solution
Content
Mathematical Techniques III (PHY 317) Solutions to Problem Set 6
Solution to Problem 1. Let z (t) = x(t) + i y (t) describe the position of a particle moving in the complex plane. Using the equivalence between the complex plane and 42 we see that z (t) carries the same information as the vector (x(t), y (t)). Therefore we see that the expression z ˙ (t) corresponds to the vector (x ˙ (t), y ˙ (t)) which is precisely the (instantaneous) velocity of the particle. The quantity |z ˙ (t)| = x ˙ (t)2 + y ˙ (t)2 is the norm of the velocity vector (x ˙ (t), y ˙ (t)) , whence it is the speed (the magnitude of the velocity) of the particle. Finally the integral
1 1
|z ˙ (t)| dt =
0 0
x ˙ (t)2 + y ˙ (t)2 dt
is nothing but the integral along the contour of the infinitesimal line element dx2 + dy 2 , hence the integral is the length of the contour. Solution to Problem 2. Consider the integral f (z ) dz =
Γ Γ
(3z 2 − 5z + i) dz
where Γ is the straight line segment from z = i to z = 1. This contour can be parametrised as z (t) = t + i (1 − t) for t ∈ [0, 1]. Then after some algebra f (z (t)) = 3z (t)2 − 5z (t) + i = (t − 3) − i (6t2 − 11t + 4). The velocity is given by z ˙ (t) = 1 − i. Therefore, (3z 2 − 5z + i) dz =
Γ 0 1 1
(t − 3) − i (6t2 − 11t + 4) (1 − i) dt (−6t2 + 12t − 7) − i (6t2 − 10t + 1) dt
=
0
= −3 + 2 i . Now, although this is the right answer, this is not the most efficient way to solve this problem. The best way to do this problem is to realise that the integrand f (z ), being a polynomial, is entire and hence, by the Cauchy Integral Theorem, its integral only depends on the endpoints: f (z ) dz = F (1) − F (i) ,
Γ
1
where F (z ) is an antiderivative, which is guaranteed to exist, but must be found. Being a polynomial, we can guess this easily: F (z ) = z 3 − 5 z2 + i z 2 Therefore, f (z ) dz = F (1) − F (i) = −3 + 2 i ,
Γ
=⇒
F (z ) = 3z 2 − 5z + i = f (z ) .
the same result at a fraction of the cost. Consider now the integral ez dz
Γ
where Γ is the upper half of the circle |z | = 1 from z = 1 to z = −1. This time the parametrised integral would be quite messy, so we use the Cauchy Integral Theorem directly. The exponential function is again entire and moreover is its own antiderivative, so that
1 −1 ez dz = ez |− − e1 = −2 sinh(1) . 1 = e
Γ
Next, we compute the integral 1 dz z
Γ
where Γ is any contour in the right half-plane from z = −3i to z = 3i. The function f (z ) = 1/z has an antiderivative F (z ) = Log(z ) in the domain D consisting of all points in the complex plane except for the origin and the negative real axis. Since the contour Γ lies in this domain, we can use this branch. Therefore, we find 1 π π dz = Log(3i) − Log(−3i) = Log(3) + i − Log(3) − i z 2 2 Log(z )2 dz ,
Γ
= iπ .
