1. f ( z ) = u( x , y ) + iv ( x , y ) for z = x + iy
f ( z + ∆z ) − f ( z )
f ' ( z ) = lim [ ] exists
∆z →0 ∆z
Its value does not depend on the direction.
Ex : Show that the function f ( z ) = x 2 − y 2 + i 2 xy is
differentiable for all values of z .
for ∆z = ∆x + i∆y
f ( z + ∆z ) − f ( z )
f ' ( z ) = lim
∆z →0 ∆z
( x + ∆x ) 2 − ( y + ∆y ) 2 + 2i ( x + ∆x )( y + ∆y ) − x 2 + y 2 − 2ixy
=
∆x + i∆y
( ∆x ) 2 − ( ∆y )2 + 2i∆x∆y
= 2x + i2 y +
∆x + i∆y
(1) choose ∆y = 0, ∆x → 0 ⇒ f ' ( z ) = 2 x + i 2 y
(2) choose ∆x = 0, ∆y → 0 ⇒ f ' ( z ) = 2 x + i 2 y

2. * * Another method :
f ( z ) = ( x + iy ) 2 = z 2
' ( z + ∆z )2 − z 2 ( ∆z )2 + 2 z∆z
f ( z ) = lim [ ] = lim [ ]
∆z →0 ∆z ∆z → 0 ∆z
= lim ∆z + 2 z = 2 z
∆z → 0
Ex : Show that the function f ( z ) = 2 y + ix is not
differentiable anywhere in the complex plane.
f ( z + ∆z ) − f ( z ) 2 y + 2∆y + ix + i∆x − 2 y − ix 2∆y + i∆x
= =
∆z ∆x + i∆y ∆x + i∆y
if ∆z → 0 along a line thriugh z of slope m ⇒ ∆y = m ∆x
f ( z + ∆z ) − f ( z ) 2 ∆ y + i∆ x 2m + i
f ' ( z ) = lim = lim [ ]=
∆z →0 ∆z ∆x ,∆y →0 ∆x + i∆y 1 + im
The limit depends on m (the direction), so f ( z )
is nowhere differentiable.

3. Ex : Show that the function f ( z ) = 1 /(1 − z ) is analytic everywhere
except at z = 1.
f ( z + ∆z ) − f ( z ) 1 1 1
f ' ( z ) = lim [ ] = lim [ ( − )]
∆z →0 ∆z ∆z →0 ∆z 1 − z − ∆z 1 − z
1 1
= lim [ ]=
∆z →0 (1 − z − ∆z )(1 − z ) (1 − z ) 2
Provided z ≠ 1, f ( z ) is analytic everywhere such that
f ' ( z ) is independent of the direction.

4. 20.2 Cauchy-Riemann relation
A function f(z)=u(x,y)+iv(x,y) is differentiable and analytic,
there must be particular connection between u(x,y) and v(x,y)
f ( z + ∆z ) − f ( z )
L = lim [ ]
∆z →0 ∆z
f ( z ) = u( x , y ) + iv ( x , y ) ∆z = ∆x + i∆y
f ( z + ∆z ) = u( x + ∆x , y + ∆y ) + iv ( x + ∆x , y + ∆y )
u( x + ∆x , y + ∆y ) + iv ( x + ∆x , y + ∆y ) − u( x , y ) − iv ( x , y )
⇒ L = lim [ ]
∆x ,∆y →0 ∆x + i∆y
(1) if suppose ∆z is real ⇒ ∆y = 0
u( x + ∆x , y ) − u( x , y ) v ( x + ∆x , y ) − v ( x , y ) ∂u ∂v
⇒ L = lim [ +i ]= +i
∆x →0 ∆x ∆x ∂x ∂x
(2) if suppose ∆z is imaginary ⇒ ∆x = 0
u( x , y + ∆y ) − u( x , y ) v ( x , y + ∆y ) − v ( x , y ) ∂u ∂v
⇒ L = lim [ +i ] = −i +
∆y →0 i ∆y i ∆y ∂y ∂y
∂u ∂v ∂v ∂u
= and =- Cauchy - Riemann relations
∂x ∂y ∂x ∂y

5. Ex : In which domain of the complex plane is
f ( z ) =| x | − i | y | an analytic function?
u( x , y ) =| x |, v ( x , y ) = − | y |
∂u ∂v ∂ ∂
(1) = ⇒ | x |= [− | y |] ⇒ (a) x > 0, y < 0 the fouth quatrant
∂x ∂y ∂x ∂y
(b) x < 0, y > 0 the second quatrant
∂v ∂u ∂ ∂
(2) =− ⇒ [− | y |] = − | x |
∂x ∂y ∂x ∂y
z = x + iy and complex conjugate of z is z * = x − iy
⇒ x = ( z + z * ) / 2 and y = ( z − z * ) / 2i
∂f ∂f ∂x ∂f ∂y 1 ∂u ∂v i ∂v ∂u
⇒ = + = ( − )+ ( + )
∂z * ∂x ∂z * ∂y ∂z * 2 ∂x ∂y 2 ∂x ∂y
If f ( z ) is analytic , then the Cauchy - Riemann relations
are satisfied. ⇒ ∂f / ∂z * = 0 implies an analytic fonction of z contains
the combination of x + iy , not x − iy

