6. SERIES   

Convergence of Sequences

A sequence z1, z2, ... , zn, ... has limit z if for all > 0 there exists N : n > N | zn – z | < .

That is, we can make zn arbitrarily close to  z  by taking n sufficiently large.

This definition is formally the same as for the real case. Of course, here the points of the sequence lie in the complex plane, rather than on the real line.

The limit, if it exists, is unique.

We say the sequence converges to z0 and write  zn z  or   zn = z.

If there is no limit, we say that the sequence diverges.

A First Convergence Theorem

Theorem 6.1  If zn = xn + iyn (n = 1, 2, ...)  and  z = x + iy, then

 zn  = z     xn  = x  and  yn  = y.

Proof

() Given > 0 there exists N : n > N    | xn + iyn – (x + iy) | < .
Hence n > N  | xn – x | <  and  | yn – y | < .
i.e.  xn  = x   and   yn  = y.

() Given > 0 there exist  N1, N2 :

n > N1    | xn – x | < /2 ,  n > N2    | yn – y | < /2.

So n > max(N1, N2)    | xn – x | + | yn – y | < .
Now | xn + iyn – (x + iy) | | xn – x | + | yn – y | < .
So  zn  = z.

Infinite Series and Partial Sums

The expression

z1 + z2 + z3 + ...

is an infinite series.

The set

SN = z1 + z2 + ... + zN

is a partial sum of the series.

If the sequence S1, S2, ... , SN, ... converges to limit S we write zn = S, and S is the sum of the series.

The sum, when it exists, is unique.

When a series does not converge, it diverges.

Convergence of Complex Series

Theorem 6.2  Suppose that zn = xn + iyn (n = 1, 2, ...) and S = X + iY. Then

zn = S    xn = X  and yn = Y.

Proof  Let SN  =  XN + iYN denote the Nth partial sum, where XN  = xn and YN = yn.

Now

zn = S     SN = S      XN = X   and  YN = Y

by Theorem 6.1. Since XN, YN are the partial sums of xn and yn, the result follows.

Remainder and Power Series

In establishing that a given series has sum S, we define the remainder after N terms to be:

RN  =  S – SN.

Since | S – SN | = | RN – 0 |,  SN S  iff  RN 0  as  N .

Hence, a series converges to sum S   the sequence of remainders converges to 0.

We shall be particularly concerned with power series.
A power series is a series of the form

a0 + an(z – z0)n  =  an(z – z0)n

where z0 and the an are complex constants, and z is any number (variable) in a stated region. We will use the notation S(z), SN(z), RN(z) for the sum, partial sum and remainder respectively.




Taylor Series

Theorem 6.3 (Taylor)  Let  f  be analytic everywhere inside the circle C0 : | z – z0 | = r0. Then at each point z inside C0

f(z)  =  f(z0) + f '(z0)(z – z0) + f ''(z0)/2! .(z – z0)2 + ... + f (n)(z0)/n! .(z – z0)n + ... .

That is, the power series converges to f(z) when | z – z0 | < r0.

Notes

(1) This is the Taylor series expansion about point z0.

(2) If all terms are real, we get the real Taylor series.

(3) The proof of Taylor's Theorem we give is remarkably ‘natural’, and is one of the rewards in our study of complex functions.

Proof of Taylor's Theorem (I)

Proof   Let z  be any fixed point inside C0 and set | z – z0 | = r  (so  r < r0). Let C1 be a circle centred at z0 and having radius r1 : 0 < r < r1 < r2. Let be any point on this circle; i.e. | – z0 | = r1.

Now z lies inside C1, and f is analytic in and on C1, so by the Cauchy integral formula      where C1 is taken in the positive sense. Now

Also for any complex c 1,    

[Continued]

Proof of Taylor's Theorem (II)

So   

Hence

We next integrate each term anticlockwise around C1, divide by 2i, and substitute the Cauchy integral formula and the Derivative formulae:

, n = 0, 1, 2, ... .

[Continued]

Proof of Taylor's Theorem (III)

So      

where                 (*)

Recall that | z – z0 | = r, | – z0 | = r1 (> r), and | – z | | – z0 | – | z – z0 | = r1 – r.

Thus if M denotes the maximum value of | f() | on C1, (*)

Since r/r1 < 1, RN(z) = 0.

So for each z interior to C0, the Taylor series for f converges to f(z).

Special Case and Observations

Let us seek the Maclaurin expansion for f(z) = ez.
We have f (n)(z) = ez, so f (n)(0)=1. Also, ez is analytic for all z, so

ez = 1 + z + z2/2! + ... + zn/n! + ... = zn/n! ,  |z| < .

