Complex Analysis: Residue
Md. Aquil Khan
Z
f (z) dz
C
Md. Aquil Khan
Complex Analysis: Singularities
Z
f (z) dz
C
If f (z) has no singularity inside the closed contour C , then
Md. Aquil Khan
Complex Analysis: Singularities
Z
f (z) dz
C
If f (z) has no singularity inside the closed contour C , then
Z
f (z) dz = 0
C
Md. Aquil Khan
Complex Analysis: Singularities
Z
f (z) dz
C
If f (z) has no singularity inside the closed contour C , then
Z
f (z) dz = 0
C
How do we deal with the case where singularities of f are included
inside the closed contour C ?
Md. Aquil Khan
Complex Analysis: Singularities
Residue
z0 is an isolated singular point of the function f
Md. Aquil Khan
Complex Analysis: Singularities
Residue
z0 is an isolated singular point of the function f
⇓
f is analytic in the annulus 0 < |z − z0 | < r for some r > 0
Md. Aquil Khan
Complex Analysis: Singularities
Residue
z0 is an isolated singular point of the function f
⇓
f is analytic in the annulus 0 < |z − z0 | < r for some r > 0
⇓
Laurent Series valid in 0 < |z − z0 | < r :
Md. Aquil Khan
Complex Analysis: Singularities
Residue
z0 is an isolated singular point of the function f
⇓
f is analytic in the annulus 0 < |z − z0 | < r for some r > 0
⇓
Laurent Series valid in 0 < |z − z0 | < r :
function analytic at z0
z
}|
{
∞
∞
X
X
n
f (z) =
an (z − z0 )
+
bn (z − z0 )−n ,
n=0
Md. Aquil Khan
|n=1
{z
}
principal part
Complex Analysis: Singularities
Residue
z0 is an isolated singular point of the function f
⇓
f is analytic in the annulus 0 < |z − z0 | < r for some r > 0
⇓
Laurent Series valid in 0 < |z − z0 | < r :
function analytic at z0
z
}|
{
∞
∞
X
X
n
f (z) =
an (z − z0 )
+
bn (z − z0 )−n ,
n=0
|n=1
{z
}
principal part
⇓
1
Co-efficient of z−z
,
that
is,
b
1 is called the residue of f at the
0
isolated singularity z0 (notation: Res(f ; z0 ))
Md. Aquil Khan
Complex Analysis: Singularities
Residue
Res(f ; z0 ) = b1 =
Md. Aquil Khan
Complex Analysis: Singularities
Residue
Res(f ; z0 ) = b1 =
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1
2πi
Z
f (z) dz,
C
Complex Analysis: Singularities
Residue
Res(f ; z0 ) = b1 =
1
2πi
Z
f (z) dz,
C
where C is any circle |z − z0 | = ρ < r
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
=
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
=1
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
I Res
1
,z
(z−z0 )k 0
=1
, (k 6= 1) =
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
I Res
1
,z
(z−z0 )k 0
=1
, (k 6= 1) = 0
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
I Res
1
,z
(z−z0 )k 0
=1
, (k 6= 1) = 0
1
I Res(e z , 0) =?
Md. Aquil Khan
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
I Res
1
,z
(z−z0 )k 0
=1
, (k 6= 1) = 0
1
I Res(e z , 0) =?
h 1
1 1
e z = 1 + 1!
z +
1 1
2! z 2
+
1 1
3! z 3
Md. Aquil Khan
+ ··· ,
i
|z| > 0
Complex Analysis: Singularities
Example
I Res
1
(z−z0 ) , z0
I Res
1
,z
(z−z0 )k 0
=1
, (k 6= 1) = 0
1
I Res(e z , 0) = 1
h 1
1 1
e z = 1 + 1!