Γ
Next we have
where Γ is the straight line segment from z = 1 to z = i. The principal branch Log(z ) of the logarithm function is analytic in the domain D defined above and Γ lies in D. Therefore, we can compute this integral by finding an antiderivative F (z ) for f (z ) = Log(z )2 , which is analytic in this domain. It is easy to check that F (z ) = z Log(z )2 − 2 z Log(z ) + 2 z 2
is an antiderivative. Therefore, Log(z )2 dz = F (i) − F (1)
Γ
= i Log(i)2 − 2i Log(i) + 2i − Log(1)2 − 2 Log(1) + 2 = (π − 2) + i 2 − 1 π2 . 4 Finally, we consider the integral
Γ
1 dz 1 + z2
where Γ is the straight line segment from z = 1 to z = 1 + i. Let us use partial fractions to rewrite the integrand as a sum of two terms, i i 1 2 2 = − , 1 + z2 z+i z−i each of which has an antiderivative in a domain containing the contour of integration: d 1 Log(z ± i) = . dz z±i Therefore, 1 i i i i dz = Log(z + i)|1+ − Log(z − i)|1+ 1 1 2 2 2 Γ 1+z i i = (Log(1 + 2i) − Log(1 + i)) − (Log(1) − Log(1 − i)) 2 2 i i i = Log(1 + 2i) + Log(1 − i) − Log(1 + i) . 2 2 2 where we have used that Log(1) = 0. In order to compute the principal values of the logarithms, it is necessary to write the complex numbers involved in polar form: √ 1 + 2i = 5eiθ where tan θ = 2 √ 1 − i = 2e−iπ/4 √ 1 + i = 2eiπ/4 . Therefore, Log(5) + i Arctan(2) Log(1 + 2i) = 1 2 π 1 Log(1 − i) = 2 Log(2) − i 4 π 1 Log(1 + i) = 2 Log(2) + i , 4 3
whence 1 i i i dz = Log(1 + 2i) + Log(1 − i) − Log(1 + i) 2 1+z 2 2 2 i π = Log(5) − 1 Arctan(2) + . 2 2 4
Γ
Solution to Problem 3. Let Γ be the contour parametrised by z (t) = 2 cos(t) + i sin(2t) for t ∈ [0, 2π ]. The periodicity of the trigonometric functions guarantee that this is a closed curve starting and ending at z = 2. It is clear that as the x-coordinate goes from 2 to −2 and back to 2 once, the y -coordinate goes from 0 to 1 to −1 and back to 0 twice. Therefore the resulting curve will be a horizontal figure 8 (more like a ∞) crossing itself at the origin at t = π/2 and t = 3π/2, as in the figure below. The “wrong” way to do this problem would be to plug in the explicit parametrisation and perform the t integral by brute force. Instead we shall use Cauchy’s Integral Theorem in order to evaluate them. The four functions that we are asked to integrate along this contour are analytic everywhere except at z = ±1. Therefore we can use Cauchy’s Integral Theorem to argue that the result of the contour integral along Γ is the same as the sum of the contour integrals around the small circles Γ+ and Γ− surrounding the singular points z = ±1, as in the figure below. Notice, however, that the contour Γ− is negatively oriented.
−1 •
Γ
+1 •
Γ− −1 • +1 •
Γ+
©
In other words, for each of the functions f (z ) to be integrated, Cauchy’s Integral Theorem tells us that f (z ) dz =
Γ Γ+
f (z ) dz +
Γ−
f (z ) dz .
Let us then use this to compute the integrals. First we compute 1 dz = z−1 1 dz + z−1 4 1 dz . z−1
Γ
Γ+
Γ−
This function is analytic everywhere but at z = +1, hence by Cauchy’s Integral Theorem its integral along Γ− vanishes. Therefore, 1 dz = z−1 1 dz = 2π i , z−1
Γ
Γ+
as we discussed in the lecture. Next we compute 1 dz = z+1 1 dz + z+1 1 dz . z+1
Γ
Γ+
Γ−
This function is now analytic everywhere but at z = −1, hence by Cauchy’s Integral Theorem its integral along Γ+ vanishes. Therefore, 1 dz = z+1 1 dz . z+1
Γ
Γ−
This contour is negatively oriented, so now 1 dz = − z+1 1 dz = −2π i . z+1
Γ
−Γ−
The next integral is
Γ
z2
1 dz . −1
Using partial fractions, we can write
1 1 1 2 = − 2 , z2 − 1 z−1 z+1
whence
Γ
z2
1 dz = −1
1 2
Γ
1 dz − z−1
1 2
Γ
1 dz , z+1