6. If Cauchy - Riemann relations are satisfied
∂ ∂u ∂ ∂v ∂ ∂v ∂ ∂u ∂ 2u ∂ 2u
(1) ( ) = ( )= ( ) = − ( )⇒ 2 + 2 2 = 0
∂x ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x ∂ y
∂ 2v ∂ 2v
(2) the same result for function v ( x , y ) ⇒ 2
+ 2 2
=0
∂x ∂ y
⇒ u( x , y ) and v ( x , y ) are solutions of Laplace' s
equation in two dimension.
For two families of curves u( x , y ) = conctant and v ( x , y ) = constant,
the normal vectors corresponding the two curves, respectively, are
 ∂u ˆ ∂u ˆ  ∂v ˆ ∂v ˆ
∇u( x , y ) = i+ j and ∇v ( x , y ) = i+ j
∂x ∂y ∂x ∂y
  ∂u ∂v ∂u ∂v ∂u ∂u ∂u ∂u
∇u ⋅ ∇v = + =− + = 0 orthogonal
∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x

7. 20.3 Power series in a complex variable
∞ ∞
f (z) = ∑ an z n
= ∑ an r n exp(inθ )
n= 0 n=0
∞
if ∑ | an | r n is convergent ⇒ f ( z ) is absolutely convergent
n=0
∞
Is ∑ | a n | r n convergent or not, can be justisfied by" Cauchy root test".
n= 0
1
The radius of convergence R ⇒ = lim | a n |1 / n ⇒ (1) | z |< R absolutely convergent
R n→ ∞
(2) | z |> R divergent
(3) | z |= R undetermined
∞
zn 1
(1) ∑ ⇒ lim ( )1 / n = 0 ⇒ R = ∞ converges for all z
n= 0 n! n→∞ n!
∞
(2) ∑ n! z n ⇒ lim ( n! )1 / n = ∞ ⇒ R = 0 converges only at z = 0
n= 0 n →∞

8. 20.4 Some elementary functions
∞
zn
Define exp z = ∑
n = 0 n!
Ex : Show that exp z1 exp z 2 = exp( z1 + z 2 )
∞
( z1 + z 2 ) n
exp( z1 + z 2 ) = ∑
n=0 n!
∞
1 n n
= ∑ n!(C 0 z1 + C1n z1n−1 z2 + C 2 z1n−2 z2 + C rn z1n−r z2 + ... + C n z2 )
n 2 r n n
n= 0
n
s r Cr 1 n! 1
set n = r + s ⇒ the coeff. of z1 z 2 is = =
n! n! ( n − r )! r ! s! r !
∞
zs ∞ zr ∞ ∞
1 s r
exp z1 exp z 2 = ∑ ∑ = ∑∑ z1 z 2
s =0 s! r = 0 r ! s =0 r = 0 s! r !
s r
There are the same coeff. of z1 z 2 for the above two terms.

9. Define the complex comonent of a real number a > 0
∞
z n (ln a ) n
a = exp( z ln a ) = ∑
z
n=0 n!
(1) if a = e ⇒ e z = exp( z ln e ) = exp z the same as real number
iy y 2 iy 3
(2) if a = e, z = iy ⇒ e = exp(iy ) = 1 − − + ...
2! 3!
y2 y4 y3
= 1− + + ..... + i ( y − + ...) = cos y + i sin y
2! 4! 3!
(3) if a = e , z = x + iy ⇒ e x + iy = e x e iy = exp( x )(cos y + i sin y )

10. Set exp w = z
Write z = r exp iθ for r is real and − π < θ ≤ π
⇒ z = r exp[i (θ + 2nπ )] ⇒ w = Lnz = ln r + i (θ + 2nπ )
Lnz is a multivalued function of z .
Take its principal value by choosing n = 0
⇒ ln z = ln r + iθ -π < θ ≤ π
If t ≠ 0 and z are both complex numbers, we define
t z = exp( zLnt )
Ex : Show that there are exactly n distinct nth roots of t .
1
1
tn = exp( Lnt ) and t = r exp[i (θ + 2kπ )]
n
1 1
1 (θ + 2kπ ) (θ + 2kπ )
⇒ tn = exp[ ln r + i ] = r n exp[i ]
n n n

11. 20.5 Multivalued functions and branch cuts
A logarithmic function, a complex power and a complex root are all
multivalued. Is the properties of analytic function still applied?
Ex : f ( z ) = z 1 / 2 and z = r exp(iθ ) (A) C
y
(A) z traverse any closed contour C that
dose not enclose the origin, θ return r
to its original value after one complete θ
x
circuit.
(B) y
(B) θ → θ + 2π enclose the origin
C' r
1/ 2 1/ 2
r exp(iθ / 2) → r exp[i (θ + 2π ) / 2]
θ
= − r 1 / 2 exp(iθ / 2) x
⇒ f (z) → − f (z)
z = 0 is a branch point of the function f ( z ) = z 1 / 2

12. Branch point: z remains unchanged while z traverse a closed contour C
about some point. But a function f(z) changes after one complete circuit.
Branch cut: It is a line (or curve) in the complex plane that we must cross ,
so the function remains single-valued.
y
Ex : f ( z ) = z 1 / 2
restrict θ ⇒ 0 ≤ θ < 2π
⇒ f ( z ) is single - valued 0
x

13. Ex : Find the branch points of f ( z ) = z 2 + 1 , and hence sketch
suitable arrangements of branch cuts.
f ( z ) = z 2 + 1 = ( z + i )( z − i ) expected branch points : z = ± i
set z − i = r1 exp(iθ1 ) and z + i = r2 exp(iθ 2 )
⇒ f ( z ) = r1r2 exp(iθ1 / 2) exp(iθ 2 / 2)
= r1r2 exp[i (θ1 + θ 2 )]
If contour C encloses
(1) neither branch point, then θ1 → θ1 , θ 2 → θ 2 ⇒ f ( z ) → f ( z )
(2) z = i but not z = − i , then θ1 → θ1 + 2π , θ 2 → θ 2 ⇒ f ( z ) → − f ( z )
(3) z = − i but not z = i , then θ1 → θ1 , θ 2 → θ 2 + 2π ⇒ f ( z ) → − f ( z )
(4) both branch points, then θ1 → θ1 + 2π ,θ 2 → θ 2 + 2π ⇒ f ( z ) → f ( z )

