Page 132 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 12: Fundamental Theorem of Algebra

Hence the result. Notes
Deduction. (i) Replacing m by a and a by 1 in the above example, we get

 cosax   a
0  x  1 dx  2 e
2
Putting a = 1, we get

 cosx   1 
0  x  1  2 e  2e
2
(ii) Taking m = 1, a = 2, we get

 cosx 
0  x  4 dx  4e 2
2
3
Example: Prove that    x sinmx   e  ma / 2 cos    ma   m > 0, a > 0
a
2
4
4
2 
x 
Solution. Consider the integral f(z) dz, where

C
z e
3 imz
f(z) =
4
z  a 4
and C is the closed contour .
z 3
Since 0 as |z| = R, so by
4
z  a 4
Jordan lemma,
3 imz
lim z e dz = 0 (2)
4 
R z  a 4
T
Poles of f(z) are given by
z + a = 0
4
4
or z = –a = e 2ni e a 4
i
4
4
or z = a e (2n+1)i/4 , n = 0, 1, 2, 3.
Out of these four simple poles, only
z = ae i/4 , a e i3/4 lie within C.

If f(z) =  (z) , then Res (z = ) = lim  (z) , a being simple pole.
 (z) z  '(z)

 For the present case,
z e e imz
3 imz
Res (z = ) = lim  lim .
z 4z 3 z 4
Thus, Res (z = ae i/4 ) + Res (z = a e iz/4 )

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