1000 Solved Problems in Modern Physics

1.3 Solutions 73

∮

c

dz

(z−a)n

= 2 πiifn= 1

=0ifn> 1

Method 2

By Cauchy’s integral formula

f(n)(a)=

n!

2 πi

∮

c

f(z)

(z−a)n+^1

dz

Ifn=2 andf(z)= 4 z^2 − 3 z+1, thenf′′(1)=8. Hence

8 =

2!

2 πi

∮

c

4 z^2 − 3 z+ 1

(z−1)^3

or

∫

4 z^2 − 3 z+ 1

(z−1)^3

= 8 πi

1.85 z=3 is a pole of order 2 (double pole);
z=iandz=− 1 + 2 iare poles of order 1 (simple poles).
1.86 z=1 is a simple pole,z=−2 is a pole of order 2 or double pole.

Residue atz=1 is limz→ 1 (z−1)

{

1

(z−1)(z+2)^2

}

=^19

Residue atz=−2 is limz→− (^2) ddz

{

(z+2)^2

(z−1)(z+2)^2

}

=

d

dz^2

{

1

z− 1

}

=

2

(z−1)^2

=

2

9

1.87 The singularity is atz= 2
Letz− 2 =U. Thenz= 2 +U.

ez

(z−1)^2

=

e^2 +U

U^2

=e^2.

eU

U^2

=

e^2

U^2

[

1 +U+

U^2

2!

+

U^3

3!

+···

]

=

e^2

(z−2)^2

+

e^2

z− 2

+

e^2

2!

+

e^2 (z−2)

3!

+

e^2 (z−2)^2

4!

+···

The series converges for all values ofz

= 2

1.88 Consider

∮

c

dz

z^4 + 1 , whereCis the closed contour consisting of line from−R

toRand the semi-circleΓ, traversed in the counter clock-wise direction.

The poles forZ^4 + 1 = 0, arez =exp(πi/4), exp(3πi/4), exp(5πi/4),

exp(7πi/4). Only the poles exp(πi/4) and exp(3πi/4) lie within C. Using

L’Hospital’s rule