z +
1 1
2! z 2
+
1 1
3! z 3
Md. Aquil Khan
+ ··· ,
i
|z| > 0
Complex Analysis: Singularities
Previous example shows that sometimes the residue value can be
found in a straightforward manner
Md. Aquil Khan
Complex Analysis: Singularities
Previous example shows that sometimes the residue value can be
found in a straightforward manner
It is not always the case
Md. Aquil Khan
Complex Analysis: Singularities
Previous example shows that sometimes the residue value can be
found in a straightforward manner
It is not always the case
We now give some results which are useful to determine residues
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
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Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
⇓
Res(f ; z0 ) = lim (z − z0 )f (z)
z→z0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
⇓
Res(f ; z0 ) = lim (z − z0 )f (z)
z→z0
f (z) =
∞
X
an (z − z0 )n +
n=0
Md. Aquil Khan
b1
z − z0
Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
⇓
Res(f ; z0 ) = lim (z − z0 )f (z)
z→z0
f (z) =
∞
X
an (z − z0 )n +
n=0
⇒ (z − z0 )f (z) =
∞
X
b1
z − z0
an (z − z0 )n+1 + b1
n=0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
⇓
Res(f ; z0 ) = lim (z − z0 )f (z)
z→z0
f (z) =
∞
X
an (z − z0 )n +
n=0
⇒ (z − z0 )f (z) =
∞
X
b1
z − z0
an (z − z0 )n+1 + b1
n=0
⇒
lim (z − z0 )f (z) =
z→z0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Simple Pole
f has a simple pole at z0
⇓
Res(f ; z0 ) = lim (z − z0 )f (z)
z→z0
f (z) =
∞
X
an (z − z0 )n +
n=0
⇒ (z − z0 )f (z) =
∞
X
b1
z − z0
an (z − z0 )n+1 + b1
n=0
⇒
lim (z − z0 )f (z) = b1
z→z0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
∞
f (z) =
X
b2
bm
b1
+
+ ··· +
+
an (z − z0 )n , bm 6= 0
2
m
z − z0
(z − z0 )
(z − z0 )
n=0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
∞
f (z) =
X
b2
bm
b1
+
+ ··· +
+
an (z − z0 )n , bm 6= 0
2
m
z − z0
(z − z0 )
(z − z0 )
n=0
m−1
⇒ (z − z0 )m f (z) = b1 (z − z0 )
+ b2 (z − z0 )m−2 + · · · + bm +
∞
X
an (z − z0 )n+m
n=0
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
∞
f (z) =
X
b2
bm
b1
+
+ ··· +
+
an (z − z0 )n , bm 6= 0
2
m
z − z0
(z − z0 )
(z − z0 )
n=0
m−1
⇒ (z − z0 )m f (z) = b1 (z − z0 )
+ b2 (z − z0 )m−2 + · · · + bm +
∞
X
an (z − z0 )n+m
n=0
⇒
i
dm−1 h
m
(z
−
z
)
f
(z)
= b1 (m − 1)! + terms involving (z − z0 )
0
dz m−1
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
⇒
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
i
dm−1 h
m
(z
−
z
)
f
(z)
= b1 (m − 1)! + terms involving (z − z0 )
0
dz m−1
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
i
dm−1 h
m
(z
−
z
)
f
(z)
= b1 (m − 1)! + terms involving (z − z0 )
0
dz m−1
i
dm−1 h
(z − z0 )m f (z) = b1 (m − 1)!
⇒ lim
m−1
z→z0 dz
⇒
Md. Aquil Khan
Complex Analysis: Singularities
Residue at a Pole of Order m
f has a pole of order m at z0
⇓
Res(f ; z0 ) =
i
dm−1 h
1
m
lim
(z
−
z
)
f
(z)
0
(m − 1)! z→z0 dz m−1
i
dm−1 h
m
(z
−
z
)
f
(z)
= b1 (m − 1)! + terms involving (z − z0 )
0
dz m−1
i
dm−1 h
(z − z0 )m f (z) = b1 (m − 1)!
