1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

2 6 0 CHAPTER 7 • TAYLOR AND LAURENT SERIES

be at least equal to S. We could then make f differentiable at z 0 by redefining

f (zo) to equal the value of the series at zo, thus contradicting the fact that zo

is a nonremovable singular point. •

00

• EXAMPLE 7.3 Show that (l!zJ' = n~O (n + 1) zn is valid for z E D1 (O).

Solution In Example 4.24 we established this identity with the use of Theorem

4.17. WenowdosoviaT heorem7.4. Uf(z)= (i!.)2,thenastandardinduction

argument (which we leave as an exercise) will show that f (n } (z) = cf:~~l., for

z E D1 (0). Thus, f (n) (O) = (n + 1)!, and Taylor's theorem gives

f(z)= 1 2 = f /(n)/O)z"= f (n+/)!zn= f(n+l)z",

(1 - z) n=O rt. n=O n. n=O

and since f is analytic in Di (0), this series expansion is valid for all z E D i (0).

• EXAMPLE 7.4 Show that, for z E D1 (0),

1 00 1 00

-- 2 = LZ2n and --2 = ~(-l}"z2n.

I - z n=O l + z f:::o

Solution For z E D 1 (0),

1 00

1 - z = Lz".

n=O

(7- 12 }

(7-13)

00

If we let z^2 take the role of z in Equation (7-13), we get that 1 !,2 = L: (z^2 t =

n=O

00

L: z^2 " for z^2 E D1 (0). But z^2 E D1 (0) iff z E D1 (0). Letting - z^2 take the

n=O

role of z in Equation (7-13) gives the second part of Equations (7-12).

R e mark 7.1 Corollary 7.3 clears up what often seems to be a mystery when

series are first introduced in calculus. The calculus analog of Equations (7-12) is

1 00 1 00

--= ~x^2 " and --= ~(-l}"x^2 ",

1-x2 L l+x2 L

n=O n=O

for x E (- 1, 1). (7-14}