which can be read off from the two integrals we have just performed, so that 1 dz = 1 2π i − 1 (−2π i) = 2π i . 2 2 2 z − 1 Γ Finally we compute 2z 2 − z + 1 dz . (z − 1)2 (z + 1) 5
Γ
Again we use partial fractions to write 2z 2 − z + 1 1 1 1 = + + ; 2 2 (z − 1) (z + 1) z − 1 (z − 1) z+1 whence 2z 2 − z + 1 dz = (z − 1)2 (z + 1) 1 dz + z−1 1 dz + (z − 1)2 1 dz . z+1
Γ
Γ
Γ
Γ
We have down the first and the last integrals before and we see that they cancel, so that all we have to do is the middle integral. Using the fact that the integrand is analytic everywhere but at z = 1, we can write this as an integral over the small circle Γ+ : 2z 2 − z + 1 dz = (z − 1)2 (z + 1) 1 dz . (z − 1)2
Γ
Γ+
This integral can be seen to be zero in two ways: either appealing to the generalised Cauchy Integral formula: f (z ) dz = 2π i f (1) , (z − 1)2
Γ+
and using that f (z ) = 1 is a constant in this case; or directly by computing the contour integral. We can parametrise Γ+ by z = 1 + r exp(2π i t) for t ∈ [0, 1], so that z ˙ (t) = 2π i r exp(2π i t). Therefore 1 2π i dz = 2 (z − 1) r
1 0
Γ+
1 e−2π i t dt = − e−2π i t r
1 0
=0,
by the periodicity of the exponential. Therefore any way one looks at it, one finds 2z 2 − z + 1 dz = 0 . 2 Γ (z − 1) (z + 1) Solution to Problem 4. We are asked to evaluate the integral eiz dz (z 2 + 1)2
Γ
where Γ is the circle |z | = 3 traversed once counterclockwise. The integrand is analytic everywhere but at z = ± i. Therefore by the 6
Γ
I
•i •
Γ+
I
−i •
Γ−
R
•
Cauchy Integral Theorem, the integral around Γ gives the same result as the integral around two small circles Γ± centred about z = ± i, respectively, as in the figure. In other words, eiz dz = (z 2 + 1)2 eiz dz + (z 2 + 1)2 eiz dz . (z 2 + 1)2
Γ
Γ+
Γ−
We now use partial fractions to write
i 1 i 1 1 4 4 4 4 = − − − . (z 2 + 1)2 z + i (z + i)2 z − i (z − i)2
The first two terms only contribute to the integral along Γ− , whereas the last two terms only contribute to the integral along Γ+ . For each of these we can use the (generalised) Cauchy Integral Formula: eiz dz = 2π i e z+i
z =−i
Γ−
Γ−
d eiz dz = 2π i eiz 2 (z + i) dz
Γ+
= −2π e
eiz dz = 2π i e−1 z−i
z =i
Γ+
eiz d dz = 2π i eiz 2 (z − i) dz
= −2π e−1 .
Collating all the terms we find that eiz π i i dz = 4 2π i e − 1 (−2π e) − 4 2 π i e −1 − 1 (−2π e−1 ) = . 4 4 2 2 (z + 1) e
Γ
Next we consider the integral
Γ
z3 7
z+i dz + 2z 2
Γ2 Γ1 • −2
I I •0
• −2 •0 • −2 Γ3 •0
I
where Γ is each of the contours Γ1 , Γ2 and Γ3 depicted in the figure. The integrand is analytic everywhere except for z = 0 and z = −2, as can be seen easily if we use partial fractions to rewrite it in the following form:
i z+i 2 = + z 3 + 2z 2 z2 1 2
− z
i 4
−
i −4 . z+2
1 2
We can now use the (generalised) Cauchy Integral Formula to evaluate the integral of each of the above terms along the contours Γk , for k = 1, 2, 3: 1 dz = 0 z2 1 dz = 0 z2 1 dz = 0 z2 1 dz = 2π i z 1 dz = 0 z 1 dz = 0 z 1 dz = 0 z+2 1 dz = 2π i z+2 1 dz = 0 . z+2
Γ1
Γ1
Γ1
Γ2
Γ2
Γ2
Γ3
Γ3
Γ3
Collating the terms, we find z+i dz = 1 π + iπ 2 2 + 2z z+i 1 dz = − 2 π − iπ 3 2 z + 2z z+i dz = 0 . 3 z + 2z 2 z3
Γ1
Γ2
Γ3
Solution to Problem 5. In this problem we are asked to integrate several functions along the 8
same contour: the positively oriented circle of radius 2 centred about the origin. In the figure we have depicted this contour in relation to the points at which the integrands below fail to be analytic.
• 3i
•i −1 • •0
I
3 2
••π
2
Γ
−3i •
We start with
Γ
sin(3z ) dz . z−1 π 2
The integrand is analytic everywhere but at z = 1 π which lies in the 2 interior of the contour. Therefore by the Cauchy Integral Formula,
Γ
sin(3z ) dz = 2π i sin z−1 π 2
3 π 2
= −2π i .