14. f ( z ) changes value around loops containing
either z = i or z = − i . We choose branch cut as follows :
(A) y
(B) y
i
i
x x
−i
−i

15. 20.6 Singularities and zeros of complex function
g( z )
Isolated singularity (pole) : f ( z ) =
( z − z0 ) n
n is a positive integer, g( z ) is analytic at all points in
some neighborhood containing z = z0 and g( z0 ) ≠ 0,
the f ( z ) has a pole of order n at z = z0 .
* * An alternate definition for that f ( z ) has a pole of
order n at z = z0 is
lim [( z − z0 ) n f ( z )] = a
z → z0
f ( z ) is analytic and a is a finite, non - zero complex number
(1) if a = 0, then z = z0 is a pole of order less than n.
(2) if a is infinite, then z = z0 is a pole of order greater than n.
(3) if z = z0 is a pole of f ( z ) ⇒| f ( z ) |→ ∞ as z → z0
(4) from any direction, if no finite n satisfies the limit ⇒ essential singularity

16. Ex : Find the singularities of the function
1 1
(1) f ( z ) = −
1− z 1+ z
2z
⇒ f (z) = poles of order 1 at z = 1 and z = −1
(1 − z )(1 + z )
(2) f ( z ) = tanh z
sinh z exp z − exp(− z )
= =
cosh z exp z + exp(− z )
f ( z ) has a singularity when exp z = − exp(− z )
⇒ exp z = exp[i ( 2n + 1)π ] = exp(− z ) n is any integer
1
⇒ 2 z = i ( 2n + 1)π ⇒ z = ( n + )πi
2
Using l' Hospital' s rule
[ z − ( n + 1 / 2)πi ] sinh z [ z − ( n + 1 / 2)πi ] cosh z + sinh z
lim { }= lim { }=1
z →( n+1 / 2 )πi cosh z z →( n+1 / 2 )πi sinh z
each singularity is a simple pole (n = 1)

17. Remove singularties :
Singularity makes the value of f ( z ) undetermined, but lim f ( z )
z→ z0
exists and independent of the direction from which z0 is approached.
Ex : Show that f ( z ) = sin z / z is a removable singularity at z = 0
Sol : lim f ( z ) = 0 / 0 undetermined
z →0
1 z3 z5 z3 z5
f (z) = (z − + − ........) = 1 − + − ....
z 3! 5! 3! 5!
lim f ( z ) = 1 is independent of the way z → 0, so
z →0
f ( z ) has a removable singularity at z = 0.

18. The behavior of f ( z ) at infinity is given by that of
f (1 / ξ ) at ξ = 0, where ξ = 1 / z
Ex : Find the behavior at infinity of (i) f ( z ) = a + bz −2
(ii) f ( z ) = z (1 + z 2 ) and (iii) f ( z ) = exp z
(i) f ( z ) = a + bz − 2 ⇒ set z = 1 / ξ ⇒ f (1 / ξ ) = a + bξ 2
is analytic at ξ = 0 ⇒ f ( z ) is analytic at z = ∞
(ii) f ( z ) = z (1 − z 2 ) ⇒ f (1 / ξ ) = 1 / ξ + 1 / ξ 3 has a pole of
order 3 at z = ∞
∞
(iii) f ( z ) = exp z ⇒ f (1 / ξ ) = ∑ (n! )−1ξ −n
n= 0
f ( z ) has an essential singularity at z = ∞

19. If f ( z0 ) = 0 and f ( z ) = ( z − z0 )n g ( z ), if n is
a positive integer, and g( z0 ) ≠ 0
(i) z = z0 is called a zero of order n.
(ii) if n = 1, z = z0 is called a simple zero.
(iii) z = z0 is also a pole of order n of 1 / f ( z )

20. 20.10 Complex integral y
B
A real continuous parameter t , for α ≤ t ≤ β
C2
x = x ( t ), y = y( t ) and point A is t = α ,
C1
point B is t = β
⇒ ∫ f ( z )dz = ∫ ( u + iv )(dx + idy ) x
C C
A C3
= ∫ udx − ∫ vdy + i ∫ udy + i ∫ vdx
C C C C
β dx β dy β dy β dx
=∫ u dt − ∫ v dt + i ∫ u dt + i ∫ v dt
α dt α dt α dt α dt

21. Ex : Evaluate the complex integral of f ( z ) = 1 / z , along
the circle |z| = R, starting and finishing at z = R. y
z ( t ) = R cos t + iR sin t , 0 ≤ t ≤ 2π C1
dx dy 1 x − iy R
= − R sin t , = R cos t , f ( z ) = = 2 = u + iv , t
dt dt x + iy x + y 2
x
x cos t −y − sin t
u= 2 = ,v = 2 =
x + y2 R x + y2 R
1 2π cos t 2π − sin t
∫C1 z dz = ∫0 R (− R sin t )dt − ∫0 ( R ) R cos tdt
2π cos t 2π − sin t
+ i∫ R cos tdt + i ∫ ( )( − R sin t )dt
0 R 0 R
= 0 + 0 + iπ + iπ = 2πi
* * The integral is also calculated by
dz 2π − R sin t + iR cos t 2π
∫C1 z 0 R cos t + iR sin t
=∫ dt = ∫ idt = 2πi
0
The calculated result is independent of R.