Similarly

sin z = z – z3/3! + z5/5! – ... ,   |z| <

cos z = 1 – z2/2! + z4/4! – ... ,   |z| <

sinh z = z + z3/3! + z5/5! + ... ,   |z| <

cosh z = 1 + z2/2! + z4/4! + ... ,   |z| <


Geometric Series

Let us try to find the Maclaurin series for the function

Now

This is not a Maclaurin series: the first two terms are unexpected, and function f has a singularity at z = 0.

Question  Perhaps there are other interesting series to investigate?




Laurent’s Theorem

Let C1, C2 be concentric circles, centre z0, with radii r1, r2 (r1 > r2).

Theorem 6.4 (Laurent) If f is analytic on C1 and C2 and throughout the annulus between these two circles, then at each point in this domain, f(z) is represented by the expansion

      (*)

where

     (**)

each path of integration taken counter-clockwise.

The series (*) is a Laurent series.

Notes on Laurent's Theorem (I)

(1) If f is analytic at all points in and on C1 except at z0, we can take r2 to be arbitrarily small.
Then (*) is valid for 0 < | z – z0 | < r1.

(2) If f is analytic at all points in and on C1, then
f(z)/(z – z0)– n+1 is analytic in and on C2 (since –n + 1 0). So integral (**) is zero, and (*) reduces to the Taylor series.







Notes on Laurent's Theorem (II)

(3) Since f(z)/(z – z0)n+1 and f(z)/(z – z0)– n+1 are analytic throughout the annular region r2 | z – z0 | r1, we can replace and by where C is any closed contour around the annulus in the positive direction.

This means that (*) can be written

  r2 < | z – z0 | < r1

where

  (***)

Notes on Laurent's Theorem (III)

(4) In practice, some, or even many of the coefficients may be zero.

Example Consider f(z) = 1/(z – 1)2 where | z – 1| > 0.

Here z0 = 1, c– 2 = 1, and all the other coefficients are zero.

Using (***), we observe that




Notes on Laurent's Theorem (IV)

(5) For particular examples we usually do not find the coefficients of an expansion using the formula. In other words, the general formula is useful more as an existence formula.

Example Find the Laurent series (z0 = 0) for f(z) = ez/z2.

Using the Maclaurin expansion for ez, we obtain:
                   
Example Find the Laurent series (z0 = 0) for f(z) = e(1/z).

Using the Maclaurin expansion for ez with a change of variable,we obtain:
              


Proof of Laurent's Theorem (I)

If z lies in the annular region, then

  (†)

This is the Cauchy integral formula extended to a multiply-connected domain.

To show this explicitly in this special case, we take a small anti-clockwise directed circle K with centre z, lying within domain. Then

Notice that by the Cauchy integral formula,

Comparing this equation with equation (†), the validity of (†) is confirmed.

Proof of Laurent's Theorem (II)

The first integral of (†) will give the Taylor series part, so as before

For the second integral of (†) we note that

Multiplying through by f(), and expanding the last quotient as a geometric series gives

[Continued]

Proof of Laurent's Theorem (III)
So from (†),         
where an, bn are as given in the statement of the theorem, RN(z) is as before, and RN(z) 0 as N .

Also,            

If r = | z – z0 |, and r2 is the radius of C2, then r2 < r.

Let M  be the maximum of | f()| on C2. Then

    0 as N .

This completes the proof of the theorem.


QUIZ 6.3  


Further Properties of Series

There are many parallels with real series. This follows from the fact that zn is convergent if and only if xn andyn are convergent.

Now if xn andyn are convergent, then xn 0, yn 0. We deduce that
zn  convergent    zn 0.  Of course, the converse is false!!

So the terms of a convergent complex sequence are bounded:

that is, there exists M: | zn | < M for all n.

We say that zn is absolutely convergent (AC) if the series
| zn | = (xn2 + yn2) is convergent. So by the Comparison Test for real series, | xn | and | yn | are convergent. Thus xn, yn are AC and so convergent. It follows that zn is convergent.

Thus zn absolutely convergent zn is convergent.

We now prove an important result for power series. Analogues of the next results hold for an(z – z0)n, but we give this proof for z0 = 0.

Theorem 6.5 If a power series anzn converges when z = z1 ( 0) then it is A.C. for all  z : | z | < | z1 |.

Proof  Since anz1n is convergent, for some M we have | anz1n | < M for all n.
We write | z | / | z1 | = k   ( < 1 ).

Then

| anzn | = | anz1n |.| z / z1 |n < Mkn.

