Mathematical Methods for Physics and Engineering : A Comprehensive Guide

APPLICATIONS OF COMPLEX VARIABLES

whereφ(z) is analytic and non-zero atz=zjandmjis positive for a zero and

negative for a pole. Then the integrandf′(z)/f(z) takes the form

f′(z)

f(z)

=

mj

z−zj

+

φ′(z)

φ(z)

. (25.16)

Sinceφ(zj)= 0, the second term on the RHS is analytic; thus the integrand has a

simple pole atz=zj, with residuemj. For zerosmj=Njand for polesmj=−Pj,

and thus (25.14) follows from the residue theorem.

(ii) Iff(z) is analytic insideCand not zero at any point on it then

2 π

∑

j

Nj=∆C[argf(z)], (25.17)

where ∆C[x] denotes the variation inxaround the contourC.

Sincefis analytic, there are noPj; further, since

f′(z)

f(z)

=

d

dz

[Lnf(z)], (25.18)

equation (25.14) can be written

2 πi

∑

Nj=

∮

C

f′(z)

f(z)

dz=∆C[Lnf(z)]. (25.19)

However,

∆C[Lnf(z)] = ∆C[ln|f(z)|]+i∆C[argf(z)], (25.20)

and, sinceCis a closed contour, ln|f(z)|must return to its original value; so the

real term on the RHS is zero. Comparison of (25.19) and (25.20) then establishes

(25.17), which is known as theprinciple of the argument.

(iii) Iff(z)andg(z) are analytic within and on a closed contourCand

|g(z)|<|f(z)|onCthenf(z)andf(z)+g(z) have the same number of zeros

insideC;thisisRouch ́e’s theorem.

With the conditions given, neitherf(z)norf(z)+g(z) can have a zero onC.

So, applying theorem (ii) with an obvious notation,

2 π

∑

jNj(f+g)=∆C[arg(f+g)]

=∆C[argf]+∆C[arg(1 +g/f)]

=2π

∑

kNk(f)+∆C[arg(1 +g/f)]. (25.21)

Further, since|g|<|f|onC,1+g/falways lieswithina unit circle centred

onz= 1; thus its argumentalwayslies in the range−π/ 2 <arg(1 +g/f)<π/ 2

and cannot change by any multiple of 2π. It must therefore return to its

original value whenzreturns to its starting point having traversedC.Hence

the second term on the RHS of (25.21) is zero and the theorem is estab-

lished.

The importance of Rouche’s theorem is that for some functions, in particular ́