zez dz , Γ 2z − 3 whose integrand is analytic except at z = 3 . Since this point lies inside 2 Γ, we can use the Cauchy Integral Formula to obtain zez dz = 2z − 3
1 zez 2 3 Γ z− 2 3 3/2 e =i dz = 2π i 1 22
Next we have
Γ
3π 3/2 e . 2
We continue with
Γ
cos(z ) dz = z 3 + 9z
Γ
cos(z ) dz , z (z + 3i)(z − 3i)
whose integrand fails to be analytic at z = 0 and z = ±3i. Of these points only z = 0 lies in the interior of Γ, hence by the Cauchy Integral Formula 2π cos(z ) cos(z ) =i dz = 2π i 2 . z + 9 z=0 9 Γ z (z + 3i)(z − 3i) 9
5z 2 + 2z + 1 dz , (z − i)3 Γ whose integrand fails to be analytic at z = i, which lies in the interior of Γ. We can appeal to the generalised Cauchy Integral Formula to obtain 2 5z 2 + 2z + 1 1 d = i 10 π . dz = 2 π i 5z 2 + 2z + 1 2 3 2 (z − i) dz Γ z =i Alternatively, we can rewrite the integrand as 5z 2 + 2z + 1 −4 + 2i 2 + 10i 5 = + + . 3 3 2 (z − i) (z − i) (z − i) z−i Using Cauchy’s Integral Theorem, the contour integral along Γ gives the same result as a contour integral along a small circle of radius r about z = i, parametrised by z (t) = i + r exp(2π i t). Doing the (easy) t integrals we see that the only the last term contributes, and it contributes 10π, i. Finally, we have the integral e−z dz , 2 Γ (z + 1) whose integrand is analytic everywhere but for z = −1, which lies inside Γ. We can appeal directly to the generalised Cauchy Integral Formula and write down the answer e−z d −z dz = 2π i e = −2π ie . 2 dz Γ (z + 1) z =−1 Alternatively, we notice that we can rewrite the integrand as follows e−z e−z d = − − 2 (z + 1) z + 1 dz e−z z+1 .
Next we have
The second term has as antiderivative the function e−z /(z + 1) which is analytic in a domain containing Γ but excluding z = −1; say the open ε-annulus 1 − ε < |z | < 1 + ε. Therefore its contour integral along Γ vanishes, and we have e−z e−z dz = − dz , 2 Γ z+1 Γ (z + 1) which can be evaluated using the Cauchy Integral Formula: e−z e−z dz = − dz = −2π i e . 2 Γ (z + 1) Γ z +1 10
Solution to Problem 1. Let z (t) = x(t) + i y (t) describe the position of a particle moving in the complex plane. Using the equivalence between the complex plane and 42 we see that z (t) carries the same information as the vector (x(t), y (t)). Therefore we see that the expression z ˙ (t) corresponds to the vector (x ˙ (t), y ˙ (t)) which is precisely the (instantaneous) velocity of the particle. The quantity |z ˙ (t)| = x ˙ (t)2 + y ˙ (t)2 is the norm of the velocity vector (x ˙ (t), y ˙ (t)) , whence it is the speed (the magnitude of the velocity) of the particle. Finally the integral
1 1
|z ˙ (t)| dt =
0 0
x ˙ (t)2 + y ˙ (t)2 dt
is nothing but the integral along the contour of the infinitesimal line element dx2 + dy 2 , hence the integral is the length of the contour. Solution to Problem 2. Consider the integral f (z ) dz =
Γ Γ
(3z 2 − 5z + i) dz
where Γ is the straight line segment from z = i to z = 1. This contour can be parametrised as z (t) = t + i (1 − t) for t ∈ [0, 1]. Then after some algebra f (z (t)) = 3z (t)2 − 5z (t) + i = (t − 3) − i (6t2 − 11t + 4). The velocity is given by z ˙ (t) = 1 − i. Therefore, (3z 2 − 5z + i) dz =
Γ 0 1 1
(t − 3) − i (6t2 − 11t + 4) (1 − i) dt (−6t2 + 12t − 7) − i (6t2 − 10t + 1) dt
=
0
= −3 + 2 i . Now, although this is the right answer, this is not the most efficient way to solve this problem. The best way to do this problem is to realise that the integrand f (z ), being a polynomial, is entire and hence, by the Cauchy Integral Theorem, its integral only depends on the endpoints: f (z ) dz = F (1) − F (i) ,
Γ
1
where F (z ) is an antiderivative, which is guaranteed to exist, but must be found. Being a polynomial, we can guess this easily: F (z ) = z 3 − 5 z2 + i z 2 Therefore, f (z ) dz = F (1) − F (i) = −3 + 2 i ,
Γ
=⇒
F (z ) = 3z 2 − 5z + i = f (z ) .
