Assignment 6
Arfken 6.1.7
Do both parts together:
N
−1
X
(cos nx + i sin nx) =
n=0
N
−1
X
einx
n=0
=
N
−1
X
eix
n
n=0
1 − eiN x
1 − eix
e−iN x/2 − eiN x/2 eiN x/2
=
e−ix/2 − eix/2
eix/2
x
x
−2i sin N x/2
=
cos N − 1 + i sin N − 1
2
2
−2i sin x/2
=
Taking the real and imaginary parts, we have
N
−1
X
x
sin N x/2
cos N − 1
cos nx =
2
sin x/2
n=0
N
−1
X
x
sin N x/2
sin N − 1
sin nx =
2
sin x/2
n=0
Arfken 6.1.8
For −1 < p < 1, do both parts together:
∞
X
n
p (cos nx + i sin nx) =
n=0
=
∞
X
n=0
∞
X
pn einx
peix
n
n=0
1
1 − peix
1 − pe−ix
=
(1 − peix )(1 − pe−ix )
1 − p cos x + i p sin x
=
1 + p2 − p eix + e−ix
1 − p cos x + i p sin x
=
1 + p2 − 2p cos x
=
Taking the real and imaginary parts, we have
∞
X
n=0
∞
X
n=0
pn cos nx =
1 − p cos x
1 + p2 − 2p cos x
pn sin nx =
p sin x
1 + − 2p cos x
p2
1
Arfken 6.1.16
(a) Show sin−1 z = −i ln(iz ±
√
1 − z 2 ). Take the sine of both sides and demonstrate equality
p
z = sin −i ln(iz ± 1 − z 2 )
√
1 i·−i ln(iz±√1−z2 )
2
e
=
− e−i·−i ln(iz± 1−z )
2i
p
p
1 (iz ± 1 − z 2 ) − (iz ± 1 − z 2 )−1
=
2i
!
√
p
−iz ± 1 − z 2
1
2
√
√
=
(iz ± 1 − z ) −
2i
(iz ± 1 − z 2 )(−iz ± 1 − z 2 )
p
p
1 =
(iz ± 1 − z 2 ) − (−iz ± 1 − z 2 )
2i
=z
which demonstrates what we wanted to show.
1+z
. Do as before:
(f) Show tanh−1 z = 21 ln 1−z
z = tanh
1
ln
2
1+z
1−z
1/2
1/2
1+z
1+z
eln( 1−z ) − e− ln( 1−z )
1+z 1/2
1+z 1/2
eln( 1−z ) + e− ln( 1−z )
1/2 1/2
1−z
1+z
−
1−z
1+z
=
1/2 1/2
1+z
1−z
+
1−z
1+z
1/2 1/2 1/2 1/2
1−z
1−z
1+z
1+z
−
−
1−z
1+z
1−z
1+z
=
1/2 1/2 1/2 1/2
1−z
1−z
1+z
1+z
+
−
1−z
1+z
1−z
1+z
1+z
1−z
1−z + 1+z − 2
= 1+z
1−z
1−z − 1+z
1 − z2
2 + 2z 2
=
−2
1 − z2
4z
=
=z
Arfken 6.1.21
(a) In general we have
eln z = eln r+iθ+i2πn
= r · eiθ · ei2πn
= r iθ
=z
(b) To do this problem it is important to realize that we can write a complex function, f (z), in polar-like
coordinates exactly as we do the complex variable z = reiθ . For example, we can write f (z) = R(x, y)eiΘ(x,y) .
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In particular
ln ez = ln f (z)
= ln R(x, y)eiΘ(x,y)
= ln R + ln eiΘ+i2πn
= ln R + iΘ + i2πn
= z + i2πn
6= z
Arfken 6.2.3
If w(z) = u(x, y) + iv(x, y) is analytic, then u and v each satisfy Laplace’s equation. (This is done in
6.2.1. Take partial derivatives of the Cauchy Riemann conditions and equate.) This means
∂2v
∂2v
∂2u ∂2u
+
=
0
=
+
∂x2
∂y 2
∂x2
∂y 2
Recall that for a function to have a maximum or a minimum, the second derivatives of that function must
be of the same sign (positive for a minimum and negative for a maximum). Because the functions u and
v satisfy the Laplace equation, they cannot satisfy this condition for any point within the region in which
w(z) is analytic. Therefore, u and v (and hence w(z)) cannot have a maximum or a minimum within the
region of analyticity.
