The argument principle and Rouché’s theorem

Published

18 October 2026

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The argument principle is a consequence of the residue theorem. It relates the winding number of a curve to the number of zeros and poles inside it.

1 Windings of a curve around a point

Let \(\gamma\) be a closed curve, \(z_0 \not\in \gamma\). Then the index of the point \(z_0\) with respect to the curve \(\gamma\) (also called the winding number) is

\[ \operatorname {Ind}\left(\gamma , z_{0}\right) := \frac{1}{2 \pi i} \int _{\gamma } \frac{1}{z-z_{0}} d z \]

Informally: this is the number of times the curve winds around the point counterclockwise. Correspondingly, clockwise windings are counted with a minus sign.

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: the index of a point with respect to a curve.

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/240/010_winding_index_fig.tex

1.1 Mapping curves: \(f \circ \gamma\)

One of the key notions in this topic is the image of one curve under a map to another. That is, if \(z=\gamma (t)\) is a curve and \(w=f(z)\) is a function, then \(w=f \circ \gamma (t)=f(\gamma (t))\) is another curve. We say that \(f\) maps \(\gamma\) to \(f \circ \gamma\).

Example 1

Let \(\gamma (t)=\mathrm{e}^{i t}\), where \(0 \leq t \leq 2 \pi\) (the unit circle). Let \(f(z)=z^{2}\). Describe the curve \(f \circ \gamma\).

Solution

\(f \circ \gamma (t)=\mathrm{e}^{2 i t}\) traces the unit circle twice as \(t\) ranges from 0 to \(2 \pi\).

Example 2

Let \(\gamma (t)=\) it, where \(-\infty <t<\infty\) (the \(y\)-axis). Let \(f(z)=1 /(z+1)\). Describe the curve \(f \circ \gamma (t)\).

Solution

\(f(z)\) is a fractional-linear (Möbius) transformation, which maps the line given by \(\gamma\) to a circle passing through the origin, centered at the point \(1 / 2\). Checking at several points: \[ f(-i)=\frac{1}{-i+1}=\frac{1+i}{2}, \quad f(0)=1, \quad f(i)=\frac{1}{i+1}=\frac{1-i}{2}, \quad f(\infty )=0. \] We see that the circle is traversed clockwise as \(t\) ranges from \(-\infty\) to \(\infty\).

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: traversal of a circle and the change in the argument.

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/240/020_circle_winding_fig.tex

2 The argument principle

Let \(D\) be a simply connected bounded domain, i.e. \(\gamma = \partial D\) is a simple closed curve, oriented counterclockwise. Let the function \(f\) be holomorphic in \(\overline{D}\) (i.e. in some domain containing \(\overline{D}\)), except possibly at finitely many points that are poles. Suppose \(f\) has neither zeros nor poles on the boundary \(\gamma\) itself. Let

\[ p_{1}, \ldots , p_{m} \]

be the poles of the function \(f\) in \(D\). Let

\[ z_{1}, \ldots , z_{n} \]

be the zeros of the function \(f\) in \(D\). Let \(\operatorname {mult}\left(z_{k}\right)\) be the multiplicity of the zero at the point \(z_{k}\). Similarly, \(\operatorname {mult}\left(p_{k}\right)\) is the order of the pole at the point \(p_{k}\).

Denote by \(Z_{f, \gamma }\) the sum of the multiplicities of the zeros of the function \(f\) inside \(\gamma\). Similarly for \(P_{f, \gamma }\):

\[ Z_{f, \gamma } = \sum _{i=1}^{n}\operatorname {mult}(z_i), \qquad P_{f, \gamma } = \sum _{i=1}^{m}\operatorname {mult}(p_i) \]

Theorem 1 (The argument principle) Under the assumptions above, \[ \frac{1}{2\pi i}\int _{\gamma } \frac{f^{\prime }(z)}{f(z)} d z= Z_{f, \gamma }-P_{f, \gamma } = \operatorname {Ind}(f \circ \gamma , 0) \]

Example 3

Let \(f(z)=z^{2}+z\). Find the winding number of the curve \(f \circ \gamma\) about the point \(0\) for the curves \(\gamma_{0.5}\), \(\gamma_{1}\), \(\gamma_{2}\), where \(\gamma_{R}\) is the circle of radius \(R\) about the origin (oriented counterclockwise).

Solution

\(f(z)\) has zeros at the points \(0,-1\). It has no poles.

So, \(f\) has no poles and has two zeros inside \(\gamma_{2}\). The argument principle states:

\[ \operatorname {Ind}\left(f \circ \gamma _{2}, 0\right)=Z_{f, \gamma _{2}}- P_{f, \gamma _{2}}=2 \]

Similarly, \(f\) has no poles and one zero inside \(\gamma_{0.5}\), so \(\operatorname {Ind}\left(f \circ \gamma_{0.5}, 0\right)=1-0=1\)

For \(\gamma_{1}\) a zero of \(f\) lies on the curve, i.e. \(f(-1)=0\), so the argument principle does not apply. The image \(f \circ \gamma_{1}\) is shown in the figure below — it passes through \(0\).

