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7. Cauchy’s Residue Theorem
§7.1 Introduction
One of the more remarkable applications of integration in the complex plane in general, and Cauchy’s theorem in particular, is that it gives a method for calculating real integrals that, up until now, would have been difficult or even impossible (assuming that you only had the tools of 1st year calculus or A-level mathematics to hand). As another application: you may remember from MATH10242 Sequences and Series that you studied whether an infinite series
n=0 an^ converged or not. However, in only very few examples were you able to say what the limit actually is! Using complex analysis, it becomes very easy to prove formulæ such as
n=1 1 /n (^4) = π (^4) /90.
§7.2 Zeros and poles of holomorphic functions
Recall that a function f has a singularity at z 0 if f is not differentiable at z 0. We will only consider the case when f has poles as singularities, and in the examples we have seen so far f (z) has a pole at z 0 because we can write f (z) = p(z)/q(z) and q(z 0 ) = 0 (so that f is not even defined at z 0 ). Thus it makes sense to first study zeros of functions.
Definition. A holomorphic function f on a domain D has a zero at z 0 if f (z 0 ) = 0.
By Taylor’s theorem, we can expand f as a Taylor series in some neighbourhood around z 0. That is we can wrote
f (z) =
∑^ ∞
n=
an(z − z 0 )n^ (7.2.1)
for all z in some disc that contains z 0.
Definition. We say that f has a zero of order m at z 0 if a 0 = a 1 = · · · = am− 1 = 0 but am 6 = 0. We say that z 0 is a simple zero if it is a zero of order 1.
Example. (i) Let f (z) = z^2. Then f has a zero of order 2 at 0.
(ii) Let f (z) = z(z + 2i)^3. Then f has a zero of order 1 at 0 and a zero of order 3 at − 2 i.
Remark. The coefficients an in the Taylor expansion are given by an = f (n)(z 0 )/n!. Thus f has a zero of order m at z 0 if and only if f (k)(z 0 ) = 0 for 0 ≤ k ≤ m − 1 but f (m)(z 0 ) 6 = 0. In partiocular, if f (z 0 ) = 0 but f ′(z 0 ) 6 = 0 then z 0 is a simple zero.
Example. (i) Let f (z) = sin z. Then f (z) has zeros at kπ, k ∈ Z. Note that f ′(kπ) = cos kπ = (−1)k^6 = 0. Hence all the zeros are simple zeros.
(ii) Let f (z) = 1 − cos z. Then f (z) has zeros at 2kπ, k ∈ Z. Now f ′(z) = sin z and f ′(2kπ) = 0, but f ′′(2kπ) = cos 2kπ = 1 6 = 0. Hence all the zeros have order 2.
Lemma 7.2. Suppose that the holomorphic function f has a zero of order m at z 0. Then, on some disc centred at z 0 , we can write f (z) = (z − z 0 )mg(z)
where g is a holomorphic function on an open disc centred on z 0 and g(z 0 ) 6 = 0.
Proof. By (7.2.1) we can write
f (z) = am(z − z 0 )m^ + am+1(z − z 0 )m+1^ + · · · = (z − z 0 )m
∑^ ∞
n=
an+m(z − z 0 )n.
Take g(z) =
n=0 an+m(z^ −^ z^0 ) n. 2
We can now link poles of a function f (z) = p(z)/q(z) with zeros of the function q.
Lemma 7.2. Suppose that f (z) = p(z)/q(z) where
(i) p is holomorphic and p(z 0 ) 6 = 0,
(ii) q is holomorphic and q has a zero of order m at z 0.
Then f has a pole of order m at z 0.
Proof. By Lemma 7.2.1, we can write q(z) = (z − z 0 )mr(z) where r is holomorphic and r(z 0 ) 6 = 0. Define g(z) = p(z)/r(z). Then g(z) is holomorphic at z 0 , and so we can expand it as a Taylor series at z 0 :
g(z) =
∑^ ∞
n=
an(z − z 0 )n
and this expression is valid in some disc |z − z 0 | < R, for some R > 0. Then
f (z) = p(z) q(z)
= p(z) (z − z 0 )mr(z)
=
g(z) (z − z 0 )m
=
(z − z 0 )m
∑^ ∞
n=
an(z − z 0 )n
a 0 (z − z 0 )m^
a 1 (z − z 0 )m−^1
a 2 (z − z 0 )m−^2
However, a 0 = g(z 0 ) = p(z 0 )/r(z 0 ) 6 = 0, as p(z 0 ) 6 = 0. Hence f has a pole of order m at z 0. 2
Example. (i) Let
f (z) = sin z (z − 3)^2
Then f has a pole of order 2 at z = 3. This is because sin z 6 = 0 when z = 3 and (z − 3)^2 has a zero of order 2 at z = 3.
