# Chapter 7. Argument Principle and Rouche's Theorem – Engineering Mathematics, Volume III

## Argument Principle and Rouche’s Theorem

#### 7.1 Introduction

In this chapter, we state and prove the argument principle and Rouche’s theorem, which are consequences of Cauchy’s residue theorem.

We shall also prove the fundamental theorem of algebra, which we have been using without proof since the beginning of the study of algebra.

The argument principle helps us find the number of zeros or poles of a function in a given region and is useful in determining the stability criteria of linear systems.

The argument principle applies to a class of complex functions called meromorphic functions defined in a domain D whose only singularities in D are poles and so we define meromorphic functions.

#### 7.2 Meromorphic Function

Definition     A function f of the complex variable z in a domain D is said to be meromorphic in D if it is analytic in D except for poles, i.e., the only singularities of f are poles in D.

Examples

1. Rational function of the form where p and q are polynomials such that q is not a constant function. 2. Trigonometric functions: tan z, cot z, sec z and cosec z
3. Hyperbolic functions: tanh z, coth z, sech z and cosech z

#### 7.3 Argument Principle (Repeated Single Pole/Zero)

Theorem 7.1     Let f (z) be meromorphic within and on a simple closed curve C except for a pole z = a of order (multiplicity) p inside C. Suppose that f(z) ≠ 0 on C but has a zero z = b of order n inside C. Then  Figure 7.1

Proof:     If we enclose z = a and z = b in small non-overlapping circles C1 and C2, respectively, then f (z) has a pole of order p at z = a. So, we can write where F (z) is analytic and non-vanishing on and inside C1. Taking logs on both sides of Eq. (7.3) and differentiating w.r.t z, we obtain  Now, is analytic in C1 since F and hence F are analytic in C1, and by Cauchy’s integral theorem Therefore, Eq. (7.5) reduces to Now f(z) has a zero of order n at z = b. So, we can write

f (z) = (zb)n G(z)           (7.7)

where G(z) is analytic and non-vanishing on and inside C. Taking logs on both sides of Eq. (7.7) and differentiating w.r.t z, we obtain By a similar argument as above, we obtain Substituting Eqs. (7.6) and (7.9) in Eq. (7.2), we get #### 7.4 Generalised Argument Theorem

Theorem 7.2     Let f (z) be meromorphic within and on a simple closed curve C except for a finite number of poles a1, a2,…, aj repeated p1, p2,…, pj times respectively inside C, and f(z) ≠ 0 on C, but has a finite number of zeros b1, b2, …, bk repeated n1, n2,…, nk times respectively inside C. Then where are the total number of zeros and poles respectively of f(z) inside C, counted according to their multiplicities. Figure 7.2

Proof:     If we enclose ar (r = 1, 2,…,j) and br (r = 1, 2,…, k) in small non-overlapping circles Cr and γr respectively, and apply the above argument theorem, we obtain Principle of Argument (Angle)

By argument theorem, [Variation of ln f(z) in going completely once around C] (change in the angle f(z) as C traversed completely once) ∵ ln |f| is the same at the beginning and at the end of one full circuit around C.

Corollary     If f(z) is analytic everywhere so that p = 0, then Geometrically, this means that the number of zeros of f(z) is equal to the number of times the locus C* of w = f(z) encircles the origin.

Note:

If w = 0 and C* does not pass through the origin, then N = net variation of arg f(z) is zero.

The following theorem due to Rouche gives a simple procedure to find the number of zeros of an analytic function in a given domain.

#### 7.5 1 Rouche’s Theorem

Theorem 7.3     Let f(z) and g(z) be analytic within and on a simple closed curve C. If |g(z)| < | f(z)| at every point z on C, then f(z) and f(z) + g(z) both have the same number of zeros inside C.

Proof:     First, we observe that f(z) and f(z) + g(z) have no zeros on C.

For, if f(a) = 0 for some point z = a on C then g(a) = 0 on C since g(a) < f(a). This implies that g(a) = f(a) = 0 on C, which contradicts the hypothesis that |g(z)| < | f(a)| for all z on C.