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: the images of 3 circles under f(z)=z^2.

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/240/030_circles_under_z2_fig.tex

3 Rouché’s theorem

Theorem 2 (Rouché) Let

\(\gamma\) be a simple closed curve
\(f, h\) be analytic on \(\gamma\) and inside it, except for finitely many poles.
Neither \(f\) nor \(h\) have poles on \(\gamma\).
\(|h|<|f|\) everywhere on \(\gamma\).

Then \[ \operatorname {Ind}(f \circ \gamma , 0)=\operatorname {Ind}((f+h) \circ \gamma , 0) \] That is, \[ Z_{f, \gamma }-P_{f, \gamma }=Z_{f+h, \gamma }-P_{f+h, \gamma } \tag {4} \]

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: an illustration for Rouché’s theorem (the source had a typo

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/240/040_rouche_illustration_fig.tex

Corollary 1 Under the same assumptions, if \(h\) and \(f\) are analytic (with no poles), then \[ Z_{f, \gamma }=Z_{f+h, \gamma } \]

Proof

Since the functions are analytic, \(P_{f, \gamma }\) and \(P_{f+h, \gamma }\) are both equal to 0. Thus equation 4 shows that \(Z_{f}=Z_{f+h}\).

Remark

We think of \(h\) as a small perturbation of \(f\).

Example 4

Show that all 5 zeros of \(z^{5}+3 z+1\) lie inside the curve \(\gamma_{2}:|z|=2\).

Solution

Let \(f(z)=z^{5}\) and \(h(z)=3 z+1\). Clearly, all \(5\) roots of the equation \(f\) (actually one root of multiplicity \(5\)) lie inside \(\gamma_{2}\). We also have \(\left|h\right| < 7 < 32 = \left|f\right|\) on \(\gamma_{2}\). By Rouché’s theorem, all \(5\) roots of the equation \(f+h=z^{5}+3 z+1\) must also lie inside the curve.

Example 5

How many roots does the equation \(z+3+2 \mathrm{e}^{z}\) have in the left half-plane (\(\operatorname {Re}z < 0\))?

Solution

Let \(f(z)=z+3, h(z)=2 \mathrm{e}^{z}\). Consider the contour from \(-i R\) to \(i R\) along the \(y\)-axis, followed by the left semicircle of radius \(R\) back to \(-i R\). That is, the contour \([-iR, iR]+C_{R}\), shown below.

Note

Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).

Plan: TikZ: an example for Rouché’s theorem, f(z)=z+3, h(z)=2e^z.

Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/240/050_rouche_example_fig.tex

To apply the corollary to Rouché’s theorem, we need to verify that (for large \(R\)) \(|h|<|f|\) on \([-iR, iR]+C_{R}\). On \([-iR,iR]\) we have \(z=i y\), so

\[ \left|f\right| = \left|3 + iy\right| \geq 3, \quad \left|h\right| = 2\left|e^{iy}\right| = 2 \]

So, \(\left|h\right| < \left|f\right|\) on \([-iR,iR]\).

On \(C_{R}\) we have \(z=x+i y\), where \(x<0\), and \(|z|=R\). So, \(\left|f\right| > R-3\) for large \(R\),

\[ \left|h\right| = 2\left|e^{x + iy}\right| = 2e^{x} < 2 \]

since \(x < 0\). So, \(\left|h\right| < \left|f\right|\) on \(C_{R}\).

The only zero of the function \(f\) is at the point \(z=-3\), which lies inside the contour. Hence, by Rouché’s theorem, \(f+h\) has the same number of roots inside the contour as \(f\), namely \(1\).

Letting \(R\) tend to infinity, we see that \(f+h\) has only one root in the entire half-plane.

Answer

\(1\)

Example 6

Using Rouché’s theorem, prove the fundamental theorem of algebra: a polynomial \[ P(z)=z^{n}+a_{n-1} z^{n-1}+\ldots +a_{0} \] over the field of complex numbers has exactly \(n\) roots (counted with multiplicity).