⇒ lim
m−1
z→z0 dz
i
1
dm−1 h
m
lim
(z
−
z
)
f
(z)
= b1 = Res(f ; z0 )
⇒
0
(m − 1)! z→z0 dz m−1
⇒
Md. Aquil Khan
Complex Analysis: Singularities
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) =
Md. Aquil Khan
Complex Analysis: Singularities
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) = limz→2 (z − 2)f (z) = limz→2
Md. Aquil Khan
1
(z 2 +4)
=
Complex Analysis: Singularities
1
8
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) = limz→2 (z − 2)f (z) = limz→2
1
(z 2 +4)
=
Res(f ; 2i) =
Md. Aquil Khan
Complex Analysis: Singularities
1
8
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) = limz→2 (z − 2)f (z) = limz→2
1
(z 2 +4)
Res(f ; 2i) = limz→2i (z − 2i)f (z) = limz→2i
=
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=
1
8
1
(z−2)(z+2i)
1
(2i−2)4i
Complex Analysis: Singularities
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) = limz→2 (z − 2)f (z) = limz→2
1
(z 2 +4)
Res(f ; 2i) = limz→2i (z − 2i)f (z) = limz→2i
=
=
1
(z−2)(z+2i)
1
(2i−2)4i
Res(f ; −2i) =
Md. Aquil Khan
1
8
Complex Analysis: Singularities
Example
f (z) =
1
(z−2)(z 2 +4)
Res(f ; 2) = limz→2 (z − 2)f (z) = limz→2
1
(z 2 +4)
Res(f ; 2i) = limz→2i (z − 2i)f (z) = limz→2i
=
=
1
(z−2)(z+2i)
1
(2i−2)4i
Res(f ; −2i) = limz→−2i (z + 2i)f (z) = limz→−2i
=
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1
8
1
(2i+2)4i
Complex Analysis: Singularities
1
(z−2)(z−2i)
Example
I f (z) = cot z
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) = cot z
I Type of singularity at z = 0 −→
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) = cot z
I Type of singularity at z = 0 −→
Md. Aquil Khan
Pole
Complex Analysis: Singularities
Example
I f (z) = cot z
I Type of singularity at z = 0 −→
Pole
I Order of Pole at z = 0 −→
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Complex Analysis: Singularities
Example
I f (z) = cot z
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Md. Aquil Khan
Pole
1
Complex Analysis: Singularities
Example
I f (z) = cot z
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
I Res(f ; 0) = limz→0
z
sin z
Pole
1
cos z = 1
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
Md. Aquil Khan
Pole
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
Pole
I Order of Pole at z = 0 −→
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Md. Aquil Khan
Pole
3
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Res(f ; 0) =
Pole
3
1
d2 3
lim
z f (z)
2
2! z→0 dz
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Res(f ; 0) =
=
Pole
3
1
d2
lim
2! z→0 dz 2
1
d2
lim
2! z→0 dz 2
Md. Aquil Khan
3
z f (z)
1
1−z
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Res(f ; 0) =
=
=
Pole
3
1
d2 3
lim
z f (z)
2
2! z→0 dz
1
d2
1
lim
2! z→0 dz 2 1 − z
1
2
lim
z→0
2!
(1 − z)3
Md. Aquil Khan
Complex Analysis: Singularities
Example
I f (z) =
1
z 3 −z 4
I Type of singularity at z = 0 −→
I Order of Pole at z = 0 −→
Pole
3
1
d2 3
lim
z f (z)
2
2! z→0 dz
1
d2
1
=
lim
2! z→0 dz 2 1 − z
1
2
=
lim
z→0
2!