Also, if f(a) + g (a) = 0 for some point z = a on C then |f(a)| = | − g(a)| = |g(a)|, which is again a contradiction.

Therefore, f(z) and f(z) + g(z) have no zeros on C. Thus, the meromorphic function has no zeros or poles on C. By the principle of the argument for F Also, for all z on C. Thus, the point w = f(z) lies inside the circle of unit radius with center at 1, as shown in Fig. 7.3 Figure 7.3

Now, it follows that for z on C.

In general,

arg w = arg F (z) = 2πk              (7.15)

(k : an integer)

If k ≠ 0, then Eq. (4) is violated. Hence, k = 0.

∴ arg w = arg F (z) = 0              (7.16)

Eq. (2) now gives

N = P                             (7.17)

Since the poles of F are the zeros of f and the zeros of F are the zeros of f(z) + g(z), Eq. (7) implies that the number of zeros of f(z) and f(z) + g(z) are equal.

#### 7.6 2Liouville Theorem

Theorem 7.4     If f (z) is entire and |f (z)| is bounded for all z, then f (z) is constant.

Proof:     By Cauchy’s generalised integral formula, where C is a circle with centre at ‘a’ and radius ‘r’.

Since |f (z)| is bounded, |f (z)| ≤ M. Applying ML-inequality to Eq. (7.12), which is known as Cauchy’s inequality.

If we put n = 1 in Eq. (7.13) Since f is entire (analytic throughout C) and | f | is bounded for all z Eq. (7.14) is true for any r. Letting r → ∞ we have | f ′(a) | → 0.

Since ‘a’ is arbitrary f ′(z) → 0 for all z and hence f (z) is constant.

#### 7.7 Fundamental Theorem of Algebra

Theorem 7.5     Every polynomial of degree n in the complex plane has at least one zero.

Proof:     (By Liouville’s theorem) Let p(z) be a polynomial in z of degree n > 1, given by

p(z) = a0 + a1z + a2z2 + ⋯ + anzn                 (7.21)

(where an ≠ 0)

Suppose that p(z) has no zero. Then p(z) ≠ 0 for all z. This implies that is analytic everywhere. Also, |f (z)| → 0 as |z| → ∞ so that |f (z)| is bounded for all z. By Liouville’s theorem, a function which is analytic everywhere and bounded must be constant. This leads to contradiction since p(z) is not constant. Hence, p(z) is zero for at least one value of z.

Alternative form of the theorem: Every polynomial of degree n in the complex plane has n zeros, counting the multiplicities.vs10

Proof:     (By Rouche’s theorem) Let p(z) = a0 + a1z + a2z2 + … + anzn (an ≠ 0) be the polynomial. Choose f (z) = anzn and g(z) = a1 + a1z + … + an−1zn−1. Let C be a circle with centre at the origin and radius R > 1. Then on C replacing each power of R by Rn−1, which is larger for n > 1 For any specified values of the coefficients a0, a1, a2, …, an, we can choose R such that By Rouche’s theorem, the total number of zeros of the polynomial p(z) = f (z) + g (z) in the circle C is the same as the number of zeros of f (z) = anzn in C.

But f (z) = anzn has n zeros (n-fold zero) all located at z = 0. Hence p(z) = f (z) + g(z) has n zeros in C. Since R is arbitrary, this is true for the entire complex plane.

#### 7.8 Maximum Modulus Theorem for Analytic Functions

Theorem 7.6     Let f (z) be a non-constant, analytic and bounded function in a domain D and also on its boundary. Then its absolute value |f (z)| cannot have a maximum at an interior point of D. Consequently, the maximum of |f (z)| is attained on the boundary of D. Similarly, if f (z) ≠ 0 the minimum of |f (z)| is also attained on the boundary of D. Figure 7.4

Proof:     We prove the theorem by the method of contradiction. Assume that |f (z)| has a maximum M at an interior point ‘a’ of D, i.e., f (a) = M and show that it leads to contradiction. Since f (z) is not constant, |f (z)| is also not constant. Describe a circle C of radius r with centre ‘a’ such that the interior of C is in D and |f (z)| < M at some point P of C.