Theorem 7.3.1 (Cauchy’s Residue Theorem) Let D be a domain containing a simple loop γ and the points inside γ. Suppose that f is meromorphic on D with finitely many isolated poles at z 1 , z 2 ,... , zn inside γ. Then
∫
γ
f (z) dz = 2πi
∑^ n
r=
Res(f, zr ).
Remark. A word of warning: you will have noticed that many expressions in complex analysis have a factor of 2πi in them. A very common mistake is to either miss a 2πi out, or put one in by mistake.
We shall defer the proof of the Residue Theorem until later.
§7.4 Calculating residues
In order to use the Residue Theorem we need to be able to easily calculate residues. In some cases, ad hoc manipulations have to be used to calculate the Laurent series, but there are many cases where one can calculate them more systematically. First recall that if f (z) has Laurent series
f (z) = bm (z − z 0 )m^
b 1 (z − z 0 )
∑^ ∞
n=
an(z − z 0 )n
then we say that f has a pole of order m at z 0. We say that a pole of order 1 is a simple pole.
Remark. If we can write f (z) = p(z)/q(z) where p and q are differentiable and p(z) 6 = 0 when q(z) = 0 then the poles of f occur at the zeros of q. Moreover f has a pole of order m at z 0 if q has a zero of order m at z 0.
It is easy to calculate the residue at a simple pole.
Lemma 7.4. (i) If f has a simple pole at z 0 then
Res(f, z 0 ) = lim z→z 0 (z − z 0 )f (z).
(ii) If f (z) = p(z)/q(z) where p(z 0 ) 6 = 0, q(z 0 ) = 0 but q′(z 0 ) 6 = 0, then
Res(f, z 0 ) =
p(z 0 ) q′(z 0 )
Proof. (i) If f has a simple pole at z 0 then it has Laurent expansion
f (z) =
b 1 z − z 0
∑^ ∞
n=
an(z − z 0 )n.
Hence (z − z 0 )f (z) = b 1 +
∑^ ∞
n=
an(z − z 0 )n+
so that limz→z 0 (z − z 0 )f (z) = b 1.
(ii) The hypotheses imply that f has a simple pole at z 0. By part (i) and the fact that q(z 0 ) = 0, the residue is
zlim→z 0
(z − z 0 )p(z) q(z) = lim z→z 0
p(z) ( q(z)−q(z 0 ) z−z 0
p(z 0 ) q′(z 0 )
Example. For example, let
f (z) =
cos πz (1 − z^976 )
This has a simple pole at z = 1 and satisfies the hypothesis of Lemma 7.4.1. Hence
Res(f, 1) = cos π (−976) × 1975
We can generalise Lemma 7.4.1 to poles of order m.
Lemma 7.4. Suppose that f has a pole of order m at z 0. Then
Res(f, z 0 ) = lim z→z 0
(m − 1)!
dm−^1 dzm−^1 ((z − z 0 )mf (z))
Proof. If f has a pole of order m at z 0 then it has a Laurent series
f (z) = bm (z − z 0 )m^
b 1 z − z 0
∑^ ∞
n=
an(z − z 0 )n
valid for 0 < |z − z 0 | < R, for some R > 0. Hence
(z − z 0 )mf (z) = bm + (z − z 0 )bm− 1 + · · · + (z − z 0 )m−^1 b 1 +
∑^ ∞
n=
an(z − z 0 )m+n.
Differentiating this m − 1 times gives
dm−^1 dzm−^1 (z − z 0 )mf (z) = (m − 1)!b 1 +
∑^ ∞
n=
(m + n)! (n + 1)! an(z − z 0 )n+1.
Dividing by (m − 1)! and letting z → z 0 gives the result. 2
Example. Let
f (z) =
z + 1 z − 1
This has a pole of order 3 at z = 1. To calculate the residue we note that (z − 1)^3 f (z) = (z + 1)^3. Hence 1 2!
d^2 dz^2
(z − 1)^3 f (z)
(z + 1) →
× 2 = 6
as z → 1. Hence Res(f, 1) = 6.