(1 − z)3
= 1
Res(f ; 0) =
Md. Aquil Khan
Complex Analysis: Singularities
Cauchy Residue Theorem
Hypothesis
C : Simple closed contour
f (z): Analytic within and on C except finitely many points
z1 , z2 , · · · , zn lying within C
Md. Aquil Khan
Complex Analysis: Singularities
Cauchy Residue Theorem
Hypothesis
C : Simple closed contour
f (z): Analytic within and on C except finitely many points
z1 , z2 , · · · , zn lying within C
Conclusion
Z
h
i
f (z) dz = 2πi Res(f , z1 ) + Res(f , z2 ) + · · · + Res(f , zn ) ,
C
integral being taken counter clock wise
Md. Aquil Khan
Complex Analysis: Singularities
Cauchy Residue Theorem
zn
C
z2
z1
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Complex Analysis: Singularities
Cauchy Residue Theorem
zn
Cn
C
z2
C2
z1
C1
Md. Aquil Khan
Complex Analysis: Singularities
Z
Z
f (z) dz
C
=
Z
C1
C2
Md. Aquil Khan
Z
f (z) dz + · · · +
f (z) dz +
f (z) dz
Cn
Complex Analysis: Singularities
Z
Z
f (z) dz
C
=
Z
C1
Z
f (z) dz + · · · +
f (z) dz +
C2
f (z) dz
Cn
= 2πi Res(f ; z1 ) + 2πi Res(f ; z2 ) + · · · + 2πi Res(f ; zn )
h
i
= 2πi Res(f ; z1 ) + Res(f ; z2 ) + · · · + Res(f ; zn ) .
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are inside C
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are inside C
Z
C
h
i
4 − 3z
dz
=
2πi
×
Res(f
;
0)
+
Res(f
;
1)
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are inside C
4 − 3z
= −4
lim z 2
z −z
Z
i
h
z
}|
{
4 − 3z
Res(f
;
0)
+
Res(f
;
1)
dz
=
2πi
×
2
C z −z
z→0
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are inside C
Z
C
4 − 3z
dz
z2 − z
4 − 3z
lim z 2
= −4
z→0 z − z
h
z }| {
= 2πi ×
Res(f ; 0)
+
Res(f ; 1)
| {z }
limz→1 (z−1) 4−3z
=1
2
z −z
Md. Aquil Khan
Complex Analysis: Singularities
i
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are inside C
Z
C
4 − 3z
dz
z2 − z
4 − 3z
lim z 2
= −4
z→0 z − z
h
z }| {
= 2πi ×
Res(f ; 0)
+
Res(f ; 1)
| {z }
limz→1 (z−1) 4−3z
=1
2
z −z
= −6πi
Md. Aquil Khan
Complex Analysis: Singularities
i
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 is inside and 1 is outside C
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 is inside and 1 is outside C
Z
C
4 − 3z
dz = 2πi Res(f ; 0)
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
dz counterclockwise around any simple closed
Evaluate C 4−3z
z 2 −z
contour C such that 0 is inside and 1 is outside C
−4
Z
C
z }| {
4 − 3z
dz
=
2πi
Res(f ; 0) =
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
dz counterclockwise around any simple closed
Evaluate C 4−3z
z 2 −z
contour C such that 0 is inside and 1 is outside C
−4
Z
C
z }| {
4 − 3z
dz
=
2πi
Res(f ; 0) = − 8πi
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are outside C
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are outside C
Z
C
4 − 3z
dz =
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
R
Evaluate C 4−3z
dz counterclockwise around any simple closed
z 2 −z
contour C such that 0 and 1 are outside C
Z
C
4 − 3z
dz = 0
z2 − z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities:
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
Md. Aquil Khan
πn
,
n!z n−1
|z| > 0
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
Res(f ; 0) =
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
Res(f ; 0) = b1 =
π2
2!
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
Res(f ; 0) = b1 =
R
π
z
−C ze dz =
π2
2!
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
2
Res(f ; 0) = b1 = π2!
R
π
3
z
−C ze dz = 2πi × Res(f , 0) = π i
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Evaluate
π
R
C
ze z dz where C is the ellipse 9x 2 + y 2 = 9 (clockwise)
Singularities: 0
P∞ πn P∞
π
ze z = z
n=0
n=0 n!z n =
πn
,
n!z n−1
|z| > 0
Essential Singularity
2
Res(f ; 0) = b1 = π2!