the same result at a fraction of the cost. Consider now the integral ez dz
Γ
where Γ is the upper half of the circle |z | = 1 from z = 1 to z = −1. This time the parametrised integral would be quite messy, so we use the Cauchy Integral Theorem directly. The exponential function is again entire and moreover is its own antiderivative, so that
1 −1 ez dz = ez |− − e1 = −2 sinh(1) . 1 = e
Γ
Next, we compute the integral 1 dz z
Γ
where Γ is any contour in the right half-plane from z = −3i to z = 3i. The function f (z ) = 1/z has an antiderivative F (z ) = Log(z ) in the domain D consisting of all points in the complex plane except for the origin and the negative real axis. Since the contour Γ lies in this domain, we can use this branch. Therefore, we find 1 π π dz = Log(3i) − Log(−3i) = Log(3) + i − Log(3) − i z 2 2 Log(z )2 dz ,
Γ
= iπ .
Γ
Next we have
where Γ is the straight line segment from z = 1 to z = i. The principal branch Log(z ) of the logarithm function is analytic in the domain D defined above and Γ lies in D. Therefore, we can compute this integral by finding an antiderivative F (z ) for f (z ) = Log(z )2 , which is analytic in this domain. It is easy to check that F (z ) = z Log(z )2 − 2 z Log(z ) + 2 z 2
is an antiderivative. Therefore, Log(z )2 dz = F (i) − F (1)
Γ
= i Log(i)2 − 2i Log(i) + 2i − Log(1)2 − 2 Log(1) + 2 = (π − 2) + i 2 − 1 π2 . 4 Finally, we consider the integral
Γ
1 dz 1 + z2
where Γ is the straight line segment from z = 1 to z = 1 + i. Let us use partial fractions to rewrite the integrand as a sum of two terms, i i 1 2 2 = − , 1 + z2 z+i z−i each of which has an antiderivative in a domain containing the contour of integration: d 1 Log(z ± i) = . dz z±i Therefore, 1 i i i i dz = Log(z + i)|1+ − Log(z − i)|1+ 1 1 2 2 2 Γ 1+z i i = (Log(1 + 2i) − Log(1 + i)) − (Log(1) − Log(1 − i)) 2 2 i i i = Log(1 + 2i) + Log(1 − i) − Log(1 + i) . 2 2 2 where we have used that Log(1) = 0. In order to compute the principal values of the logarithms, it is necessary to write the complex numbers involved in polar form: √ 1 + 2i = 5eiθ where tan θ = 2 √ 1 − i = 2e−iπ/4 √ 1 + i = 2eiπ/4 . Therefore, Log(5) + i Arctan(2) Log(1 + 2i) = 1 2 π 1 Log(1 − i) = 2 Log(2) − i 4 π 1 Log(1 + i) = 2 Log(2) + i , 4 3
whence 1 i i i dz = Log(1 + 2i) + Log(1 − i) − Log(1 + i) 2 1+z 2 2 2 i π = Log(5) − 1 Arctan(2) + . 2 2 4
Γ
Solution to Problem 3. Let Γ be the contour parametrised by z (t) = 2 cos(t) + i sin(2t) for t ∈ [0, 2π ]. The periodicity of the trigonometric functions guarantee that this is a closed curve starting and ending at z = 2. It is clear that as the x-coordinate goes from 2 to −2 and back to 2 once, the y -coordinate goes from 0 to 1 to −1 and back to 0 twice. Therefore the resulting curve will be a horizontal figure 8 (more like a ∞) crossing itself at the origin at t = π/2 and t = 3π/2, as in the figure below. The “wrong” way to do this problem would be to plug in the explicit parametrisation and perform the t integral by brute force. Instead we shall use Cauchy’s Integral Theorem in order to evaluate them. The four functions that we are asked to integrate along this contour are analytic everywhere except at z = ±1. Therefore we can use Cauchy’s Integral Theorem to argue that the result of the contour integral along Γ is the same as the sum of the contour integrals around the small circles Γ+ and Γ− surrounding the singular points z = ±1, as in the figure below. Notice, however, that the contour Γ− is negatively oriented.
−1 •
Γ
+1 •
Γ− −1 • +1 •
Γ+
©
In other words, for each of the functions f (z ) to be integrated, Cauchy’s Integral Theorem tells us that f (z ) dz =
Γ Γ+
f (z ) dz +
Γ−
f (z ) dz .