22. Ex : Evaluate the complex integral of f ( z ) = 1 / z along
(i) the contour C 2 consisting of the semicircle | z |= R in
the half - plane y ≥ 0
(ii) the contour C 3 made up of two straight lines C 3a and C 3b
(i) This is just as in the previous example, but for
y
0 ≤ t ≤ π ⇒ ∫ dz / z = πi iR
C2
C 3b
(ii) C 3a : z = (1 − t ) R + itR for 0 ≤ t ≤ 1 C 3a
C 3b : − sR + i (1 − s ) R for 0 ≤ s ≤ 1 s=1 t=0
dz 1 − R + iR 1 − R − iR −R R x
∫C3 z 0 R + t (− R + iR) 0 iR + s(− R − iR)
=∫ dt + ∫ dt
1 −1+ i 1 2t − 1 1 1
1st term ⇒ ∫ dt = ∫ dt + i ∫ dt
0 1 − t + it 0 1 − 2 t + 2t 2 0 1 − 2t + 2t 2
1 i t −1/ 2 1
= [ln(1 − 2t + 2t 2 )] |1 + [2 tan −1 (
0 )] |0
2 2 1/ 2
i π π πi a x
= 0 + [ − ( − )] = ∫ a2 + x2 dx = tan −1 ( ) + c
2 2 2 2 a

23. 1 1+ i 1 (1 + i )[ s − i ( s − 1)]
2nd term ⇒ ∫ ds = ∫ ds
0 s + i ( s − 1) 0 2
s + ( s − 1) 2
1 2s − 1 1 1
=∫ ds + i ∫ ds
0 2s 2 − 2s + 1 0 2s 2 − 2s + 1
1 s −1/ 2 1
= [ln( 2 s 2 − 2 s + 1)] |1 + i tan −1 (
0 ) |0
2 1/ 2
π π πi
= 0 + i[ − ( − )] =
4 4 2
dz
⇒ ∫C3 z = πi
The integral is independent of the different path.

24. Ex : Evaluate the complex integral of f ( z ) = Re( z ) along
the path C1 , C 2 and C 3 as shown in the previous examples.
2π
(i) C1 : ∫ R cos t ( − R sin t + iR cos t )dt = iπR 2
0
π iπ 2
(ii) C 2 : ∫ R cos t ( − R sin t + iR cos t )dt = R
0 2
(iii) C 3 = C 3a + C 3b :
1 1
∫0 (1 − t ) R( − R + iR )dt + ∫ ( − sR )( − R − iR )ds
0
1 1
= R 2 ∫ (1 − t )( −1 + i )dt + R 2 ∫ s(1 + i )ds
0 0
1 2 1
= R ( −1 + i ) + R 2 (1 + i ) = iR 2
2 2
The integral depends on the different path.

25. 20.11 Cauchy theorem
If f ( z ) is an analytic function, and f ' ( z ) is continuous
at each point within and on a closed contour C
⇒ ∫ f ( z )dz = 0
C
∂p( x , y ) ∂q( x , y )
If and are continuous within and
∂x ∂y
on a closed contour C, then by two - demensional
∂p ∂q
divergence theorem ⇒ ∫∫ ( + )dxdy = ∫ ( pdy − qdx )
R ∂x ∂y C
f ( z ) = u + iv and dz = dx + idy
I = ∫ f ( z )dz = ∫ ( udx − vdy ) + i ∫ (vdx + udy )
C C C
∂ ( − u) ∂ ( − v ) ∂ ( − v ) ∂u
= ∫∫ [ + ]dxdy + i ∫∫ [ + ]dxdy = 0
R ∂y ∂x R ∂y ∂x
f ( z ) is analytic and the Cauchy - Riemann relations apply.

26. Ex : Suppose twos points A and B in the complex plane are joined
by two different paths C1 and C 2 . Show that if f ( z ) is an
analytic function on each path and in the region enclosed by the
two paths then the integral of f ( z ) is the same along C1 and C 2 .
y
B
∫C 1
f ( z )dz − ∫
C2
f ( z )dz = ∫
C1 − C 2
f ( z )dz = 0 C1
path C1 − C 2 forms a closed contour enclosing R R
⇒∫ f ( z )dz = ∫ f ( z )dz x
C1 C2
A

27. Ex : Consider two closed contour C and γ in the Argand diagram, γ being
sufficiently small that it lies completely with C . Show that if the function
f ( z ) is analytic in the region between the two contours then ∫ f ( z )dz = ∫ f ( z )dz
C γ
the area is bounded by Γ, and
f ( z ) is analytic C
y
∫Γ f ( z )dz = 0 γ
C1
= ∫ f ( z )dz + ∫ f ( z )dz + ∫ f ( z )dz + ∫ f ( z )dz
C γ C1 C2
If take the direction of contour γ as that of C2
contour C ⇒ ∫ f ( z )dz = ∫ f ( z )dz
C γ
x
Morera' s theorem :
if f ( z ) is a continuous function of z in a closed domain R
bounded by a curve C , for ∫ f ( z )dz = 0 ⇒ f ( z ) is analytic.
C

28. 20.12 Cauchy’s integral formula
If f ( z ) is analytic within and on a closed contour C
1 f (z)
2πi ∫C z − z0
and z0 is a point within C then f ( z0 ) = dz
f (z) f (z) C
I=∫ dz = ∫ dz
C z − z0 γ z−z
0
for z = z0 + ρ exp(iθ ), dz = iρ exp(iθ )dθ γ
2π f ( z0 + ρe iθ ) z0
I=∫ iρe iθ dθ
0 ρe iθ
2π ρ →0
iθ
= i∫ f ( z0 + ρe )dθ = 2πif ( z0 )
0

29. The integral form of the derivative of a complex function :
1 f (z)
f ' ( z0 ) =
2πi ∫C ( z − z )2 dz
0
f ( z 0 + h) − f ( z 0 )
f ' ( z0 ) = lim
h→ 0 h
1 f (z) 1 1
= lim [
h→0 2πi
∫C h z − z0 − h z − z0 )dz ]
( −
1 f (z)
= lim [
h→ 0 2πi ∫C ( z − z0 − h)( z − z0 )dz ]
1 f (z)
=
2πi ∫C ( z − z )2 dz
0
n! f (z)
2πi ∫C ( z − z0 )n+1
For nth derivative f ( n ) ( z0 ) = dz

