COMPLEX ALGEBA 5 Laurent Series

PHY310 - MATHEMATICAL METHODS FOR PHYSICISTS I

Odd Term 2019 Dr. Anosh Joseph, IISER Mohali

LECTURE 17

September 17, 2019

COMPLEX ALGEBA 5

Topics: Laurent series, Singularities.

Laurent series

In the last lecture we saw that the n-th derivative of an analytic function f(z) at a point z0 can be written as the Cauchy integral I (n) n! f(z) f (z0) = dz n+1 . (1) 2πi C (z − z0)

This formula is telling us that if we know the value of f(z) on contour C then we also know not only the value of the function at any interior point z0 but also the value of all its derivatives.

Taylor’s theorem for complex functions

We can make use of Eqn. (1) to establish Taylor’s theorem for functions of a complex variable. Consider the function f(z), which is analytic inside and on a circle C of radius R centered on the point z = z0, and z is a point inside C. Then we can perform a Taylor expansion of the function about the point z0 ∞ X n f(z) = an(z − z0) , (2) n=0 where the complex coeﬃcients an are given by

1 a = f (n)(z ). (3) n n! 0

The Taylor expansion of f(z) is valid inside the region of analyticity and, for any particular z0, can be shown to be unique. PHY310 - MATHEMATICAL METHODS FOR PHYSICISTS I Odd Term 2019

Let us prove Taylor’s theorem Eq. (2). We note that, since f(z) is analytic inside and on C, we may use Cauchy’s formula to write

1 I f(t) f(z) = dt 2πi C (t − z) 1 I f(t) = dt , (4) 2πi C (t − z0) − (z − z0) where t lies on C. Since z lies inside the circle we have

|t − z0| > |z − z0|, (5) and we can expand the denominator,

1 1 1 = z−z (t − z0) − (z − z0) t − z0 1 − 0 t−z0 ∞ n 1 1 X z − z0 = . (6) (t − z ) t − z t − z 0 0 n=0 0

This series converges absolutely and uniformly for t on the circle and z ﬁxed inside. Substituting the above series in Cauchy’s integral formula for f(z)

1 I f(t) f(z) = dt 2πi C (t − z) I ∞ n 1 f(t) X z − z0 = dt 2πi (t − z ) t − z C 0 n=0 0 ∞ 1 X I f(t) = (z − z )n dt . (7) 2πi 0 (t − z )n+1 n=0 C 0

Applying Cauchy’s integral formula for the derivatives of f(z) we get

∞ " (n) # 1 X f (z0) f(z) = (z − z )n 2πi . (8) 2πi 0 n! n=0

Cancelling the factors of 2πi, we thus get the result

∞ X n f(z) = an(z − z0) , (9) n=0

(n) with an = f (z0)/n!. The Taylor series given above will converge inside a circle having radius equal to the distance from z0 to the nearest singularity and diverge outside such a circle.

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y R

η0

C2 x

Figure 1: The cicle of convergence for the Taylor series of f(z) about a point z = z0. The point η0 is the closest singularity of f to z0 and ρ = |z0 − η0| is the radius of convergence. The Taylor series converges within the circle of convergence, may or may not converge on the circle of convergence and diverges outside the circle of convergence.

As an example of Taylor series let us consider the function

1 f(z) = . (10) 1 − z

This function is analytic except at z = 1. Expanding in Taylor series about the origin,

1 f(z) = = 1 + z + z2 + ··· , (11) 1 − z converges only for |z| < 1. We may also obtain a Taylor series about another point say, z = −1. We have

1 1 f(z) = = 1 − z 2 − (z + 1) 1 1 = z+1 2 1 − 2 " # 1 z + 1 z + 12 = 1 + + . (12) 2 2 2

The above series converges inside a circle of radius 2, centered about z = −1. Note that in the above two cases the singularity at z = 1 lies on the circle of convergence.

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Laurent series

So far we have assumed that f(z) is analytic inside and on the (circular) contour C. Suppose f(z) has a singularity inside C at the point z = z0. Then it cannot be expanded in a Taylor series.

Suppose that f(z) has a pole of order p at z = z0 and it is analytic at every other point inside and on C. Then the function p g(z) = (z − z0) f(z) (13) is analytic at z = z0, and we may be expanded it as a Taylor series about z = z0

∞ X n g(z) = bn(z − z0) . (14) n=0

Thus, for all z inside C, f(z) will have a power series representation of the form

a−p a−1 2 f(z) = p + ··· + + a0 + a1(z − z0) + a2(z − z0) + ··· , (15) (z − z0) (z − z0) with a−p 6= 0.