Solution

Let \(f(z)=z^{n}\) and \(h=P-f\). Choose \(R\) such that

\[ R > \max \left\{ 1, n\left|a_{n-1}\right|, \ldots , n\left|a_{0}\right|\right\} \]

Then on the curve \(\gamma_{R} = \left\{ \left|z\right| = R\right\}\) we have

\[ \left|h\right| \leq \left|a_{n-1}\right|R^{n-1} + \left|a_{n-2}\right|R^{n-2} + \ldots + \left|a_{1}\right|R + \left|a_0\right| \leq \frac{R}{n}R^{n-1} + \frac{R}{n}R^{n-2} + \ldots + \frac{R}{n}R + \frac{R}{n} < R^{n} \]

On \(|z|=R\) we have \(|f(z)|=R^{n}\), so we have shown that \(\left|h\right| < \left|f\right|\) on the curve \(\gamma_{R}\). Thus, by Rouché’s theorem, \(f+h\) and \(f\) have the same number of zeros inside \(|z|=R\). Since we know that \(f\) has exactly \(n\) zeros inside the curve, the same is true for the polynomial \(f+h\). Now let \(R\) tend to infinity. We have shown that \(f+h\) has exactly \(n\) zeros in the entire plane.

Remark. The proof gives a simple bound on the magnitude of the roots: all of them have modulus not exceeding \(\max \left(1, n\left|a_{n-1}\right|, \ldots , n\left|a_{0}\right|\right)\).

Example 7

Find the number of roots of the equation \[ F(z) = z^{4}+4 z^{3}-8 z-2=0 \]

in the domain \(\{ z: 1<|z|<3\}\),
in the domain \(\{ z:|z|>3\}\).

(2017/2018 academic year, semester exam in complex analysis)

Solution

Let us first consider the domain \(\left|z\right| < 1\). Let \(f(z) = -8z\), \(h(z) = z^{4}+4 z^{3}-2\). Compare their values on the boundary of the domain \(\gamma_{1}\). For \(|z|=1\) we have \(\left|f(z)\right| = 8\),

\[ \left|h(z)\right| = \left|z^{4}+4 z^{3}-2\right| \leq \left|z\right|^4 + 4\left|z\right|^3 + 2 = 7 < 8 \]

Since \(\left|f\right| > \left|h\right|\) on \(\gamma_{1}\), by Rouché’s theorem the function \(F = f + h\) has, in the domain \(\left|z\right| < 1\), the same number of zeros (with multiplicity) as the function \(f(z)\), namely \(1\) zero. Moreover, using the remark to Rouché’s theorem, we see that the function \(F\) does not vanish on \(\gamma_1\), so \(F(z)\) has \(1\) zero in the disk \(|z| \leq 1\).
Now consider the domain \(|z|<3\) and determine how many zeros the function \(F\) has there. To this end let us define the functions \(f\) and \(h\) differently:

\[ \begin{aligned} & f(z)=z^{4}+4 z^{3} \\ & h(z)=-8 z-2 \end{aligned} \]

On \(\gamma_{3}\) (for \(\left|z\right| = 3\)) we get

\[ \begin{align} \left|f(z)\right| & = \left|z^{4}+4 z^{3}\right|=|z|^{3}|z+4|=27|z+4| \geq 27| | z|-4|=27(4-3)=27\\ \left|h(z)\right| & = |-8 z-2| \leq |8 z|+2=24+2=26 \end{align} \]

Since the inequality \(\left|f\right| > \left|h\right|\) holds on \(\gamma_{3}\), by Rouché’s theorem the function \(F\) has, in the domain \(|z|<3\), the same number of zeros (with multiplicity) as the function \(f\). The function \(f(z)\) has a zero \(z=0\) of multiplicity \(3\), lying in the domain \(\left|z\right| < 3\), and a zero \(z=-4\), which lies outside this domain. Hence the function \(F\) has three zeros inside the domain \(\left|z\right| < 3\). Since one of these zeros lies in the disk \(|z| \leq 1\), in the domain \(1<|z|<3\) the function \(F(z)\) has two zeros.

Since \(F\) is a degree-4 polynomial, by the fundamental theorem of algebra it has \(4\) zeros in the entire complex plane. As we have just proved using Rouché’s theorem, in the domain \(|z|<3\) the polynomial \(F(z)\) has 3 zeros. By the remark to Rouché’s theorem, on the boundary of the domain, at \(|z|=3\), the polynomial \(F(z)\) has no zeros. Hence, in the domain \(|z|>3\) the polynomial \(F(z)\) has exactly 1 zero. Thus, \(F\) has one root in the domain \(|z|>3\).

Answer

\(2\) roots,

\(1\) root.

Problem 15.1

Find the number of roots of the equations in the domains given in parentheses:

\(z^{7}-5 z^{4}+z^{2}-2 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} )\);
\(z^{3}-12 z+2 = 0 \quad (\left\{ z:\left|z\right| < 2\right\} )\);
\(z^{6}-6 z+10 = 0 \quad (\left\{ z:\left|z\right| > 1\right\} )\);
\(z^{4}+z^{3}-4 z+1 = 0 \quad (\left\{ z: 1 < \left|z\right| < 2\right\} )\).