30. Ex : Suppose that f ( z ) is analytic inside and on a circleC of radius
R centered on the point z = z0 . If | f ( z ) |≤ M on the circle, where
Mn!
M is some constant, show that | f ( n ) ( z0 ) |≤ n
.
R
n! f ( z )dz n! M Mn!
| f ( n ) ( z0 ) |= |∫ |≤ 2πR = n
2π C ( z − z0 )n+1 2π R n+1 R
Liouville' s theorem : If f ( z ) is analytic and bounded for all
z then f ( z ) is a constant.
Mn!
Using Cauchy' s inequality : | f ( n ) ( z0 ) |≤
Rn
set n = 1 and let R → ∞ ⇒ | f ' ( z0 ) |= 0 ⇒ f ' ( z0 ) = 0
Since f ( z ) is analytic for all z , we may take z0 as any
point in the z - plane. f ' ( z ) = 0 for all z ⇒ f ( z ) = constant

31. 20.13 Taylor and Laurent series
Taylor’s theorem:
If f ( z ) is analytic inside and on a circle C of radius R centered
on the point z = z0 , and z is a point inside C, then
∞ ∞
f ( n ) ( z0 )
f (z) = ∑ a n ( z − z0 ) n
= ∑ n!
( z − z0 ) n
n= 0 n= 0
1 f (ξ )
2πi ∫C ξ − z
f ( z ) is analytic inside and on C, so f ( z ) = dξ where ξ lies on C
∞
1 z − z0 1 1 z − z0 n
expand
ξ −z
as a geometric series in
ξ − z0
⇒ =
ξ − z ξ − z0
∑( ξ − z0
)
n=0
1 f (ξ ) ∞ z − z0 n 1 ∞ f (ξ )
⇒ f (z) = ∑ ( ξ − z ) dξ = 2πi ∑ ( z − z0 )n ∫C (ξ − z )n+1 dξ
2πi ∫C ξ − z0 n= 0 0 n= 0 0
1 ∞ n 2πif
( n)
( z0 ) ∞ n f
(n)
( z0 )
= ∑
2πi n=0
( z − z0 )
n!
= ∑ ( z − z0 )
n!
n=0

32. If f ( z ) has a pole of order p at z = z0 but is analytic at every other point inside
and on C . Then g ( z ) = ( z − z0 ) p f ( z ) is analytic at z = z0 and expanded as a Taylor
∞
series g( z ) = ∑ bn ( z − z0 )n .
n=0
Thus , for all z inside C f ( z ) can be exp anded as a Laurent series
a− p a − p +1 a
f (z) = + + ....... + −1 + a0 + a1 ( z − z0 ) + a 2 ( z − z0 )2
( z − z0 ) p ( z − z0 ) p−1 z − z0
g ( n ) ( z0 ) 1 g( z )
a n = bn+ p and bn =
n!
=
2πi ∫ ( z − z )n+1 dz C2
0
R
1 g( z ) 1 f (z)
2πi ∫ ( z − z0 ) n+1+ p 2πi ∫ ( z − z0 ) n+1
⇒ an = dz = dz
z 0 C1
∞
f (z) = ∑ an ( z − z0 )n is analytic in a region R between
n= −∞
two circles C1 and C 2 centered on z = z0

33. ∞
f (z) = ∑ a n ( z − z0 ) n
n = −∞
(1) If f ( z ) is analytic at z = z0 , then all an = 0 for n < 0.
It may happen a n = 0 for n ≥ 0, the first non - vanishing
term is a m ( z-z0 ) m with m > 0, f ( z ) is said to have a zero
of order m at z = z0 .
(2) If f ( z ) is not analytic at z = z0
(i) possible to find a − p ≠ 0 but a − p− k = 0 for all k > 0
f ( z ) has a pole of order p at z = z0 , a −1 is called the residue of f ( z )
(ii) impossible to find a lowest value of − p ⇒ essential singularity

34. 1
Ex : Find the Laurent series of f ( z ) = 3
about the singularities
z ( z − 2)
z = 0 and z = 2. Hence verify that z = 0 is a pole of order 1 and z = 2 is a
pole of order 3, and find the residue of f ( z ) at each pole.
(1) point z = 0
−1 −1 − z ( −3)( −4) − z 2 ( −3)( −4)( −5) − z 3
f (z) = 3
= [1 + ( −3)( ) + ( ) + ( ) + ...]
8 z (1 − z / 2) 8z 2 2! 2 3! 2
1 3 3 5z 2
=− − − z− − ... z = 0 is a pole of order 1
8 z 16 16 32
(2) point z = 2 ⇒ set z − 2 = ξ ⇒ z( z − 2) 3 = ( 2 + ξ )ξ 3 = 2ξ 3 (1 + ξ / 2)
1 1 ξ ξ ξ ξ
f (z) = 3 = 3 [1 − ( ) + ( )2 − ( )3 + ( )4 − ...]
2ξ (1 + ξ / 2) 2ξ 2 2 2 2
1 1 1 1 ξ 1 1 1 1 z−2
= − + − + − .. = − + − + −
2ξ 3
4ξ 2 8ξ 16 32 2( z − 2) 3
4( z − 2) 2 8( z − 2) 16 32
z = 2 is a pole of order 3, the residue of f ( z ) at z = 2 is 1 / 8.

