xx

xxf

0,

0,0)(

22

1

)(01

)(2

2

2

0

2

0

0

0

xx

dxxdx

dxxfa

T

Desarrollaen serie de Fourier:

22

00

0

0

0

)1(11cos

cos1sin

1sin)(

1

cos)(01

cos)(1

nn

n

n

nx

ndxnx

nn

nxx

dxnxxdxdxnxxfa

n

n

nnxdxxbn

1sin)(

10

1

2 sin1

cos)1(1

4)(

n

n

nxn

nxn

xf

La función f es continua en (−, ) excepto en x = 0. Así su

serie de Fourier converge en x = 0 a:

La serie es una extensión periodica de la

función f. Las discontinuidades en x = 0,

2, 4, … convergen a:22

)0()0( ff

Y las discontinuidades en

x = , 3, … convergen a:

220

2)0()0( ff

02

)0()( ff

xxxSxxSS 2sin2

1sincos

2

4,sincos

2

4 ,

4 321

Secuencia de sumas parciales y su representación gráfica

Evaluar donde el contorno C

es el círculo |z|= 2.

C

z

dzzz

e34 5

iz

ezzi

zz

ez

dz

di

zfidzzz

e

z

z

z

z

C

z

12517

)5(

)178(lim

)5(lim

!21

2

)0,)((Res25

3

2

0

33

2

2

0

34

A Question

• In part (b) of Example 2 in Sec.19.3, we showed that the Laurent series of f(z) = 1/z(z – 1) valid for 1 < |z| is

The point z = 0 is an isolated singularity of f and the Laurent series contains an infinite number of terms involving negative inter powers of z. Does it mean that z = 0 is an essential singularity?

...111

)(432

zzzzf

• The answer is “NO”. Since the interested Laurent series is the one with the domain 0 < |z| < 1. From part (a) of that example, we have

Thus z = 0 is a simple pole for 0 < |z| < 1.

...11

)( 2 zzz

zf

C is positively oriented circle | z – 2| = 1.

Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2

Residue of f at the isolated singular point z0

C

dzzz 4)2(

1

16

1

)2(

1

2

1)( Res

41zz 0

C

dzzzi

bzf

)2|2|0()2(2

)1(

32

1

1

)2(2

1

)2(

1 4

0144

zz

zzzzn

nn

n

ALTERNATIVE METHOD

Residue of f at the isolated singular point 2 is the coefficient of 1/(z–2).

16

1)( Res

0zzzf

4

324

4

2344

)2(

)2()2()2()2(

)2(

1

)2|2|0()2()2()2()2()2(

1

zz

zEzzDzzCzBzzA

zz

zz

E

z

D

z

C

z

B

z

A

zz

Solve for A, B, C, D, E by setting coefficients of z, z2, z3, z4 equal to 0.A + E = 0(A(z – 2) – Az)(z – 2)3 = – 2A(z – 2)3

D – 2A = 0(– 2A (z – 2) + 2A z)(z – 2)2 = 4A (z – 2)2

4A + C = 0(– 4A z + 4A (z – 2)) (z – 2) = –8A(z – 2)B – 8A =0 8A z – 8A (z – 2) = 16 A = 1A = 1/16, E = – 1/16

Partial Fraction Expansion Review

BBz

zA

zz

zz

Az

BzA

zz

zz

z

B

z

A

zz

)2(

2

1

)2(

)2(,2When

)2(2

1

)2(,0When

)2()2(

1

75.1

5.025.01)1(1

1)2(

1,1Pick

)2(

)2()2(

2

1

)2(

)2(,2When

)2()2(4

1

)2(,0When

)2()2()2(

1

2

22

2

2

22

22

B

BCBA

zzz

CCz

zB

z

zA

zz

zz

Az

Cz

z

BzA

zz

zz

z

C

z

B

z

A

zz

C is positively oriented circle | z – 2| = 1.Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2

Residue of f at the isolated singular point z0

C

dzzz 4)2(

1

16

1

)2(

1

2

1)( Res

41zz 0

C

dzzzi

bzf

n

n

n

z

zzz

zzz

zzzzzzz

zzzzzzzz

)2(2

1

32

1

...2

)2(

2

)2(

2

21

32

1

2/)2(1

1

32

1

12/)2(

1

32

11

)2(

16/1

)2(

8/1

)2(

4/1

)2(

2/1

2)2(

16/1

)2(

1

)2|2|0()2(

16/1

)2(

8/1

)2(

4/1

)2(

2/116/1

)2(

1

0

3

3

2

2

2344

2344

All terms have positive exponents

b1