In complex analysis, we may come across various formulas, theorems, such as Rouche’s theorem, and functions such as harmonic function, entire function, analytic function, etc. Analytic functions play an important role in defining many related functions and theorems. Let’s recall what an analytic function is, along with some terms related to Rouche’s theorem such that we can quickly understand the use of Rouche’s theorem.

Analytic function

Consider a complex function f(z), it is said to be an analytic function at a certain point z0 if f(z) is differentiable not only at z0 but also at each point in some neighbourhood of z0. It is also known as a regular, monogenic or holomorphic function. Besides, a value of z is called the zero of an analytic function f(z) if f(z) = 0.

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Meromorphic function

A complex function f(z) is meromorphic on a domain D if and only if f(z) is an analytic function in D except at finitely many poles.

Argument principle

In complex analysis, the argument principle describes the difference between the number of zeros and poles of a meromorphic function to a contour integral of the logarithmic derivative of that particular function. This is also called Cauchy’s argument principle.

What is Rouche’s Theorem in Complex Analysis?

Rouche’s Theorem Statement: If f(z) and g(z) are two analytic functions within and on a simple closed curve C such that |f(z)| > |g(z)| at each point on C, then both f(z) and f(z) + g(z) have the same number of zeros inside C.

Rouche’s Theorem Proof

Given:

Two analytic functions f(z) and g(z) inside and on a closed contour C.

Also, |g(z)| < |f(z)| or |f(z)| > |g(z) at each point on C.

To prove:

f(z) and f(z) + g(z) have the same number of zeros inside C.

Proof:

First, let us prove neither f(z) nor f(z) + g(z) has a zero on C.

Let us assume that f(z) has a zero at a, i.e., z = a.

So, f(a) = 0

Thus, |g(z)| < |f(z)| can be written as:

|g(a)| < |f(a)|

⇒ g(a) = 0 {since f(a) = 0}

⇒ |f(a)| = |g(a)|

That means |f(z)| = |g(z)| at z = a.

This is the contradiction to the initial statement that |g(z)| < |f(z)| on C.

Now, let f(z) + g(z) has a zero at z = b on C.

So, f(b) | g(b) = 0

⇒ f(b) = -g(b)

⇒ |f(b)| = |g(b)|

Or

|f(z)| = |g(z)| at z = b.

Therefore, neither f(z) nor f(z) + g(z) has a zero on C.

Let N1 and N2 be the number of zeros of f(z) and f(z) + g(z), respectively inside C.

Now, we need to prove that N1 = N2.

As we know, f(z) and f(z) + g(z) are analytic functions within and on C.

Also, they have no pole inside C.

Thus, from the argument principle, we can write as:

\(\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z)}{f(z)}dz=N_{1}-P\end{array} \)

When P = 0,

\(\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z)}{f(z)}dz=N_{1}\end{array} \)

Similarly,

\(\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z) + g'(z)}{f(z) + g(z)}dz=N_{2}\end{array} \)

The difference of the above two equations is:

\(\begin{array}{l}\frac{1}{2\pi i}\int_{c}\left [ \frac{f'(z)+g'(z)}{f(z)+g(z)}-\frac{f'(z)}{g'(z)} \right ]dz=N_{2}-N_{1}…(1)\end{array} \)

Let Φ(z) = g(z)/f(z)

From this we can write g(z) as:

g(z) = f(z) Φ(z), i.e., g = fΦ

Consider |g(z)| < |f(z)|

That means, |g(z)/f(z)| < 1

|Φ(z)| < 1

Also,

[f’(z) + g’(z)]/ [f(z) + g(z)] = [f’(z) + f’(z) Φ(z) + f(z) Φ’(z)]/ [f(z) + f(z) Φ(z)] {since g(z) = f(z) Φ(z)}

= [(1 + Φ(z))f’(z) + f(z) Φ’(z)]/ [f(z) (1 + Φ(z))]

= [f’(z)/f(z)] + [Φ’(z)/ (1 + Φ(z))]……………(2)

From (1) and (2), we have;

\(\begin{array}{l}N_{2}-N_{1}=\frac{1}{2\pi i}\int_{c}\frac{\Phi'(z)}{1+\Phi(z)} dz=\frac{1}{2\pi i}\int_{c}\Phi'(z)(1+\Phi(z))^{-1} dz…(3)\end{array} \)

As we already got |Φ(z)| < 1 that means the binomial expansion of (1 + Φ(z))-1 is possible.