35. How to obtain the residue ?
a− m a−1
f (z) = + ...... + + a0 + a1 ( z − z0 ) + a 2 ( z − z0 )2 + ...
( z − z0 ) m ( z − z0 )
⇒ ( z − z0 )m f ( z ) = a − m + a − m +1 ( z − z0 ) + ....... + a −1 ( z − z0 )m −1 + ...
d m −1 ∞
⇒ m −1
[( z − z0 ) f ( z )] = ( m − 1)! a −1 + ∑ bn ( z − z0 )n
m
dz n =1
Take the limit z → z0
1 d m −1
R( z0 ) = a −1 = lim { [( z − z0 ) m f ( z )]} residue at z = z0
z → z0 ( m − 1)! dz m −1
(1) For a simple pole m = 1 ⇒ R( z0 ) = lim [( z − z0 ) f ( z )]
z → z0
g( z )
(2) If f ( z ) has a simple at z = z0 and f ( z ) = , g ( z ) is analytic and
h( z )
non - zero at z0 and h( z0 ) = 0
( z − z0 ) g ( z ) ( z − z0 ) 1 g( z )
⇒ R( z0 ) = lim = g ( z0 ) lim = g ( z0 ) lim ' = ' 0
z → z0 h( z ) z → z0 h( z ) z → z0 h ( z ) h ( z0 )

36. Ex : Suppose that f ( z ) has a pole of order m at the point z = z0 . By
considering the Laurent series of f ( z ) about z0 , deriving a general
expression for the residue R( z0 ) of f ( z ) at z = z0 . Hence evaluate
exp iz
the residue of the function f ( z ) = 2 2
at the point z = i .
( z + 1)
exp iz exp iz
f (z) = 2 2
= 2 2
poles of order 2 at z = i and z = − i
( z + 1) (z + i) (z − i)
for pole at z = i :
d d exp iz i 2
[( z − i ) 2 f ( z )] = [ 2
]= 2
exp iz − 3
exp iz
dz dz ( z + i ) (z + i) (z + i)
1 i 2 −i
R( i ) = [ e −1 − e −1 ] =
1! ( 2i ) 2 ( 2i ) 3 2e

37. 20.14 Residue theorem
f ( z ) has a pole of order m at z = z0 C
∞ γ
f (z) = ∑ a n ( z − z0 ) n
n= − m ρ • z0
I = ∫ f ( z )dz = ∫ f ( z )dz
C γ
set z = z0 + ρe iθ ⇒ dz = iρe iθ dθ
∞ ∞ 2π
I= ∑ an ∫ ( z − z0 ) dz =
C
n
∑ an ∫ 0
iρ n+1e i ( n+1)θ dθ
n= − m n= − m
2π n +1 i ( n+1)θ iρ n+1e i ( n+1)θ 2π
for n ≠ −1 ⇒ ∫ iρ e dθ = |0 = 0
0 i ( n + 1)
2π
for n = 1 ⇒ ∫ idθ = 2πi
0
I = ∫ f ( z )dz = 2πia −1
C

38. Residue theorem:
f ( z ) is continuous within and on a closed contour C
and analytic, except for a finite number of poles within C
∫C f ( z )dz = 2πi ∑ R j
j
∑ R j is the sum of the residues of f ( z ) at its poles within C
j
C C'

39. The integral I of f ( z ) along an open contour C
C
if f ( z ) has a simple pole at z = z0 ρ
⇒ f ( z ) = φ ( z ) + a−1 ( z − z0 ) −1 z0
φ ( z ) is analytic within some neighbour surrounding z0
| z − z0 |= ρ and θ1 ≤ arg( z − z0 ) ≤ θ 2
ρ is chosen small enough that no singularity of f ( z ) except z = z0
I = ∫ f ( z )dz = ∫ φ ( z )dz + a −1 ∫ ( z − z0 ) −1 dz
C C C
lim
ρ →0 C
∫ φ ( z )dz = 0
θ2 1
I = lim ∫ f ( z )dz = lim (a −1 ∫ iθ
iρe iθ dθ ) = ia −1 (θ 2 − θ1 )
ρ →0 C ρ →0 ρe θ1
for a closed contour θ 2 = θ1 + 2π ⇒ I = 2πia −1

40. 20.16 Integrals of sinusoidal functions
2π
∫0 F (cos θ , sin θ )dθ set z = exp iθ in unit circle
1 1 1 1
⇒ cos θ = ( z + ), sin θ = ( z − ), dθ = − iz −1dz
2 z 2i z
2π cos 2θ
Ex : Evaluate I = ∫ 2 2
dθ for b > a > 0
0 a + b − 2ab cos θ
1 n 1
cos nθ = ( z + z − n ) ⇒ cos 2θ = ( z 2 + z − 2 )
2 2
1 2 1
cos 2θ ( z + z − 2 )( − iz −1 )dz − ( z 4 + 1)idz
dθ = 2 = 2 2
a 2 + b 2 − 2ab cos θ 2 2 1 −1 z ( za 2 + zb 2 − abz 2 − ab )
a + b − 2ab ⋅ ( z + z )
2
i ( z 4 + 1)dz i ( z 4 + 1)
= = dz
2ab 2 2 a b 2ab 2 a b
z ( z − z ( − + ) + 1) z ( z − )( z − )
b a b a