Let C1 be an arc of C containing the point P (Fig. 7.4) Analyticity of f (z) implies continuity of f (z) within and on C. Therefore, |f (z)| < M on C1 containing the point P. Thus if ∊ > 0 then |f (z)| ≤ M − ∊ for all z on C1.

Let the length of C1 be l1 so that the complementary arc C2 of C has the length (2πrl1) since |za| = r. Now, we have by Cauchy’s integral formula which is a contradiction. Hence our assumption is false. This proves the first part of the theorem.

To prove the last part, we observe that f (z) ≠ 0 in D implies that is analytic in D.

By the above result, attains the maximum on the boundary of D so that |f (z)| attains its minimum on the boundary of D. Hence the theorem.

Maximum principle for harmonic functions

Let D be a simply-connected, bounded domain with C as its boundary curve. Then, if ϕ(x, y) is a nonconstant harmonic function in a domain containing domain D and curve C then ϕ(x, y) has neither a maximum nor a minimum in D.

Consequently, the maximum and the minimum of ϕ(x, y) occur only on the boundary C of D.

Example 7.1

Evaluate using the argument principle, where C is a simple closed curve and f(z) = z5 − 8z3i + 2z − 3 + 5i.

Solution     By the fundamental theorem of algebra

f (z) = z5 − 8z3i + 2z − 3 + 5i

being a polynomial and hence analytic everywhere of degree 5 has 5 zeros and has no poles in the complex plane, i.e., N = 5, P = 0

∴ By the argument principle Example 7.2

Evaluate using the argument principle, where C is the simple closed curve |z| = 2 and .

Solution has two zeros of order 1 at z = ±1 and two poles each of order 2 at z = 0 and −1; otherwise, it is analytic inside C : |z| = 2 (circle of radius 2 and centre 0).

∴ By the argument principle Example 7.3

Prove that all the roots of the equation z7 − 5z3 + 12 = 0 lie between the circles |z| = 1 and |z| = 2.

Solution     Let f (z) = 12 and g (z) = z7 − 5z3. Both f (z) and g(z) are analytic within and on C1 : |z| = 1. By Rouche’s theorem, both f (z) and f (z) + g(z) have the same number of zeros inside C1. But f (z) = 12 has no zeros inside C1. Thus, f (z) + g(z) = z7 − 5z3 + 12 has no zeros inside C1.

Next, let f (z) = z7 and g(z) = −5z3 + 12. We observe that f (z) and g(z) are analytic within and on C2 : |z| = 2. By Rouche’s theorem, both f (z) and f (z) + g(z) have the same number of zeros inside C2. But f (z) = z7 has 7 zeros at z = 0 inside C2. Consequently, f (z) + g(z) = z7 − 5z3 + 12 also has 7 zeros inside C2.

Therefore, all the 7 zeros of z7 − 5z3 + 12 lie between the two circles C1 : |z| = 1 and C2 : |z| = 2.

Example 7.4

Evaluate where f (z) = tan πz and C : |z| = π.

Solution     Since , the zeros of tan πz are those of sin πz, which are z = ±n; n = 0, 1, 2, … and the poles of tan πz are the zeros of cos πz, which are z = ; n = 0, 1, 2, ….

The zeros of tan πz lying inside C are z = 0, ±1, ±2, ±3. ⇒ N = 7 and the poles of tan πz lying inside C are . ⇒ P = 6.

Now the radius of C is π = 3.1416 (approximately).

∴ By argument principle, Example 7.5

If the real number a > e prove, by using Rouche’s theorem, that the equation e = azn has n roots inside the unit circle.

[JNTU 2001S]

Solution     Letf (z) = azn and g(z) = −ez so that f (z) + g(z) = aznez = 0 is the given equation. Let C denote the unit circle |z| = 1.