Problem 15.4

Prove that for \(\lambda > 1\) the equation \[ z = \lambda -e^{-z} \] has exactly one root (and moreover a real one) in the half-plane \(\left\{ z: \operatorname {Re} z \geqslant 0\right\}\).

Problem T.7

Find the number of roots of the polynomial \(4 z^{6}+4 z^{3}+9 z-4\) in the disk \(\left|z\right| < 1\).

Problem T.8

Find the number of roots of the polynomial \(z^{2 n}-\alpha z^{2 n-1}+1\) in the left half-plane for \(\alpha > 0, n \in \mathbb {N}\).

Problem T.9

Prove that there exists \(\delta > 0\) such that for all values \(\alpha , \beta \in B_{\delta }(0)\) the equation \(e^{z}-1 = \alpha z+\beta\) has exactly one root in the disk \(B_{\pi }(0)\).

Problem T.11

Prove that the polynomial \[ P(z) = z^{8}+a z^{3}+b z+c, \quad a \geq 0, \quad b \geq 0, \quad c > 0, \] has exactly two roots in the first quadrant.

4 Evaluating integrals

Example 8

Using residue theory and Rouché’s theorem, compute the integral over the positively oriented contour \[ I = \oint _{|z|=1} z^{-2} \exp \left(\frac{81}{4 z^{5}+4 z^{3}-4 z+9}\right) d z \]

(2019/2020 academic year, semester exam in complex analysis)

Solution

Let us determine where the singular points of the function

\[ g(z)=\frac{1}{z^{2}} \exp \left(\frac{81}{4 z^{5}+4 z^{3}-4 z+9}\right) \]

are located. To do this, let us find the number of roots of the equation

\[ F(z) = 4 z^{5}+4 z^{3}-4 z+9=0 \tag {2} \]

in the domain \(|z|<1\). In this case, trying to define functions \(f(z)\) and \(h(z)\) for applying Rouché’s theorem by grouping the terms on the left-hand side of the equation, and then comparing their moduli at \(|z|=1\) using only the triangle inequality for the modulus of a sum and difference of numbers, does not lead to the desired result.

The solution to this and similar problems relies on the special form of the polynomial on the left-hand side of equation (2). This polynomial contains three terms whose powers of \(z\) form a decreasing arithmetic progression (in our case these are the terms \(4 z^{5}, 4 z^{3}\), and \(-4 z\)), with the coefficients of the highest and lowest powers being real and equal in absolute value.

Using the special form of the polynomial, let us define the functions as follows: \(f(z) = 9\),

\[ h(z) = 4 z^{5}+4 z^{3}-4 z \]

and compare their values at \(|z|=1\). Then \(\left|f(z)\right| = 9\),

\[ \left|h(z)\right| = \left|4 z^{5}+4 z^{3}-4 z\right| = 4\left|z^3\right|\left|z^2 + 1 - z^{-2}\right| = 4\left|z^2 + 1 - z^{-2}\right| \]

Let us parametrize the circle: \(z=e^{i \varphi }, \quad 0 \leq \varphi <2 \pi\). We get

\[ \left|h(z)\right| = 4\left|e^{2 i \varphi }+1-e^{-2 i \varphi }\right| = 4\left|2 i \sin 2 \varphi +1\right| = 4\sqrt{4 \sin ^{2} 2 \varphi +1} \leq 4 \sqrt{5}=\sqrt{80}<9 \]

Thus, the function

\[ g(z)=\frac{1}{z^{2}} \exp \left(\frac{81}{4 z^{5}+4 z^{3}-4 z+9}\right) \]

has only one singular point inside the disk \(|z|<1\), namely \(z=0\), which is a pole of the second order.

Using Cauchy’s residue theorem, we obtain

\[ \begin{align} I & = 2 \pi i \cdot \mathop{\mathrm{res}}_{0}g= 2\pi i \cdot \left.\left(z^2g(z)\right)'\right|_{z = 0} = 2 \pi i \cdot \left.\left[\exp \left(\frac{81}{4 z^{5}+4 z^{3}-4 z+9}\right)\right]^{\prime }\right|_{z=0}= \\ & =\left.2 \pi i\left[\exp \left(\frac{81}{4 z^{5}+4 z^{3}-4 z+9}\right) \cdot \frac{(-81)}{\left(4 z^{5}+4 z^{3}-4 z+9\right)^{2}} \cdot \left(20 z^{4}+12 z^{2}-4\right)\right]\right|_{z=0}= \\ & =2 \pi i e^{9}(-1)(-4)=8 \pi i e^{9} \end{align} \]

Answer

\(8 \pi i e^{9}\).

Problem T.10

Using Rouché’s theorem and residue theory, compute the integral \[ I = \oint _{\left|z\right| = 1} \frac{z^{6} d z}{3 z^{4}+z+1} \]