41. i z4 + 1
2ab ∫C
I= dz double poles at z = 0 and z = a / b within the unit circle
a b
z 2 ( z − )( z − )
b a
1 d m −1
Residue : R( z0 ) = lim { [( z − z0 ) m f ( z )]
z → z0 ( m − 1)! dz m −1
(1) pole at z = 0, m = 2
1 d 2 z4 + 1
R(0) = lim { [z 2 ]}
z →0 1! dz z ( z − a / b )( z − b / a )
4z 3 ( z 4 + 1)( −1)[2 z − (a / b + b / a )]
= lim { + }= a/b+b/a
z →0 ( z − a / b )( z − b / a ) 2
( z − a / b) ( z − b / a ) 2
(2) pole at z = a / b, m = 1
z4 + 1 (a / b)4 + 1 − (a 4 + b 4 )
R(a / b) = lim [( z − a / b ) 2
]= 2
=
z →a / b z ( z − a / b )( z − b / a ) (a / b) (a / b − b / a ) ab(b 2 − a 2 )
i a 2 + b2 a 4 + b4 2πa 2
I = 2πi × [ − ]= 2 2
2ab ab ab(b − a ) b (b − a 2 )
2 2

42. 20.17 Some infinite integrals
∞
∫−∞ f ( x )dx
f ( z ) has the following properties :
(1) f ( z ) is analytic in the upper half - plane, Im z ≥ 0, except for
a finite number of poles, none of which is on the real axis.
(2) on a semicircle Γ of radius R, R times the maximum of y
| f | on Γ tends to zero as R → ∞ (a sufficient condition
is that zf ( z ) → 0 as | z |→ ∞ ). Γ
0 ∞
(3) ∫ f ( x )dx and ∫0 f ( x )dx both exist
−∞ x
∞
⇒ ∫− ∞ f ( x )dx = 2π i ∑ R j −R 0 R
j
for | ∫ f ( z )dz |≤ 2π R × (maximum of | f | on Γ ), the integral along Γ
Γ
tends to zero as R → ∞ .

43. ∞ dx
Ex : Evaluate I = ∫0 2
(x + a ) 2 4
a is real
dz R dx dz
∫C ( z 2 + a 2 )4 = ∫− R ( x 2 + a 2 )4 ∫Γ ( z 2 + a 2 )4
+ as R → ∞
Γ
dz dz ∞ dx ai
⇒∫ →0⇒ ∫C ( z 2 + a 2 )4 = ∫− ∞ ( x 2 + a 2 )4
Γ ( z 2 + a 2 )4
( z 2 + a 2 )4 = 0 ⇒ poles of order 4 at z = ± ai , −R 0 R
only z = ai at the upper half - plane
1 1 1 iξ − 4
set z = ai + ξ , ξ → 0 ⇒ 2 2 4
= 2 4
= 4
(1 − )
(z + a ) ( 2aiξ + ξ ) ( 2aiξ ) 2a
1 ( −4)( −5)( −6) − i 3 − 5i
the coefficient of ξ −1 is ( ) =
( 2a ) 4 3! 2a 32a 7
∞ dx − 5i 10π 1 10π 5π
∫0 ( x 2 + a 2 )4
= 2πi (
32a 7
)=
32a 7
⇒I = ×
2 32a 7
=
32a 7

44. For poles on the real axis: y
Γ
Principal value of the integral, defined as ρ → 0
R z0 − ρ R
P∫ f ( x )dx = ∫− R f ( x )dx + ∫z + ρ f ( x )dx γ ρ
−R 0
-R 0 z0 R x
for a closed contour C
z0 − ρ R
∫C f ( z )dz = ∫− R f ( x )dx + ∫γ f ( z )dz + ∫z + ρ
0
f ( x )dx + ∫Γ f ( z )dz
R
= P∫ f ( x )dx + ∫γ f ( z )dz + ∫Γ f ( z )dz
−R
(1) for ∫ f ( z )dz has a pole at z = z0 ⇒ ∫ f ( z )dz = −πia1
γ γ
iθ
(2) for ∫ f ( z )dz set z = Re dz = i Re iθ dθ
Γ
⇒ ∫Γ f ( z )dz = ∫Γ f (Re iθ )i Re iθ dθ
If f ( z ) vanishes faster than 1 / R 2 as R → ∞, the integral is zero

45. Jordan’s lemma
(1) f ( z ) is analytic in the upper half - plane except for a finite
number of poles in Im z > 0
( 2) the maximum of | f ( z ) |→ 0 as | z |→ ∞ in the upper half - plane
(3) m > 0, then
∫Γ e
imz
IΓ = f ( z )dz → 0 as R → ∞, Γ is the semicircular contour
for 0 ≤ θ ≤ π / 2, 1 ≥ sin θ / θ ≥ π / 2 f (θ ) f = 2θ / π
| exp(imz )| = | exp(− mR sin θ )|
π
IΓ ≤ ∫Γ |e imz f ( z )||dz| ≤ MR ∫ e − mR sin θ dθ
0
f = sin θ
π/ 2 − mR sin θ
= 2 MR ∫ e dθ
0
M is the maximum of |f ( z )| on |z| = R, R → ∞ M → 0 π /2 θ
πM
π / 2 − mR ( 2θ / π ) πM
I Γ < 2 MR ∫ e (1 − e − mR ) <
dθ =
0 m m
as R → ∞ ⇒ M → 0 ⇒ I Γ → 0

46. cos mx
∞
Ex : Find the principal value of ∫− ∞ x − a dx a real, m > 0 Γ
e imz
Consider the integral I = ∫C z−a
dz = 0 no pole in the
γ ρ
−1
upper half - plane, and |( z − a ) | → 0 as |z| → ∞
-R 0 a R
e imz
I = ∫C z−a
dz
a−ρ e imx e imz R e imx e imz
= ∫− R x−a
dx + ∫γ z−a
dz + ∫a + ρ x−a
dx + ∫Γ z−a
dz = 0
e imz
As R → ∞ and ρ → 0 ⇒ ∫ dz → 0
Γ z−a
∞e imx
⇒ P∫ dx − iπa−1 = 0 and a−1 = e ima
−∞ x − a
∞ cos mx ∞ sin mx
⇒ P∫ dx = −π sin ma and P ∫ dx = π cos ma
−∞ x − a −∞ x − a