Note that |f (z)| = |a||zn| = |a| = a     ∴ |z| = 1. ∴ By Rouche’s theorem, f (z) + g(z) and f (z) have the same number of zeros inside C. But f (z) = azn has n zeros inside C. So, f (z) + g(z) = aznez has n zeros inside C.

Example 7.6

Applying Rouche’s theorem, find the number of zeros of ez − 4zn + 1 inside the unit circle |z| = 1.

Solution     Let f (z) = −4zn and g(z) = ez + 1 and C be the curve |z| = 1. On C : |z| = 1 so that |f (z)| = 4 and |g(z)| = |ez| + 1 = e + 1 < |f (z)|.

f (z) + g(z) = e2 − 4zn + 1 has the same number of zeros inside the unit circle C : |z| = 1 as f (z), which has n zeros.

Hence ez − 4zn + 1 has n zeros inside C : |z| = 1.

Example 7.7

Use Rouche’s theorem to show that the equation z5 + 15z + 1 = 0 has one root in the disk |z| = and four roots in the annulus < |z| < 2.

[JNTU 2004(2)]

Solution     First, we show that all the five roots lie within the circle C : |z| = 2. Next, we show that one root lies inside the circle C1 : |z| = and the remaining four roots lie inside the annulus A : < |z| < 2.

Let f (z) = z5 and g(z) = 15z + 1, f (z) and g(z) are polynomials and are hence analytic everywhere and, in particular, inside and on C : |z| = 2. By Rouche’s theorem, both f (z) + g (z) = z5 + 15z + 1 and f (z) = z5 have the same number of zeros inside C : |z| = 2. But f (z) = z5 has 5 zeros inside C : |z| = 2 so that z5 + 15z + 1 has all the 5 zeros inside the circle C : |z| = 2. Next consider the circle C1 : |z| = .

Let F(z) = 15z and G(z) = z5 + 1. F and G are analytic everywhere and, in particular, within and on the circle C1 : |z| = . ∴ By Rouche’s theorem, both F(z) + G(z) = 15z + z5 + 1 and F(z) = 15z have the same number of zeros inside the circle |z| = . But f (z) = 15z has only one zero (z = 0) inside |z| = . Therefore, the remaining four zeros must lie inside the annulus A : < |z| < 2.

Example 7.8

Show that the polynomial z5 + z3 + 2z + 3 has just one zero in the first quadrant of the complex plane.

[JNTU 2004 (Sets 1, 4)]

Solution     Let

f (z) = u + iv = z5 + z3 + 2z + 3              (1)

and z = Re where R → ∞ Figure 7.5

Define the first quadrant as OABO and let C be the boundary of this quadrant (Fig. 7.5). f (z) is analytic at all points except at z = ∞. So, it is analytic within and on C.

Along OA : z = x and x varies from 0 to ∞ so that Eq. (1) becomes

 u + iv = x5 + x3 + 2x + 3 arg f = = tan −1 0 = 0 for all x > 0 (∵ v = 0) ∴ OA (arg f) = Var(arg f) = 0 Along BO: z = iy and y varies from ∞ to 0. Equation (1) takes the form f = u + iv = iy5iy3 + 2iy + 3  This shows that the polynomial has just one zero in the first quadrant.

Example 7.9

Show that the equation z4 + 4(1 + i)z +1 = 0 has one root in each quadrant.

[JNTU 2004 (Sets 1, 4)]

Solution     Let f (z) = u+iv = z4 +4(1 + i)z +1 = 0 and z = Re, 0 ≤ θ ≤ where R → ∞ define the first quadrant OABO and C denotes the complete boundary of the quadrant OABO (Fig. 7.6). Figure 7.6

Along OA : z = x and x varies from 0 to ∞ and Eq. (1) becomes  Along BO: z = iy and y varies from ∞ to 0 so that Eq. (1) becomes This implies that the given equation has one root in the first quadrant. The conjugate complex root lies in the fourth quadrant. The equation being fourth degree has 4 roots; the other roots lie in each of the second and third quadrants.