Let us then use this to compute the integrals. First we compute 1 dz = z−1 1 dz + z−1 4 1 dz . z−1
Γ
Γ+
Γ−
This function is analytic everywhere but at z = +1, hence by Cauchy’s Integral Theorem its integral along Γ− vanishes. Therefore, 1 dz = z−1 1 dz = 2π i , z−1
Γ
Γ+
as we discussed in the lecture. Next we compute 1 dz = z+1 1 dz + z+1 1 dz . z+1
Γ
Γ+
Γ−
This function is now analytic everywhere but at z = −1, hence by Cauchy’s Integral Theorem its integral along Γ+ vanishes. Therefore, 1 dz = z+1 1 dz . z+1
Γ
Γ−
This contour is negatively oriented, so now 1 dz = − z+1 1 dz = −2π i . z+1
Γ
−Γ−
The next integral is
Γ
z2
1 dz . −1
Using partial fractions, we can write
1 1 1 2 = − 2 , z2 − 1 z−1 z+1
whence
Γ
z2
1 dz = −1
1 2
Γ
1 dz − z−1
1 2
Γ
1 dz , z+1
which can be read off from the two integrals we have just performed, so that 1 dz = 1 2π i − 1 (−2π i) = 2π i . 2 2 2 z − 1 Γ Finally we compute 2z 2 − z + 1 dz . (z − 1)2 (z + 1) 5
Γ
Again we use partial fractions to write 2z 2 − z + 1 1 1 1 = + + ; 2 2 (z − 1) (z + 1) z − 1 (z − 1) z+1 whence 2z 2 − z + 1 dz = (z − 1)2 (z + 1) 1 dz + z−1 1 dz + (z − 1)2 1 dz . z+1
Γ
Γ
Γ
Γ
We have down the first and the last integrals before and we see that they cancel, so that all we have to do is the middle integral. Using the fact that the integrand is analytic everywhere but at z = 1, we can write this as an integral over the small circle Γ+ : 2z 2 − z + 1 dz = (z − 1)2 (z + 1) 1 dz . (z − 1)2
Γ
Γ+
This integral can be seen to be zero in two ways: either appealing to the generalised Cauchy Integral formula: f (z ) dz = 2π i f (1) , (z − 1)2
Γ+
and using that f (z ) = 1 is a constant in this case; or directly by computing the contour integral. We can parametrise Γ+ by z = 1 + r exp(2π i t) for t ∈ [0, 1], so that z ˙ (t) = 2π i r exp(2π i t). Therefore 1 2π i dz = 2 (z − 1) r
1 0
Γ+
1 e−2π i t dt = − e−2π i t r
1 0
=0,
by the periodicity of the exponential. Therefore any way one looks at it, one finds 2z 2 − z + 1 dz = 0 . 2 Γ (z − 1) (z + 1) Solution to Problem 4. We are asked to evaluate the integral eiz dz (z 2 + 1)2
Γ
where Γ is the circle |z | = 3 traversed once counterclockwise. The integrand is analytic everywhere but at z = ± i. Therefore by the 6
Γ
I
•i •
Γ+
I
−i •
Γ−
R
•
Cauchy Integral Theorem, the integral around Γ gives the same result as the integral around two small circles Γ± centred about z = ± i, respectively, as in the figure. In other words, eiz dz = (z 2 + 1)2 eiz dz + (z 2 + 1)2 eiz dz . (z 2 + 1)2
Γ
Γ+
Γ−
We now use partial fractions to write
i 1 i 1 1 4 4 4 4 = − − − . (z 2 + 1)2 z + i (z + i)2 z − i (z − i)2
The first two terms only contribute to the integral along Γ− , whereas the last two terms only contribute to the integral along Γ+ . For each of these we can use the (generalised) Cauchy Integral Formula: eiz dz = 2π i e z+i
z =−i
Γ−
Γ−
d eiz dz = 2π i eiz 2 (z + i) dz
Γ+
= −2π e
eiz dz = 2π i e−1 z−i
z =i
Γ+
eiz d dz = 2π i eiz 2 (z − i) dz
= −2π e−1 .