§7.5 Applications
§7.5.1 Easy examples
We shall evaluate some simple integrals around the circular contours C 2 (t) = 2eit, 0 ≤ t ≤ 2 π and C 4 (t) = 4eit, 0 ≤ t ≤ 2 π. Thus C 2 is the circle of radius 2 centred at 0 described anticlockwise, and C 4 is the circle of radius 4 centred at 0 described anticlockwise. Hence both C 2 and C 4 are simple loops. Consider the function f (z) =
z − 1
Then f has a pole at z = 1 and no other poles. We can read off from the definition of f that Res(f, 1) = 3. As the pole at z = 1 lies inside C 2 , by the Residue Theorem we have that (^) ∫
C 2
f dz = 2πi Res(f, 1) = 6πi.
Similarly, the pole at z = 1 lies inside C 4 , hence ∫
C 4
f dz = 2πi Res(f, 1) = 6πi.
See Figure 7.5.1.
1 2 4
C 2
C 4
Figure 7.5.1: The function f (z) = 3/(z − 1) has a pole at z = 1 which lies inside both C 2 and C 4.
Now consider the function
f (z) =
z^2 + (i − 3) − 3 i
Then f has a pole when the denominator has a zero. To find the poles we first factorise the denominator” z^2 + (i − 3)z − 3 i = (z − 3)(z + i)
(to do this we could either use the quadratic formula or inspired guesswork). Thus f has simple poles z = 3 and z = −i. Using Lemma 7.4.1 we can calculate that
Res(f, −i) =
3 + i
, Res(f, 3) =
3 + i
See Figure 7.5.2.
2 4
C 2
C 4
3 −i
Figure 7.5.2: The function f (z) = 1/(z^2 + (i − 3)z − 3 i)) has simple poles at z = −i and z = 3.
Now consider
C 2 f dz.^ The pole^ z^ =^ −i^ is inside^ C^2 but the pole^ z^ = 3 is outside. Hence ∫
C 2
f dz = 2 πi Res(f, −i) = 2πi
3 + i
− 2 πi(3 − i) 10
=
− 2 π − 6 πi 10
−π 5 (1 + 3i).
Now consider
C 4 f dz. In this case, both the poles at^ z^ =^ −i^ and^ z^ = 3 lie inside^ C^4. Hence (^) ∫
C 4
f dz = 2πi (Res(f, −i) + Res(f, 3)) = 2πi
3 + i
3 + i
§7.5.2 Infinite real integrals
In this section we shall show how to use the Residue Theorem to calculate some infinite real integrals, i.e. integrals of the form ∫ (^) ∞
−∞
f (x) dx (7.5.1)
where f is a real-valued function defined on the real line. First we need to make precise what (7.5.1) means. Formally, we say that
−∞ f^ (x)^ dx exists if
lim A,B→∞
∫ B
−A
f (x) dx (7.5.2)
(note that we have introduced a complex variable). Let [−R, R] denote the path along the real axis that starts at −R and ends at R. This has parametrisation t, −R ≤ t ≤ R. Note that we can equate the real integral (7.5.5) with the complex integral as follows:
∫ (^) R
−R
(x^2 + 1)(x^2 + 4) dx =
[−R,R]
f (z) dz.
To use the Residue Theorem, we need a closed contour. Introduce a semi-circular path SR(t) = Reit, 0 ≤ t ≤ π and the ‘D-shaped’ contour ΓR = [−R, R] + SR (see Figure 7.5.3).
−R (^) R
Figure 7.5.3: The ‘D-shaped’ contour ΓR. It starts at −R, travels along the real axis to R, and then anticlockwise along the semicircle SR with centre 0 and radius R.
