INTRODUCTION

Suppose that z = z₀ is an isolated singularity of a function f and that

(1)

is the Laurent series representation of f valid for the punctured open disk 0 < |z − z₀| < R. We saw in the preceding section that a Laurent series (1) consists of two parts. That part of the series in (1) with negative powers of z − z₀, namely,

(2)

is the principal part of the series. In the discussion that follows we will assign different names to the isolated singularity z = z₀ according to the number of terms in the principal part.

Classification of Isolated Singular Points

An isolated singular point z = z₀ of a complex function f is given a classification depending on whether the principal part (2) of its Laurent expansion (1) contains zero, a finite number, or an infinite number of terms.

1. If the principal part is zero, that is, all the coefficients a_−k in (2) are zero, then z = z₀ is called a removable singularity.
2. If the principal part contains a finite number of nonzero terms, then z = z₀ is called a pole. If, in this case, the last nonzero coefficient in (2) is a_−n, n ≥ 1, then we say that z = z₀ is a pole of order n. If z = z₀ is a pole of order 1, then the principal part (2) contains exactly one term with coefficient a₋₁. A pole of order 1 is commonly called a simple pole.
3. If the principal part (2) contains infinitely many nonzero terms, then z = z₀ is called an essential singularity.

The following table summarizes the form of the Laurent series for a function f when z = z₀ is one of the above types of isolated singularities. Of course, R in the table could be ∞.

EXAMPLE 1 Removable Discontinuity

If a function f has a removable singularity at the point z = z₀, then we can always supply an appropriate definition for the value of f(z₀) so that f becomes analytic at the point. For instance, since the right side of (3) is 1 at z = 0, it makes sense to define f(0) = 1. With this definition, the function f(z) = (sin z)/z in Example 1 is now analytic at z = 0.

EXAMPLE 2 Poles and Essential Singularity

(a) From

0 < |z|, we see that a₋₁ ≠ 0, and so z = 0 is a simple pole of the function f(z) = (sin z)/z². The function f(z) = (sin z)/z³ represented by the series in (2) of Section 19.3 has a pole of order 2 at z = 0.

(b) In Example 3 of Section 19.3 we showed that the Laurent expansion of f(z) = 1/(z − 1)²(z − 3) valid for 0 < |z − 1| < 2 was

Since a₋₂ ≠ 0, we conclude that z = 1 is a pole of order 2.

(c) From Example 6 of Section 19.3 we see from the Laurent series that the principal part of the function f(z) = e^3/z contains an infinite number of terms. Thus z = 0 is an essential singularity. ≡

In part (b) of Example 2 in Section 19.3, we showed that the Laurent series representation of f(z) = 1/z(z − 1) valid for 1 < |z| is

The point z = 0 is an isolated singularity of f and the Laurent series contains an infinite number of terms involving negative integer powers of z. Does this mean that z = 0 is an essential singularity of f? The answer is “no.” For this particular function, a reexamination of (1) shows that the Laurent series we are interested in is the one with the annular domain 0 < |z| < 1. From part (a) of that same example we saw that

was valid for 0 < |z| < 1. Thus we see that z = 0 is a simple pole.

Zeros

Recall that z₀ is a zero of a function f if f(z₀) = 0. An analytic function f has a zero of order n at z = z₀ if

(4)

A zero of order n is also referred to as a zero of multiplicity n. For example, for we see that but Thus f has a zero of order (or multiplicity) 3 at A zero of order 1 is called a simple zero.

If an analytic function f has a zero of order n at it follows from (4) that the Taylor series expansion of f centered at must have the form

(5)

where a_n ≠ 0.

EXAMPLE 3 Order of a Zero

The analytic function f(z) = z sin z² has a zero at z = 0. By replacing z by z² in (13) of Section 19.2, we obtain

and so

Comparing the last result with (5) we see that z = 0 is a zero of order 3. ≡

A zero z₀ of a nontrivial analytic function f is isolated in the sense that there exists some neighborhood of z₀ for which f(z) ≠ 0 at every point z in that neighborhood except at z = z₀. As a consequence, if z₀ is a zero of a nontrivial analytic function f , then the function 1/f(z) has an isolated singularity at the point z = z₀. The following result enables us, in some circumstances, to determine the poles of a function by inspection.

EXAMPLE 4 Order of Poles

(a) Inspection of the rational function

shows that the denominator has zeros of order 1 at z = 1 and z = −5, and a zero of order 4 at z = 2. Since the numerator is not zero at these points, it follows from Theorem 19.4.1 that F has simple poles at z = 1 and z = −5, and a pole of order 4 at z = 2.

(b) In Example 3 we saw that z = 0 is a zero of order 3 of f(z) = z sin z². From Theorem 19.4.1, we conclude that the function F(z) = 1/(z sin z²) has a pole of order 3 at z = 0. ≡

From the preceding discussion, it should be intuitively clear that if a function has a pole at z = z₀, then |f(z)| → ∞ as z → z₀ from any direction.

19.4 Exercises Answers to selected odd-numbered problems begin on page ANS-48.

In Problems 1 and 2, show that z = 0 is a removable singularity of the given function. Supply a definition of f(0) so that f is analytic at z = 0.

1. f(z) =
2. f(z) =

In Problems 3–8, determine the zeros and their orders for the given function.

3. f(z) = (z + 2 − i)²
4. f(z) = z⁴ − 16
5. f(z) = z⁴ + z²
6. f(z) = z +
7. f(z) = e^2z − e^z
8. f(z) = sin²z

In Problems 9–12, the indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero.

9. f(z) = z(1 − cos z²); z = 0
10. f(z) = z − sin z; z = 0
11. f(z) = 1 − e^z−1; z = 1
12. f(z) = 1 − πi + z + e^z; z = πi

In Problems 13–24, determine the order of the poles for the given function.

13. f(z) =
14. f(z) = 5 −
15. f(z) =
16. f(z) =
17. f(z) = tan z
18. f(z) =
19. f(z) =
20. f(z) =
21. f(z) =
22. 
23. 
24. 
25. Determine whether z = 0 is an isolated or nonisolated singularity of f(z) = tan (1/z).
26. Show that z = 0 is an essential singularity of f(z) = z³ sin (1/z).