However, the powers of Φ(z) in the binomial expansion of (1 + Φ(z))-1 will be uniformly convergent.

Therefore,

∫c Φ’(z) (1 + Φ(z))-1 dz = ∫c Φ’(z) [1 – Φ(z) + Φ2(z) – Φ3(z) + ……] dz………….(4)

From the given,

f(z) and g(z) are analytic functions within and on C such that f(z) ≠ 0 and g(z) ≠ 0 for any point on C (which is proved above).

Therefore, Φ(z) is an analytic function since Φ(z) = g(z)/f(z), and is non-zero for any point on C.

Consequently, we can say that Φ(z) and all of its derivatives are analytic functions.

Thus, by Cauchy’s theorem, the RHS of equation (4) vanishes.

That means, ∫c Φ’(z) (1 + Φ(z))-1 dz = 0.

Then from equation (3) and the above equation, we can say that N2 – N1 = 0

Therefore, N1 = N2

i.e., f(z) and f(z) + g(z) have the same number of zeros inside C.

Hence proved.

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Applications of Rouche’s Theorem

Let’s have a look at some important applications of Rouche’s theorem.

  • We can use Rouche’s theorem to simplify an analytic function for finding the zeros.
  • Rouche’s theorem helps us to prove a short type proof for the fundamental theorem of algebra.
  • It also helps in proving the open mapping theorem for analytic functions in complex analysis.

Rouche’s Theorem Solved Example

Question:

Find the number of zeros of p(z) = z6 + 9z4 + z3 + 2z + 4 in the unit disk.

Solution:

Given,

p(z) = z6 + 9z4 + z3 + 2z + 4

Let U = ∆ that means the boundary of U is the unit circle.

Let us write p(z) as the sum of two functions such that one of them is greater than the other on the unit circle.

Also, on the unit circle, |zk| = 1, regardless of k.

Consequently, the power of z is invisible.

Now, f(z) = 9z4 (greatest term)

Therefore, g(z) = z6 + z3 + 2z + 4.

Let us find |f(z)| and |g(z)|.

|f(z)| = |9z4| = 9(1) = 9

|g(z)| = |z6 + z3 + 2z + 4|

= |z6| + |z3| + |2z| + |4|

= 1 + 1 + 2 + 4

= 8 ≤ 9 = |f(z)|

Thus , we can say that p(z) = f(z) + h(z) has the same number of zeroes as f(z) on the unit disk.

However, f(z) = 9z4 has four zeroes.

Therefore, p(z) has four zeroes on the unit disk.

Problems on Rouche’s Theorem

  1. Consider f(z) = 1 + 2z + 7z2 + 3z5, show that f has exactly two roots inside the unit disc.
  2. Show all 5 zeros of z5 + 3z + 1 are inside the curve C2 : |z| = 2.
  3. Find the number of zeros of p(z) = 2z4 – 2z3 + 2z2 = 2z + 9 in |z| < 1.

Frequently Asked Questions on Rouche’s Theorem

Q1

State Rouche’s theorem in complex analysis.

Suppose f(z) and g(z) are two analytic functions within and on a closed contour C such that |f(z)| > |g(z)| at each point on C, then both f(z) and f(z) + g(z) have the same number of zeros inside C.

Q2

Is it possible to find the number of zeros for an analytic function using Rouche’s theorem?

Yes, it is possible to find the number of zeros for an analytic function using Rouche’s theorem.

Q3

What do you mean by argument principle?

The argument principle shows a special relationship between the number of zeros and the number of poles of an analytic function f(z) in a domain D and how the function f(z) maps the boundary of the derivative of the domain.

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