Now ΓR is a simple closed loop. To use the Residue Theorem, we need to know the poles and residues of f (z). Now
f (z) =
(z^2 + 1)(z^2 + 4)
(z − i)(z + i)(z − 2 i)(z + 2i)
Hence f (z) has simple poles at z = +i, −i, +2i, − 2 i. If we take R > 2 then the poles at z = i, 2 i lie inside ΓR (note that the poles at z = −i, − 2 i lie outside ΓR). Now by Lemma 7.4.1,
Res(f, i) = lim z→i (z − i)f (z)
= lim z→i
(z + i)(z − 2 i)(z + 2i)
6 i
and
Res(f, 2 i) = lim z→ 2 i (z − 2 i)f (z)
= lim z→ 2 i
(z − i)(z + i)(z + 2i)
12 i
Hence by the Residue Theorem ∫
[−R,R]
f (z), dz +
SR
f (z) dz =
ΓR
f (z) dz
= 2 πi (Res(f, i) + Res(f, 2 i))
= 2 πi
6 i
12 i
π 6
If we can show that lim R→∞
SR
f (z) dz = 0 (7.5.6)
then we will have that ∫ (^) ∞
−∞
(x^2 + 1)(x^2 + 4) dx = lim R→∞
[−R,R]
f (z) dz =
π 6
To complete the calculation, we show that (7.5.6) holds. We shall use the Estimation Lemma. Let z be a point on SR. Note that |z| = R. Hence
|(z^2 + 1)(z^2 + 4)| ≥ (R^2 − 1)(R^2 − 4)
so that (^) ∣ ∣∣ ∣
(z^2 + 1)(z^2 + 4)
∣ ≤^
(R^2 − 1)(R^2 − 4)
Hence, by the Estimation Lemma, ∣ ∣ ∣∣
SR
f (z) dz
(R^2 − 1)(R^2 − 4)
length(SR)
πR (R^2 − 1)(R^2 − 4) → 0
as R → ∞, which is what we wanted to check.
Remark. As a general method, to evaluate
∫ (^) R
−R
f (x) dx
one uses the following steps:
(i) Construct a ‘D-shaped’ contour ΓR as in Figure 7.5.3.
(ii) Find the poles and residues of f (z) that lie inside ΓR when R is large.
(iii) Use the Residue Theorem to write down
ΓR f^ (z)^ dz. (iv) Split this integral into an integral over [−R, R] and an integral over SR. Use the Estimation Lemma to conclude that the integral over SR converges to 0 as R → ∞.
For a particular example, one may need to make small modifications to the above process, but the general method is normally as above.
C 1
z + z−^1 2
z − z−^1 2 i
dz iz
=
C 1
z^3 8
3 z 8
3 z−^1 8
z−^3 8
z^2 4
z−^2 4
dz iz
=
C 1
i
z^2 8
3 z−^2 8
z−^4 8
z 4
z−^1 2
z−^3 4
dz
The integrand has a pole of order 4 at z = 0 with residue 1/ 2 i, and no other poles. Hence
∫ (^2) π
0
(cos^3 t + sin^2 t) dt = 2πi
2 i
= π.
Example. We shall compute (^) ∫ 2 π
0
cos t sin t dt.
Again, substituting z = eit^ we have that ∫ (^2) π
0
cos t sin t dt
C 1
4 i
(z + z−^1 )(z − z−^1 ) dz iz
=
C 1
4 i (z^2 − z−^2 ) dz iz
=
C 1
z^2 −
z^3
dz.
The integrand has a pole of order 3 at z = 0 with residue 0. There are no other poles. Hence (^) ∫ (^2) π
0
cos t sin t dt = 0.
§7.5.4 Summation of series
Recall that cot πz = cos πz/ sin πz. Then cot πz has a pole whenever sin πz = 0, i.e. whenever z = n, n ∈ Z. First note that sin πz has a simple zero at z = n (as sin′^ πz = π cos πz 6 = 0 when z = n). Hence cot πz has a simple pole at z = n. By Lemma 7.4.1(ii) we have Res(cot πz, n) = cos πn π cos πn
π
This suggests a method for summing infinite series of the form
n=1 an.^ Let^ f^ (z) be a meromorphic function defined on C such that f (n) = an. Consider the function f (z) cot πz. Then Res(f (z) cot πz) =
an π
and we can use the Residue Theorem to calculate
n=1 an. For example, we will show how to use this method to calculate
n=1 1 /n
There are two technicalities to overcome. First of all, we need to choose a good contour to integrate round. We will want to use the Estimation Lemma along this contour, so we will need some bounds on |f (z) cot(πz)|. Secondly, f (z) may have poles of its own and
these will need to be taken into account. (In the above example, to calculate
n=1 1 /n 2
we will take f (z) = 1/z^2 , which has a pole at z = 0.) Instead of choosing a D-shaped contour, here we use a square contour. Let CN denote the square in C with vertices at ( N +
− i
N +
N +
N +
N +
N +
N +
− i
N +
(see Figure 7.5.4). This is a square with each side having length 2N + 1. (The factors of 1 /2 are there so that the sides of this square do not pass through the integer points on the real axis.)