Example 7.10

Prove that one root of the equation z4 + z3 + 1 = 0 lies in the first quadrant.

[JNTU 2004 (Set 3)]

Solution     Let

f (z) = u + iv
[−4ex] = z4 + z3 + 1 = 0                 (1)

We first prove that the equation has no real root.

Putting z = x, we get f (x) = x4 + x3 + 1 which does not become zero for any positive real number x. So, f (z) = 0 has no positive real root.

Let z = −x where x > 0

 Now, f (−x) = x4 − x3 + 1 = x2(x − 1) + 1 > 0 if x > 1 = x4 + (1 − x3) = x4 + (1 − x)(1 + x + x2) > 0 if 0 < x < 1

Thus f (−x) > 0 if x > 1, f (−x) > 0 if 0 < x < 1, which is a contradiction.

Hence f (z) = 0 has no negative real root also.

f (z) = 0 has no real root.

We also prove that the equation has no purely imaginary root.

Put z = iy; f (z) = y4iy3 + 1 = 0 ⇒ y4 + 1 = 0, y3 = 0 (equating the real and imaginary parts zero).

These equations are not compatible and hence f (z) = 0 has no purely imaginary roots.

We now find the number of roots in the first quadrant. Let z = Re R → ∞ define the first quadrant bounded by the straight line segments OA and BO along the x- and y-axes and the part of circular between these lines. f (z) is analytic within and on C, consisting of OA, BO and .

1. Along OA : z = x and x varies from 0 to ∞ 2. Along AB: On this circular arc z = Re, 0 ≤ θ ≤  3. Along BO: z = iy and y varies from ∞ to 0 By the argument principle, the equation has only one complex root in the first quadrant.

##### EXERCISE 7.1
1. Locate the quadrant in which the roots of the equation z4 + z3 + 4z2 + 2z + 3 = 0 are situated.

[JNTU 2003S]

Ans: 2 roots in QII and 2 in QIV

2. Find the number of roots of the equation 2z5 − 6z2 + z + 1 = 0 in the region 1 ≤ |z| < 2.

[Hint: Take f (z) = −6z2;g(z) = 2z5 + z + 1 on C1 : |z| = 1 and f (z) = 2z5; g(z) = −6z2 + z + 1 on C2 : |z| = 2.]

Ans: 3

3. Use Rouche’s theorem to show that the equation z5 + 15z + 1 = 0 has one root in the disk |z| > and four roots in the annulus < |z| < 2.

[Hint: Take f (z) = z5;g(z) = 15z + 1 on C1 : |z| = 2 and f (z) = 15z; g (z) = z5 + 1 on C2 : |z| = 3/2.]

[JNTU 2004, Nov 2008S, Nov 2008 Set 1]

4. Show that the equation z3 + |z|i = 0 has a root in each of the first, second and fourth quadrants.
5. 5. Evaluate where f (z) = z5 − 3iz2 + 2z − 1 + 1 and C is a simple closed curve.

Ans: 10πi

6. Apply Rouche’s theorem to determine the number of zeros of z9 − 2z6 + z2 − 8z − 2 that lie within the circle C : |z| = 1.

[Hint: Take f (z) = −8z,
g(z) = z9 − 2z6 + z2 − 2]

Ans: One zero

7. By Rouche’s theorem determine the number of zeros of z4 − 5z + 1 in the region 1 < |z| < 2.

[Hint: Take f (z) = −5z, g(z) = z4 + 1 on C1 : |z| = 1 and f (z) = z4; g (z) = −5z + 1 on C2 : |z| = 2.]

Ans: Three

8. Evaluate where f (z) = sin πz and L is the circle |z| = π.

Ans: 14πi

9. Use Rouche’s theorem to find the number of zeros of the polynomial z10 − 6z7 + 3z3 + 1 if |z| < 1.

[Hint: Take f (z) = −6z7, g (z) = z10 + 3z3 + 1]

Ans: Seven zeros