Such a series, containing negative as well as positive powers of z − z0 is called Laurent series. We can compare the coeﬃcients in Eq. (14) and (15), to get

an = bn+p. (16)

The coeﬃcients bn in the Taylor expansion of g(z) can be obtained using Eq. (8). We get

(n) I g (z0) 1 g(z) bn = = n+1 dz. (17) n! 2πi (z − z0)

Upon using an = bn+p we obtain the coeﬃcients an in Eq. (15) in the following form

1 I g(z) an = dz n+1+p 2πi (z − z0) 1 I f(z) = dz n+1 . (18) 2πi (z − z0)

This expression that is valid for both positive and negative n.

The terms in the Laurent series containing inverse powers of (z − z0), are collectively called the principal part of the series. The terms with n ≥ 0 are collectively called the analytic part of the series, It is possible to have the principal part to contain an inﬁnite number of terms, making the series

∞ X n f(z) = an(z − z0) . (19) n=−∞

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The analytic part will converge inside some circle centered on z0. The principal part will −1 converge only for |(z − z0) | less than some constant, that is, outside some (different) circle centered on z0. If the former circle has the greater radius then the Laurent series will converge in the region R between two circles; otherwise it does not converge at all. It is possible to show that any function f(z) that is analytic in a region R between two such circles C1 and C2 centered on z = z0 can be expressed as a Laurent series about z0 that converges in R. We note that depending on the nature of the point z = z0, the inner circle may be a point (when the principal part contains only a finite number of terms) and the outer circle may have an infinite radius.

We may use the Laurent series of a function f(z) about any point z = z0 to classify the nature of that point.

For the case when f(z) is analytic at z = z0 the series

∞ X n f(z) = an(z − z0) (20) n=−∞ must have vanishing coeﬃcients an with n < 0.

There can be a case where not only are all an zero for n < 0 but a0, a1, ··· , am−1 are all zero as well. In this case, the ﬁrst non-vanishing term in

∞ X n f(z) = an(z − z0) (21) n=−∞

m is am(z − z0) , with m > 0, and f(z) is then said to have a zero of order m at z = z0.

If fz) is not analytic at z = z0 we can have two cases for the number of negative power terms in the series. Taking integer p as positive

(i.) we can ﬁnd an integer p such that a−p 6= 0 but a−p−k = 0 for all integers k > 0;

(ii) we cannot ﬁnd such a lowest value of −p.

In case (i), f(z) is of the form (given above) Eq. (15)

a−p a−1 2 f(z) = p + ··· + + a0 + a1(z − z0) + a2(z − z0) + ··· , (22) (z − z0) (z − z0) and is said to have a a pole of order p at z = z0. In addition, the value of a−1 (not a−p) is called the residue of f(z) at the pole z = z0.

For case (ii), in which the negatively decreasing powers of z − z0 do not terminate, f(z) is said to have an essential singularity. Let us consider the following example to illustrate the Laurent series expansion. Consider the function 1 f(z) = . (23) z(z − 1)

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y R

Figure 2: The region of convergence (shaded area) for a Laurent series of f(z) about a point z = z0 where f(z) has a singularity.

This function is analytic at all points of the z plane except at the points z = 0 and z = 1, Therefor, it is possible to expand this function in a Laurent series in an annular region about z = 0 or z = 1. To expand about z = 0, we have

1 1 = − (1 − z)−1 z(z − 1) z 1 = − (1 + z + z2 + z3 + ··· ) z 1 = − − 1 − z − z2 − z3 − · · · , 0 < |z| < 1. (24) z

To expand about z = 1, write

1 1 1 = z(z − 1) z − 1 ((z − 1) + 1 1 = 1 − (z − 1) + (z − 1)2 − (z − 1)3 + ··· z1 1 = − 1 + (z − 1) − (z − 1)2 + ··· , 0 < |z − 1| < 1. (25) z − 1

Let us consider another function f(z) = e1/z. (26)

This function may be expanded in any annular region enclosing the origin in the form

