Mathematics of Physics and Engineering

232 Singularities of Complex Functions

To proceed, we need the following two equalities:

oo oo

£ ak(z - z 0 )k = Y, °n+N(z - z 0 )n+N, N>1, (4.4.24)

fc=JV n=0

and

(

oo \ / oo \ oo / fc \

]T ak(z - z 0 )k I ]T h{z - z 0 )k I = S 1 S a™^6 *-™ I (z - zo)h-

k=0 J \k=0 J k=0 \m=0 )

(4.4.25)

EXERCISE 4.4.15? (a) Verify (4.4.24). Hint: set n = k-N, so that k = n+N.

(b) Verify (4-4-25). Hint: to get the idea, look at the first few terms by writing

(a 0 + ai(z — zo) + a 2 (z - z$) + .. .)(6o + b\(z - z 0 ) + bi(z — z%) + ...) and then

multiplying through.

If the functions p, q are analytic in a neighborhood of a point ZQ, then

the point z$ is called regular for equation (4.4.23), and it is natural to

expect that all solutions of (4.4.23) are also analytic functions in some

neighborhood of ZQ. Indeed, writing

00 00 00

P(z) = ^2pk(z- z 0 )k, q(z) = ^qk(z-z 0 )k, w(z) = ^wk{z - z 0 )k

fc=0 k=0 k=0

and substituting into (4.4.23), we find:

w"(z) + p(z)w'(z) + q(z)w(z)

00 / 00 \ / 00 \

= J2 Kk - l)wk(z - z 0 )k-^2 + £>*(* ~ z^ E kw^z ~^2 °)fc_1

fc=2 \fc=0 / \fc=l /

+ £>(*-*>)* $>fc(z-z 0 )fc) =^((k + l)(k + 2)wk+2

U=0 / \k=0 J k=0 V

fc+1 k \

+ ^ mwmpk+i-m + ^ wmqk-m J (z - z 0 )k = 0,

m=l m=0 /

(4.4.26)

where the last equality follows from (4.4.24) and (4.4.25).

EXERCISE 4.4.16.B Verify (4.4.26).

By Corollary 4.1 on page 212, we conclude that the coefficients wk are