Page 130 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 12: Fundamental Theorem of Algebra

12.3 Jordan’s Inequality Notes

If 0    /2, the 2  sin   

This inequality is called Jordan inequality. We know that as  increases from 0 to /2, cos
decreases steadily and consequently, the mean ordinate of the graph of y = cos x over the range
0  x   also decreases steadily. But this mean ordinate is given by

1  sin 
 0  cosx dx  

It follows that when 0    /2,
2  sin  1
 

Jordan’s Lemma

If f(z) is analytic except at a finite number of singularities and if f(z)0 uniformly as z, then

lim e imz f(z) dz = 0 , m > 0

R
T
where T denotes the semi-circle |z| = R, I . z  0, R being taken so large that all the singularities
m
of f(z) lie within T.
Proof. Since f(z)0 uniformly as |z|, there exists > 0 such that | f(z)| <   z on T.
Also |z| = R  z = Re i  dz = Re id |dz| = Rd
i
|e | = |e im Re i | = |e imR cos  e -mR sin  |
imz
= e –mR sin

Hence, using Jordan inequality,
| e imz f(z) dz| | e imz f(z)| |dz|


T T
< 0   e  mR sin a  R d

= 2 R 0   /2 e  mR sin  d
2a
  sin
/2

= 2 R 0  e  2mR / d   2


i.e. sin


(1 e  mR )

= 2  R
2mR/
=   (1 e  mR )   

m m

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