1 1 1 1 1 e1/z = 1 + + + + ··· , z 6= 0. (27) z 2! z2 3! z3

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As another example let us consider the function

x 1 f(z) = exp z − . (28) 2 z

It is analytic except at z − 0. So it has a Laurent expansion about z = 0

x 1 f(z) = exp z − 2 z ∞ X n = anz , (29) n=−∞ where 1 I f(t) an = dt n+1 . (30) 2πi C t Let us choose C to be a circle of unit radius. Thus t = eiθ and

Z 2π 1 ix sin θ −inθ an = dθ ie e (31) 2πi 0

Noting that Z 2π sin(nθ − x sin θ)dθ = 0, (32) 0 we get

1 Z 2π an = dθ cos(nθ − x sin θ). (33) 2π 0

The function ∞ x 1 X f(z) = exp z − = J (x)zn (34) 2 z n −∞ is the generating function for the Bessel functions of integer order Jn(x). Thus, using Laurent expansion, we have derived the integral representation of the Bessel function

1 Z 2π Jn(x) = dθ cos(nθ − x sin θ). (35) 2π 0

Singularities of complex functions

A singular point or singularity of a complex function f(z) is any point in the Argand diagram at which f(z) does not have a unique derivative. They are the points at which the function fails to be analytic. For example, the function 1 f(z) = (36) (z − 1)

7 / 9 PHY310 - MATHEMATICAL METHODS FOR PHYSICISTS I Odd Term 2019 has a singularity at z − 1 = 0 or at z = 1. In the last lecture we encountered one sort of singularity - the branch point. Let us look for other types of singularities of complex functions. For the case of single-valued functions f(z), we can have two types fo singularities. They are poles or non-essential singularities and essential singularities.

Isolated singularity

If f(z) has a singular point at z = z0 but is analytic at all points in some neighborhood containing z0 but no other singularities, then z = z0 is called an isolated singularity. The opposite case is a non-isolated singularity. The function 1 f(z) = (37) (z − 1)(z − 2) has two isolated singularities, one at z = 1 and the other at z = 2. The function π −1 f(z) = sin (38) z is not analytic at the points where sin π/z = 0. They correspond to the points π/z = nπ. That is, z = 1/n with n = 1, 2, ··· . Thus 1 1 z = 1, , , ··· , 0, (39) 2 3 are singular points of the function. The point z = 0 is a non-isolated singularity of the function because in the neighborhood of z = 0, there are inﬁnite number of other singularities. Note that branch points are not isolated singularities.

Pole of order n

Poles (non-essential singularities) are also isolated singularities. If f(z) has the form g(z) f(z) = n , (40) (z − z0) where n is a positive integer, g(z) is analytic at all points in some neighborhood containing z = z0 and g(z0) 6= 0, then f(z) has a pole of order (or multiplicity) n at z = z0. When n = 1 the pole is said to be a simple pole. The function 1 f(z) = (41) (z − 1)(z − 3)2(z − 5)3 has a simple pole at z = 1, second order pole at z = 3 and a third order pole at z = 5. An alternative (an equivalent) deﬁnition is that

n lim [(z − z0) f(z)] = a, (42) z→z0

8 / 9 PHY310 - MATHEMATICAL METHODS FOR PHYSICISTS I Odd Term 2019 where a is a ﬁnite, non-zero complex number.

If the above limit is equal to zero, then z = z0 is a pole of order less than n, or f(z) is analytic there. If the limit is inﬁnite, then the pole is of an order greater than n.

It is also possible to show that if f(z) has a pole at z = z0, then |f(z)| → ∞ as z → z0 from any direction in the Argand diagram.

Removable singularity

If there are points at which the value of f(z) takes an indeterminate form such as 0/0 but limz→z0 f(z) exists, and is independent of the direction from which z0 is approached, then such points are called removable singularities. An example of a removable singularity is the singularity of the function

sin z sinc(z) = (43) z at z = 0. We can remove this singularity by deﬁning sinc(0) = 1.

Essential singularity

If no ﬁnite value of n can be found such that Eq. (42) is satisﬁed, then z = z0 is called an essential singularity. An essential singularity is neither a removable singularity nor a pole.

In other words the Laurent series of f(z) at the point z0 has infinitely many negative degree terms. That is, the principal part of the Laurent series is an infinite sum. The function f(z) = e1/z (44) has an essential singularity at z = 0. We can see that from the Laurent series expansion of the function: the series contains an infinite number of negative powers of z.

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