−(N + 1) −N − 1 0 1 N N + 1
Figure 7.5.4: The square contour CN.
Lemma 7.5. There is a bound, independent of N , on cot πz where z ∈ CN , i.e. there exists M > 0 such that for all N and all z ∈ CN , we have | cot πz| ≤ M.
Proof. Consider the square CN. This has two horizontal sides and two vertical sides, parallel to the real and imaginary axes, respectively. Consider first the horizontal sides. Let z = x + iy be a point on one of the horizontal sides of CN. Then |y| ≥ 1 /2. Hence
| cot πz| =
∣∣^ e
iπz (^) + e−iπz eiπz^ − e−iπz
∣eiπz^
∣e−iπz^
|eiπz^ | − |e−iπz^ |
z^2
(πz)^2 2!
(πz)^4 4!
(πz) − (πz)^3 3!
(πz)^5 5!
z^2
πz
(πz)^2 2!
(πz)^4 4!
(πz)^2 3!
(πz)^4 5!
z^2
πz
(πz)^2 2!
(πz)^4 4!
(πz)^2 3!
(πz)^4 5!
πz^3
(πz)^2 3
so that Res(cot πz/z^2 , 0) = −π/3. (We used the expansion (1−x)−^1 = 1+x+x^2 +· · ·.) Note that to calculate the residue we need only calculate the coefficient of the term involving 1 /z; hence we need to be very careful when manipulating these infinite sums to ensure that we account for all the possible places which may contribute towards 1/z. Alternatively, from real analysis we have the following power series expansion for cot z:
cot z =
z
z 3
z^3 45
2 z^5 945
Hence cot πz z^2
πz^3
π 3 z
π^3 z 45
2 π^5 z^3 945
from which it is clear that z = 0 is a pole of order 3 with residue −π/3. Finally, one could use Lemma 7.4.2 to calculate the residue. Now let CN be the square contour illustrated above. Note that each side of the square has length 2N + 1. Hence the length of CN is 4(2N + 1). Note that the poles that lie inside CN occur at z = 0, ± 1 , · · · , ±N. By the Residue Theorem we have that
2 πi
∑^ N
n=−N
Res
cot πz z^2
, n
CN
cot πz z^2
dz.
Recall from Lemma 7.5.2 that | cot πz| ≤ M on CN , where M is independent of N. Also note that | 1 /z^2 | ≤ 1 /N 2 for z on CN. By the Estimation Lemma we have ∣ ∣∣ ∣
CN
cot πz z^2
dz
∣ ≤^
M
N 2
length CN =
M
N 2
4(2N + 1)
which tends to 0 as N → ∞. Hence
lim N →∞
∑^ N
n=−N
Res
cot πz z^2
, n
Now
∑^ N
n=−N
Res
cot πz z^2 , n
∑^ −^1
n=−N
Res
cot πz z^2
, n
cot πz z^2
∑^ N
n=
Res
cot πz z^2
, n
∑^ N
n=
πn^2
π 3
and combining this with (7.5.8) we see that
∑^ ∞
n=
πn^2
π 3
This rearranges to give ∑∞
n=
n^2
π^2 6
§7.6 Proof of the Residue Theorem
Let us first recall the statement of the Residue Theorem:
Theorem 7.6.1 (Cauchy’s Residue Theorem) Let D be a domain containing a simple loop γ and the points inside γ. Suppose that f is holomorphic on D except for finitely many isolated poles at z 1 , z 2 ,... , zn inside γ. Then
∫
γ
f (z) dz = 2πi
∑^ n
r=
Res(f, zr ).
Proof. The proof is a simple application of the Generalised Cauchy Theorem (Theo- rem 4.5.7). Since D is open, for each r = 1,... , n, we can find circles
Sr(t) = zr + εreit, 0 ≤ t ≤ 2 π
centred at zr and of radii εr such that Sr and the points inside Sr lie in D and such that Sr contains no singularity other than zr (see Figure 7.6.1). Let D′^ = D \ {z 1 ,... , zn}. We claim that the collection of paths
−γ, S 1 ,... , Sn
satisfy the hypotheses of the Generalised Cauchy Theorem (Theorem 4.5.7) with respect to D′: i.e. their winding numbers sum to zero for every point not in D′. To see this, first note that
w(−γ, z) = w(Sr , z) = 0 for z 6 ∈ D.