R
π
3
z
−C ze dz = 2πi × Res(f , 0) = π i
R
C
π
ze z dz = −
π
R
−C
ze z dz = −π 3 i
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities:
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities: 0 and 1
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities: 0 and 1
Z
1
I
dz = 2πi Res(f , 0)
3
4
C z −z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities: 0 and 1
Z
1
I
dz = 2πi Res(f , 0)
3
4
C z −z
I f (z) =
1
z3
+ z12 + z1 + 1 + z + z 2 + · · · , 0 < |z| < 1;
− z15 − z16 − · · · ,
|z| > 1.
− z14
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities: 0 and 1
Z
1
I
dz = 2πi Res(f , 0)
3
4
C z −z
+ z12 + z1 + 1 + z + z 2 + · · · , 0 < |z| < 1;
− z15 − z16 − · · · ,
|z| > 1.
I Which Laurent series should be consider to get Res(f , 0)?
I f (z) =
1
z3
− z14
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Integrate f (z) =
C : |z| = 12 .
1
z 3 −z 4
counter clockwise around the circle
I Singularities: 0 and 1
1
Z
I
C
z }| {
1
dz
=
2πi
Res(f , 0) = 2πi
z3 − z4
+ z12 + z1 + 1 + z + z 2 + · · · , 0 < |z| < 1;
− z14 − z15 − z16 − · · · ,
|z| > 1.
I Which Laurent series should be consider to get Res(f , 0)?
I f (z) =
1
z3
Md. Aquil Khan
Complex Analysis: Singularities
Evaluation of Real Integrals by Residue Methods
Md. Aquil Khan
Complex Analysis: Singularities
Form 1
Consider a real integral of the form
Z 2π
f (cos θ, sin θ) dθ,
0
where f (x, y ) is a rational function defined inside the unit circle
|z| = 1, z = x + iy .
Md. Aquil Khan
Complex Analysis: Singularities
Form 1
Consider a real integral of the form
Z 2π
f (cos θ, sin θ) dθ,
0
where f (x, y ) is a rational function defined inside the unit circle
|z| = 1, z = x + iy .
I Convert into a contour integral around the unit circle |z| = 1
using the substitution
z = eiθ
Md. Aquil Khan
Complex Analysis: Singularities
Form 1
Consider a real integral of the form
Z 2π
f (cos θ, sin θ) dθ,
0
where f (x, y ) is a rational function defined inside the unit circle
|z| = 1, z = x + iy .
I Convert into a contour integral around the unit circle |z| = 1
using the substitution
z = eiθ
eiθ + e−iθ
1
1
I cos θ =
=
z+
2
2 z
eiθ − e−iθ
1
1
sin θ =
=
z−
2i
2i
z
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
Evaluate
0
2π
√
1
dθ using residue method
2 − cos θ
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
Evaluate
0
2π
√
1
dθ using residue method
2 − cos θ
z
= eiθ
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
Evaluate
0
2π
√
1
dθ using residue method
2 − cos θ
z
dz
= eiθ
= ieiθ dθ = izdθ
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
Evaluate
0
2π
√
1
dθ using residue method
2 − cos θ
z
= eiθ
= ieiθ dθ = izdθ
eiθ + e−iθ
1
1
cos θ =
=
z+
2
2
z
dz
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
Evaluate
0
2π
√
1
dθ using residue method
2 − cos θ
z
= eiθ
= ieiθ dθ = izdθ
eiθ + e−iθ
1
1
cos θ =
=
z+
2
2
z
1
1
√
= √
2 − cos θ
( 2 − 12 z + z1 )
dz
Md. Aquil Khan
Complex Analysis: Singularities