Now the series with terms Mkn (k < 1) is a real, convergent, geometric series.
So by the Comparison Test,

anzn is convergent.

Our previous result shows that the set of all points inside some circle centred at the origin is a region of convergence for anzn. The largest such circle is the circle of convergence.

We note that by the theorem, the series cannot converge at any point z2 outside this circle.

Similarly, if the series bn / zn converges for z = z1, then it is absolutely comvergent at every point z exterior to the circle centre O passing through z1. The exterior of some circle centred at O is therefore a region of convergence.

The theory starts to get a bit solid (boring!) here, so we settle for some stated results. Nevertheless, these results are important.

Let S(z) = anzn over some circle of convergence C1. Thus S is the function defined by the convergent power series. Then

(1) Function S(z) is continuous at each z interior to C1.

(2) Function S(z) is analytic at each z interior to C1.

(3) If C is any contour interior to C1, then the power series can be integrated term by term, i.e.

S(z) dz = an zn dz.

[If C is a closed contour, then of course we get the value 0 on both sides.]

Some Stated Results (II)

(4) The power series can be differentiated term by term.

Thus for each z inside C1,

S'(z) = nanz n –1.

(5) The Taylor/Laurent series about z0 for a given function is unique.

Illustration 

sin(z2) = z2 – z6/3! + z10/5! – ... .

Even though this series is obtained by substituting z2 in the series for sin z, it will be the same as the Maclaurin series for sin (z2).


Some Stated Results (III): Cauchy Product

(6) Let f(z) = an zn,  g(z) = bn zn.

If we formally multiply these series together and collect the coefficients of like powers of z, we get the Cauchy Product of the two series:

f(z).g(z) = a0b0 + (a0b1 + a1b0)z + (a0b2 + a1b1 + a2b0)z2 + ...
+(akbn–k ) zn + ... .

We now have:

The Cauchy Product of two power series converges to the product of their sums at all points interior to their circles of convergence.



   

Laurent’s Theorem: Example I

For the next examples, set

This function is analytic everywhere except at z = 1 and z = 2.

Find the Maclaurin series for f(z) valid in | z | < 1.

Now              

Also | z | < 1  | z/2 | < 1. Hence

valid for | z | < 1.

Thus this is the Maclaurin series for f(z) over the given domain.

Laurent's Theorem: Example II

We are given

Find the Laurent expansion for f(z) in 1 < | z | < 2.

In this case we have | 1/z | < 1 and | z / 2 | < 1.

So

valid for 1 < | z | < 2.

This is the required Laurent expansion.

We observe here that c-1 (or b1) = 1.

Evaluation of Integrals

We noted in the previous calculation that c–1 (or b1) = 1.

Recall that     

Thus

where C is any simple closed contour around the annulus (taken in the positive direction). That is,

f() d = 2i.

This suggests the use of the Laurent series for the evaluation of integrals.



Laurent's Theorem: Example III

Obtain the Laurent expansion in | z | > 2 for

Here | 2/z | < 1 and so | 1/z | < 1.

So

valid for | z | > 2.

Note that the coefficient of z-1 is zero. Hence f() d = 0 for any small contour C about 0 and exterior to circle | z | = 2.

Note It is possible to develop the whole theory of analytic functions beginning with series, but there is no great advantage. The proofs are not easy and motivation is lacking.

Zeros of an Analytic Function (I)

If f is analytic at z0, there exists a circle centre z0 within which f is represented by a Taylor series:

f(z) = a0 + an(z – z0)n   (| z - z0 | < r0)

where a0 = f(z0) and an = f (n)(z0)/n! .

If z0 is a zero of f, then a0 = 0. If in addition

f '(z0) = 0 = f ''(z0) = ... = f (m -1)(z0)

but f (m)(z0) 0, then z0 is a zero of order m. In this case,

f(z) = (z – z0)mam+n(z – z0)n   (am 0, | z – z0 | < r0)

     = (z – z0)m g(z)   say,

where g(z0) = am 0.

Zeros of an Analytic Function (II)

Thus f(z) = (z - z0)m g(z) where g(z0) = am 0.

Since g(z) is represented by a convergent power series, g is continuous at z0. That is, given > 0, there exists a > 0 such that | z - z0 | < | g(z) – g(z0) | < .

Now choose = | am | /2.
Then there exists : | z - z0 | < | g(z) – am | < | am | /2.

It follows that g(z) 0 at any point in neighbourhood | z - z0 | < .

We have proved:

Theorem 6.6 Let f be analytic at point z0 which is a zero of f. Then there exists a neighbourhood of z0 throughout which f has no other zeros, unless f 0. That is, the zeros of an analytic function are isolated.

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