + z + 3\) has \(5\) roots in \(\bar A_{1,2}(0)\)

Note that \(|z| = 2\), \(|z^5| = 32 > 5 = |z|+ 3 = |z+3|\) so that \(h\) has \(5\) zeros on \(C_2(0)\), also note that \(|h(z)| > 0\) so that no zeros on the \(\partial C_2\). In addition, note that on \(C_1(0)\), \(|z+3| \geq 3 - |z| = 2 > 1 = |z^5|\) and \(z+3\) has on zeros inside \(C_1(0)\). Therefore, all \(5\) zeros must locate in \(A_{1,2}(0)\)

Fundamental Theorem of Algebra

Theorem For a polynomial \(p(z) = \sum_{k=0}^n a_kz^k\), \(p\) has exactly \(n\) roots in \(\mathbb C\).

proof. Note that \(p(z)\) has the same number of roots as \(\frac{p(z)}{a_n}\), so wlog assume \(a_n = 1\).
Let \(f(z) = z^n, g(z) = \sum_{k=0}^{n-1}a_k z^k\) so that \(p = f+g\).
Then, \(\forall |z| \geq 1\)

\[\begin{align*} |g(z)| &\leq \sum_{k=0}^{n-1} |a_k||z|^k\\ &\leq |z|^{n-1}\sum_{k=0}^{n-1}|a_k|\\ &\leq \max\{\sum_{k=0}^{n-1}|a_k|, 1\} |z|^{n-1} \end{align*}\]

Let \(R = \max\{\sum_{k=0}^{n-1}|a_k|, 1\}\), so that on \(C_R(0)\), i.e. \(\forall |z| = R\) we have

\[|f(z)| = R^n = R |z|^{n-1}\geq |g(z)|\]

By Rouche's Theorem, the statement is proven.