47. 20.18 Integral of multivalued functions
1/ 2
y
Multivalued functions such as z , Lnz
Single branch point is at the otigin. We let R → ∞ R Γ
and ρ → 0. The integrand is multivalued, its values
γ
along two lines AB and CD joining z = ρ to z = R A B
are not equal and opposite. ρ x
C D
∞ dx
Ex : I = ∫0 ( x + a )3 x1 / 2
for a > 0
(1) the integrand f ( z ) = ( z + a ) −3 z −1 / 2 , |zf ( z )| → 0 as ρ → 0 and R → ∞
the two circles make no contribution to the contour integral
(2) pole at z = − a , and ( − a )1 / 2 = a 1 / 2 e iπ / 2 = ia 1 / 2
1 d 3 −1 1
R( − a ) = lim [( z + a )3 ]
z → − a ( 3 − 1)! dz 3 − 1 3 1/ 2
(z + a) z
1 d 2 −1 / 2 − 3i
= lim z =
z → − a 2! dz 2 8a 5 / 2

48. − 3i
∫AB dz + ∫Γ dz + ∫DC dz + ∫γ dz = 2πi (
8a 5/2
)
and ∫ dz = 0 and ∫γ dz = 0
Γ
along line AB ⇒ z = xe i 0 , along line CD ⇒ z = xe i 2π
∞ dx 0 dx 3π
∫0, A→ B ( x + a )3 x1 / 2 ∞,C → D ( xe i 2π + a )3 x1 / 2e (1 / 2×2πi ) 4a 5 / 2
+∫ =
1 ∞ dx 3π
⇒ (1 − iπ )∫ =
e 0 ( x + a )3 x1 / 2 4a 5 / 2
∞ dx 3π
⇒∫ =
0 ( x + a )3 x1 / 2 8a 5 / 2

49. ∞ x sin x
Ex : Evaluate I (σ ) = ∫− ∞ x 2 − σ 2 dx
z sin z 1 ze iz 1 ze − iz
∫C z2 − σ 2
dz =
2i ∫C 1 z2 − σ 2
dz −
2i ∫C 2
2
z −σ 2
dz = I1 + I 2
(1) for I1 , the contour is choosed on the upper half - plane C1
due to the term e iz , and only one pole at z = σ . Γ
1 ze iz 1 −σ − ρ xe ix γ1
I1 =
2i ∫C 1 z2 − σ 2
dz =
2i ∫− R x2 − σ 2
dx
ρ
1 σ −ρ xe ix 1 ∞ xe ix -R -σ σ R
+
2i ∫−σ + ρ x2 − σ 2
dx +
2i ∫σ + ρ x 2 − σ 2 dx γ2
1 ze iz 1 ze iz 1 ze iz
+
2i ∫γ 1 z2 − σ 2
dz + ∫ 2
2i γ 2 z − σ 2
dz +
2i ∫Γ z 2 − σ 2 dz
1 σe iσ π
= 2πi × Res( z = σ ) = π = e iσ
2i 2σ 2

50. As ρ → 0 and R → ∞ ⇒ ∫ dz → 0
Γ
1 ze iz 1 − π − iσ
2i ∫γ 1 ( z + σ )( z − σ )
dz =
2i
× ( −πi )Res( z = −σ ) =
4
e
1 ze iz 1 π
∫γ dz = × πiRes( z = σ ) = e iσ
2i 2 ( z + σ )( z − σ ) 2i 4
1 ∞ xe ix π π γ1
dx + (e iσ − e − iσ ) = e iσ
2i ∫− ∞ x 2 − σ 2
I1 = σ
4 2
−σ
(2) for I 2 , we choose the lower half - plane by the
− iz Γ γ2
term e , only one pole at z = −σ
−1 ze − iz − 1 −σ − ρ xe − ix
I2 =
2i ∫C2 z 2 − σ 2 dz = 2i ∫− R x 2 − σ 2 dx C2
1 σ −ρ xe − ix 1 ∞ xe − ix 1 ze − iz
−
2i ∫−σ + ρ x2 − σ 2
dx −
2i ∫σ + ρ x2 − σ 2
dx −
2i ∫γ 1
2
z −σ 2
dz
1 ze − iz 1 ze − iz −1 ( −σ )e iσ π
− ∫γ dz − ∫Γ dz = ( ) × ( −2πi ) = e iσ
2i 2 z2 − σ 2 2i z2 − σ 2 2i − 2σ 2

51. As ρ → 0, R → ∞ ⇒ ∫ dz → 0
Γ
−1 ze − iz −1 ( −σ )e iσ π
= e iσ
2i ∫γ 1 ( z + σ )( z − σ )
dz = ( )( −πi )
2i − 2σ 4
−1 ze − iz −1 σe − iσ − π − iσ
2i ∫γ 2 ( z + σ )( z − σ )dz = (
2i
)(πi )
2σ
=
4
e
− 1 ∞ xe − ix π π
dx + (e iσ − e − iσ ) = e iσ
2i ∫− ∞ x 2 − σ 2
I2 =
4 2
− 1 ∞ xe − ix π 1
dx = e iσ − (e iσ − e − iσ )
2i ∫− ∞ x 2 − σ 2
⇒
2 4
∞ x sin x 1 ∞ xe ix 1 ∞ xe − ix
I (σ ) = ∫
2i ∫− ∞ x 2 − σ 2 2i ∫− ∞ x 2 − σ 2
dx = dx − dx
−∞ x 2 − σ 2
π π π π
= e iσ − ( e iσ − e − iσ ) + e iσ − ( e iσ − e − iσ )
2 4 2 4
π π π
= πe iσ − e iσ + e − iσ = (e iσ + e − iσ ) = π cos σ
2 2 2