Hence the hypotheses of the Generalised Cauchy Theorem hold for points z not in D. It remains to consider points in D that are not in D′, i.e. the poles zr. Since each pole zr lies inside γ, we have that
w(−γ, zr ) = −w(γ, zr ) = − 1.
Moreover,
w(Sj , zr ) =
0 if j 6 = r 1 if j = r.
Hence w(−γ, zr ) + w(S 1 , zr) + · · · + w(Sn, zr) = 0.
Exercise 7. (a) Consider the following real integral: ∫ (^) ∞
−∞
x^2 + 1
dx.
(i) Explain why this integral is equal to its principal value. (ii) Use the Residue Theorem to evaluate this integral. (How would you have done this without using complex analysis?)
(b) (i) Now evaluate, using the Residue Theorem, the integral ∫ (^) ∞
−∞
e^2 ix x^2 + 1
dx.
(ii) By taking real and imaginary parts, calculate ∫ (^) ∞
−∞
cos 2x x^2 + 1 dx,
−∞
sin 2x x^2 + 1 dx.
(Why is it obvious that one of these integrals is zero?) (iii) Why does the ‘D-shaped’ contour used in the lectures for calculating such inte- grals fail when we try to integrate ∫ (^) ∞
−∞
e−^2 ix x^2 + 1
dx?
By choosing a different contour, explain how one could evaluate this integral using the Residue Theorem.
Exercise 7. Use the Residue Theorem to evaluate the following real integrals:
(i)
−∞
(x^2 + 1)(x^2 + 3) dx, (ii)
−∞
28 + 11x^2 + x^4 dx.
Exercise 7. By considering the function
f (z) = eiz z^2 + 4z + 5 integrated around a suitable contour, show that ∫ (^) ∞
−∞
sin x x^2 + 4x + 5 dx =
−π sin 2 e
Exercise 7. Evaluate the integral discussed in §1.1:
∫ (^) ∞
−∞
x sin x (x^2 + a^2 )(x^2 + b^2 ) dx
by integrating around a suitable contour.
Exercise 7. Convert the following real integrals into complex integrals around the unit circle in the complex plane. Hence use the Residue Theorem to evaluate them.
(i)
∫ (^2) π
0
2 cos^3 t + 3 cos^2 t dt, (ii)
∫ (^2) π
0
1 + cos^2 t dt.
Exercise 7. Use the method of summation of series to show that
n=1 1 /n (^4) = π (^4) /90. Why doesn’t the method work for evaluating
n=1 1 /n
Exercise 7. Suppose a 6 = 0. Consider the function
cot πz z^2 + a^2
Show that this function has poles at z = n, n ∈ Z and z = ±ia. Calculate the residues at these poles. Hence show that (^) ∞ ∑
n=
n^2 + a^2
π 2 a coth πa −
2 a^2
Exercise 7. (Sometimes, one has to be rather creative in picking the right contour...) Let 0 < a < 1. Show that (^) ∫ (^) ∞
−∞
eaz 1 + ez^
π sin aπ
using the following steps.
(i) Show that this integral exists and is equal to its principal value.
(ii) Let f (z) = eaz^ /(1 + ez^ ). Show that f is holomorphic except for simple poles at z = (2k + 1)πi, k ∈ Z. Draw a diagram to illustrate where the poles are. Calculate the residue Res(f, πi).
(iii) On the diagram from (ii), draw the contour ΓR = γ 1 ,R + γ 2 ,R + γ 3 ,R + γ 4 ,R where:
γ 1 ,R is the horizontal straight line from −R to R,
γ 2 ,R is the vertical straight line from R to R + 2πi,
γ 3 ,R is the horizontal straight line from R + 2πi to −R + 2πi,
γ 4 ,R is the vertical straight line from −R + 2πi to −R.
Which poles does Γ∫ R wind around? Use Cauchy’s Residue Theorem to calculate ΓR f^. (iv) Show, by choosing suitable parametrisations of the paths γ 1 ,R and γ 3 ,R and direct computation, that
γ 3 f^ =^ −e 2 πia ∫ γ 1 f^.
