5.5 Poles and Zeros

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We have been looking at functions which are analytic at all points except

some set of 'bad' or singular points. In the case where every point in some

neighbourhood of a singular point is analytic, we say that we have an isolated

singularity. Almost all examples have such singularities poles of the function,

places where the modulus goes through the roof no matter how high the roof

is: put more formally, points w such that

lim

z!w jf(z)j = 1

We can distinguish di_erent types of singularity: there are those that look

like 1=z, those that look like 1=z2, those that look like Log (at the origin)

and there are those that are just not de_ned at some point w but could have

been if we wanted to. The last are called removable singularities because we

can remove them. For example, if I give you the real function f(x) = x2=x,

you might in a careless mood just cancel the x and assume that it is the

same as the function f(x) = x. This is so easily done, you can do it by

accident, but strictly speaking, you have a new function. It so happens

that it agrees with the old function everywhere except at zero, where the

original function is not de_ned. Moreover, the new function is di_erentiable

everywhere, while the old function has a singularity at the origin. But not

the sort of singularity which should worry a reasonable man, the gap can be

plugged in only one way that will make the resulting function smooth and

indeed in_nitely di_erentiable. And if I'd made the x a z and said it was

144 CHAPTER 5. TAYLOR AND LAURENT SERIES

a complex function, exactly the same applies. Of course, it may take a bit

more e_ort to see that a singularity of a function is removable. But if there is

a new function which is analytic at w and which agrees with the old function

in a neighbourhood of w, the w is a removable singularity.

To be more formal in our de_nitions, we can say that if w is an isolated

singularity of a function f, then f has a Laurent expansion about w, and the

cases are as follows:

f(z) =

1 X

􀀀1

ak(z 􀀀 w)k

1. If ak = 0 for all negative k, then f has a removable singularity.

2. If ak = 0 for all negative k less than negative n, and a

􀀀n 6= 0 then we

say that w is a pole of order n. Thus 1=z has a pole of order 1. (Its

Laurent expansion has every other coe_cient zero!)

3. If there are in_nitely many negative k non-zero, then we say that w is

a pole of in_nite order. The singularity is said to be essential.

Exercise 5.5.1 Give examples of all types of poles.

Exercise 5.5.2 Verify that

sin z

z

has a removable singularity at 0, and remove it (i.e. de_ne a value1 for the

function at 0).

We can do the same kind of thing with zeros as we have done with poles. If

w is a point such that f(z) is analytic at w then if f(w) = 0 we say that w is

a zero of f. It is an isolated zero if there is a neighbourhood of w such that

f(z) is non zero throughout the neighbourhood except at w. An isolated

zero of f, w is said to be of order n if the Taylor series for f centred on w

f(z) =

1 X

0

ak(z 􀀀w)k

has ak = 0 for every k less than n, and an 6= 0.

1Never forget that cancelling sin6z

6z is a sin.

5.5. POLES AND ZEROS 145

Example 5.5.1 The function z2 cos z has a zero of order 2 at the origin.

We have the following easy theorem:

Theorem 5.5 If f is analytic in a neighbourhood of w and has a zero of

order n at w, then there is a function g which is analytic in the neighbourhood

of w, is non-zero at w and has

f(z) = (z 􀀀w)ng(z)

Proof:

Write out the Taylor expansion about w for f and divide by (z􀀀w)n to get a

Laurent expansion for some function g. This will in fact be a Taylor series,

the constant term of which is non-zero, and it must converge everywhere the

original Taylor series converged. 2

Exercise 5.5.3 Show the converse: if f can be expressed as

f(z) = (z 􀀀w)ng(z)

for some analytic function g which is non-zero at w, then f has a zero at w

of order n.

Corollary 5.5.1 If f and g are analytic and have zeros at w of n and m,

then the product function has a zero at w of order n + m. 2

Very similar to the above theorem is the corresponding result for poles. I

leave it as an exercise:

Exercise 5.5.4 Prove that if a function f has an isolated pole of order n at

w then there is a neighbourhood W of w and a function g analytic on W and

having g(w) 6= 0 such that

f(z) =

g(z)

(z 􀀀w)n

I de_ned a meromorphic function earlier as one that had isolated singularities.

I really ought to have said isolated poles, and moreover, isolated poles of _nite

order.

146 CHAPTER 5. TAYLOR AND LAURENT SERIES

Exercise 5.5.5 What is the di_erence?

Because the poles and zeros of a meromorphic function tell us a lot about

the function, it is important to be able to say something about them. (The

correspondence pages of one of the major Electrical Engineering Journals

used to be called 'Poles and Zeros'.) In Control Theory for example, knowing

the locations of poles and zeros is critical in coming to conclusions about the

stability of the system.

Example 5.5.2 Locate the poles and zeros of the function

tan z

z2

Solution The poles will be the zeros of cos z together with the origin; writing

sin z = z 􀀀

z3

3!

+

z5

5! 􀀀

z7

7!

+ _ _ _

and

cos z = 1􀀀

z2

2!

+

z4

4! 􀀀_ _ _

we get the ratio of power series:

z 􀀀

z3

3! + z5

5! 􀀀_ _ _

z2 􀀀1􀀀

z2

2! + z4

4! 􀀀_ _ _

=

1􀀀

z2

3! + z4

5! 􀀀_ _ _

z 􀀀

z3

2! + z5

4! 􀀀_ _ _

Doing the (very) long division:

z 􀀀

z3

2!

+

z5

4! 􀀀_ _ _ _1􀀀

z2

3! + z4

5! 􀀀_ _ _􀀀

1

z

1􀀀

z2

2! + z4

4! 􀀀_ _ _

_0 + z2

3 􀀀

4z4

5! + _ _ _􀀀

z

3

z2

3 􀀀

z4

3! + _ _ _

5.5. POLES AND ZEROS 147

If you can't do long division, check the result by cross multiplication and take

it on faith, or _nd out how to do long division.

This gives the _rst few terms of the Laurent Series:

1

z

+

z

3

+

16z3

5!

+ _ _ _

which tells us that there is a pole of order 1 at the origin, which should not

come as a surprise to the even moderately alive.

Away from the origin, we have zeros at the locations of the zeros of sin z,

namely at n_ where n is an integer, and poles at the zeros of cos z, i.e. at

n_=2 for n an integer. Each of these poles and zeros will have order one.

This follows by observing that if you di_erentiate sin you get cos and when

sinz = 0; cos z = _1 and vice versa.

You have probably already realised:

Theorem 5.6 If f is analytic and has an isolated zero of order n at w, then

1=f is meromorphic in a neighbourhood of w and has a pole of order n at w.

2

Exercise 5.5.6 Work out the possible poles and zeros and orders thereof,

for the ratio of two meromorphic functions with known poles and zeros and

orders thereof.

Exercise 5.5.7 (Riemann's Singularity Theorem)

f is a function known to be analytic in a punctured disk D centred on w that

has w removed, with a singularity at w, and jfj is bounded on the punctured

disk. Show that the singularity is removable.

[Hint: investigate g(z) = (z 􀀀w)2f(z) ]

In conclusion, the trick of writing out Laurent Series for functions is a smart

way of learning a lot about their local behaviour, and there are scads of

results you can do which we don't have time in this course to look at. Which

is a little sad, and I hope the other material in your engineering courses is as

interesting to explore as this stu_ is.

148 CHAPTER 5. TAYLOR AND LAURENT SERIES

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