Problem
2π
Z
√
Evaluate
0
Z
0
2π
√
1
dθ using residue method
2 − cos θ
1
dθ =
2 − cos θ
Z
C
Md. Aquil Khan
√
iz( 2 −
1
1
2
z+
1
z
dz
)
Complex Analysis: Singularities
Problem
2π
Z
√
Evaluate
0
Z
0
2π
√
1
dθ using residue method
2 − cos θ
1
dθ =
2 − cos θ
Z
√
C
= −
2
i
Md. Aquil Khan
dz
)
1
√
√
(z − 2 − 1)(z − 2 + 1)
iz( 2 −
Z
C
1
1
2
z+
1
z
Complex Analysis: Singularities
Problem
2π
Z
√
Evaluate
0
Z
0
2π
√
1
dθ using residue method
2 − cos θ
1
dθ =
2 − cos θ
Z
C
2
i
2
= −
i
= −
Md. Aquil Khan
√
1
dz
)
1
√
√
C (z − 2 − 1)(z − 2 + 1)
√
× 2πi Res(f , 2 − 1)
iz( 2 −
Z
1
2
z+
1
z
Complex Analysis: Singularities
Problem
2π
Z
√
Evaluate
0
Z
0
2π
√
1
dθ using residue method
2 − cos θ
1
dθ =
2 − cos θ
Z
C
√
1
dz
)
1
√
√
C (z − 2 − 1)(z − 2 + 1)
√
× 2πi Res(f , 2 − 1)
iz( 2 −
Z
1
2
z+
1
z
2
i
2
= −
i
2
1
= − × 2πi × (− )
i
2
= −
Md. Aquil Khan
Complex Analysis: Singularities
Problem
2π
Z
√
Evaluate
0
Z
0
2π
√
1
dθ using residue method
2 − cos θ
1
dθ =
2 − cos θ
Z
C
√
1
dz
)
1
√
√
C (z − 2 − 1)(z − 2 + 1)
√
× 2πi Res(f , 2 − 1)
iz( 2 −
Z
1
2
z+
1
z
2
i
2
= −
i
2
1
= − × 2πi × (− )
i
2
= 2π.
= −
Md. Aquil Khan
Complex Analysis: Singularities
Form 2
Z
∞
f (x) dx,
−∞
where f (x) is a rational function whose denominator
1
is different from zero for all real x
2
is of degree at least two units higher than the degree of the
numerator
Md. Aquil Khan
Complex Analysis: Singularities
Form 2
Z
∞
f (x) dx,
−∞
where f (x) is a rational function whose denominator
Z
1
is different from zero for all real x
2
is of degree at least two units higher than the degree of the
numerator
∞
f (x) dx
−∞
h
= 2πi sum of residues at the poles of f (z) in the
i
upper half-plane
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
1
dx using the residue method
1 + x2
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
f (x) satisfies both the conditions (i) and (ii) given above
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
f (x) satisfies both the conditions (i) and (ii) given above
Z ∞
h
f (x) dx = 2πi sum of residues at the poles of f (z) in
−∞
i
the upper half-plane
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
f (x) satisfies both the conditions (i) and (ii) given above
Z ∞
h
f (x) dx = 2πi sum of residues at the poles of f (z) in
−∞
i
the upper half-plane
Poles : ±i
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
f (x) satisfies both the conditions (i) and (ii) given above
Z ∞
h
f (x) dx = 2πi sum of residues at the poles of f (z) in
−∞
i
the upper half-plane
Poles : ±i
Z
∞
f (x) dx = 2πi Res(f ; i)
−∞
Md. Aquil Khan
Complex Analysis: Singularities
Problem
Z
∞
Evaluate
−∞
f (x) =
1
dx using the residue method
1 + x2
1
1+x 2
f (x) satisfies both the conditions (i) and (ii) given above
Z ∞
h
f (x) dx = 2πi sum of residues at the poles of f (z) in
−∞
i
the upper half-plane
Poles : ±i
Z
∞
1
2i
z }| {
f (x) dx = 2πi Res(f ; i) = π
−∞
Md. Aquil Khan
Complex Analysis: Singularities