Collating all the terms we find that eiz π i i dz = 4 2π i e − 1 (−2π e) − 4 2 π i e −1 − 1 (−2π e−1 ) = . 4 4 2 2 (z + 1) e
Γ
Next we consider the integral
Γ
z3 7
z+i dz + 2z 2
Γ2 Γ1 • −2
I I •0
• −2 •0 • −2 Γ3 •0
I
where Γ is each of the contours Γ1 , Γ2 and Γ3 depicted in the figure. The integrand is analytic everywhere except for z = 0 and z = −2, as can be seen easily if we use partial fractions to rewrite it in the following form:
i z+i 2 = + z 3 + 2z 2 z2 1 2
− z
i 4
−
i −4 . z+2
1 2
We can now use the (generalised) Cauchy Integral Formula to evaluate the integral of each of the above terms along the contours Γk , for k = 1, 2, 3: 1 dz = 0 z2 1 dz = 0 z2 1 dz = 0 z2 1 dz = 2π i z 1 dz = 0 z 1 dz = 0 z 1 dz = 0 z+2 1 dz = 2π i z+2 1 dz = 0 . z+2
Γ1
Γ1
Γ1
Γ2
Γ2
Γ2
Γ3
Γ3
Γ3
Collating the terms, we find z+i dz = 1 π + iπ 2 2 + 2z z+i 1 dz = − 2 π − iπ 3 2 z + 2z z+i dz = 0 . 3 z + 2z 2 z3
Γ1
Γ2
Γ3
Solution to Problem 5. In this problem we are asked to integrate several functions along the 8
same contour: the positively oriented circle of radius 2 centred about the origin. In the figure we have depicted this contour in relation to the points at which the integrands below fail to be analytic.
• 3i
•i −1 • •0
I
3 2
••π
2
Γ
−3i •
We start with
Γ
sin(3z ) dz . z−1 π 2
The integrand is analytic everywhere but at z = 1 π which lies in the 2 interior of the contour. Therefore by the Cauchy Integral Formula,
Γ
sin(3z ) dz = 2π i sin z−1 π 2
3 π 2
= −2π i .
zez dz , Γ 2z − 3 whose integrand is analytic except at z = 3 . Since this point lies inside 2 Γ, we can use the Cauchy Integral Formula to obtain zez dz = 2z − 3
1 zez 2 3 Γ z− 2 3 3/2 e =i dz = 2π i 1 22
Next we have
Γ
3π 3/2 e . 2
We continue with
Γ
cos(z ) dz = z 3 + 9z
Γ
cos(z ) dz , z (z + 3i)(z − 3i)
whose integrand fails to be analytic at z = 0 and z = ±3i. Of these points only z = 0 lies in the interior of Γ, hence by the Cauchy Integral Formula 2π cos(z ) cos(z ) =i dz = 2π i 2 . z + 9 z=0 9 Γ z (z + 3i)(z − 3i) 9
5z 2 + 2z + 1 dz , (z − i)3 Γ whose integrand fails to be analytic at z = i, which lies in the interior of Γ. We can appeal to the generalised Cauchy Integral Formula to obtain 2 5z 2 + 2z + 1 1 d = i 10 π . dz = 2 π i 5z 2 + 2z + 1 2 3 2 (z − i) dz Γ z =i Alternatively, we can rewrite the integrand as 5z 2 + 2z + 1 −4 + 2i 2 + 10i 5 = + + . 3 3 2 (z − i) (z − i) (z − i) z−i Using Cauchy’s Integral Theorem, the contour integral along Γ gives the same result as a contour integral along a small circle of radius r about z = i, parametrised by z (t) = i + r exp(2π i t). Doing the (easy) t integrals we see that the only the last term contributes, and it contributes 10π, i. Finally, we have the integral e−z dz , 2 Γ (z + 1) whose integrand is analytic everywhere but for z = −1, which lies inside Γ. We can appeal directly to the generalised Cauchy Integral Formula and write down the answer e−z d −z dz = 2π i e = −2π ie . 2 dz Γ (z + 1) z =−1 Alternatively, we notice that we can rewrite the integrand as follows e−z e−z d = − − 2 (z + 1) z + 1 dz e−z z+1 .
Next we have
The second term has as antiderivative the function e−z /(z + 1) which is analytic in a domain containing Γ but excluding z = −1; say the open ε-annulus 1 − ε < |z | < 1 + ε. Therefore its contour integral along Γ vanishes, and we have e−z e−z dz = − dz , 2 Γ z+1 Γ (z + 1) which can be evaluated using the Cauchy Integral Formula: e−z e−z dz = − dz = −2π i e . 2 Γ (z + 1) Γ z +1 10