Arfken 6.2.5
(a) For u(x, y) = x3 − 3xy 2 , the Cauchy Riemann conditions imply
Z
∂u
v(x, y) =
dy
∂x
Z
=
3x2 − 3y 2 dy
= 3x2 y − y 3 + a0 (x)
and
Z
∂u
dx
v(x, y) = −
∂y
Z
= (6xy) dx
= 3x2 y + a1 (y)
For consistency, this leads to a0 (x) = 0 and a1 (y) = −y 3 and w(z) = x3 − 3xy 2 + i(3x2 y − y 3 ).
(b) For v(x, y) = e−y sin x, the Cauchy Riemann conditions imply
Z
∂v
u(x, y) =
dx
∂y
Z
= −e−y sin x dx
= e−y cos x + a0 (y)
and
Z
∂v
u(x, y) = −
dy
∂x
Z
= −e−y cos x dy
= e−y cos x + a1 (y)
For consistency, this leads to a0 (x) = 0 and a1 (y) = 0 and w(z) = e−y cos x + ie−y sin x.
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Arfken 6.2.7
The function f (z) is analytic. The function f (z ∗ ) can be thought of as f (z) with y replaced with −y.
Likewise f ∗ (z ∗ ) can be thought of as f (z ∗ ) with i replace with −i. So, we have
f ∗ (z ∗ ) = u(x, −y) − iv(x, −y)
= û(x, y) + iv̂(x, y)
If f ∗ (z ∗ ) is to be analytic, the Cauchy Riemann conditions must be satisfied for it
∂v̂
∂ û
=
∂x
∂y
∂u
∂(−v)
= (−1)
∂x
∂y
∂u
∂v
=
∂x
∂y
and
∂ û
∂v̂
=−
∂y
∂x
∂(−v)
∂u
=−
(−1)
∂y
∂x
∂u
∂v
−
=
∂y
∂x
which are the Cauchy Riemann conditions for f (z) which we know are satisfied since it is analytic. Therefore,
f ∗ (z ∗ ) is analytic.
Arfken 6.3.2
Recall that the definition of a contour integral is
Z
f (z)dz = lim
c
n→∞
∞
X
f (ζj )(zj − zj−1 )
j=1
where ζj is a point on the contour between zj and zj−1 and the contour is assumed specified. We can now
construct the following inequality if we replace the value of the complex function f (ζj ) with the maximum
value of f (z) along the curve
Z
∞
X
f (z)dz < lim
|f |max (zj − zj−1 )
n→∞
c
j=1
= |f |max lim
n→∞
∞
X
j=1
= |f |max lim L
n→∞
= |f |max L
where L is the length of the contour, C.
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(zj − zj−1 )
Arfken 6.3.4
H
The integral c dz/(z 2 + z) cannot be evaluated for any contour enclosing the origin by using the Cauchy
integral theorem because the function being integrated is not analytic at z = 0 (or at z = −1). However, it
is zero if we consider a circle for C with |z| = R > 1:
I
dz
=
z2 + z
Z
2π
0
Z
2π
=
(Reiθ
0
Z
=
0
Reiθ idθ
+ Reiθ
R2 ei2θ
2π
iR−iθ + i
dθ
+ 1)(Re−iθ + 1)
R sin θ + i(1 + R cos θ)
dθ
R2 + 2R cos θ + 1
The real part of the integral can be shown to be zero (substitute ξ = θ − π and show that it is an odd
integral over a symmetric interval – which must be zero) while the imaginary part can be done most easily
by looking it up in a table, or better yet by using Mathematica or Maple. It is then straightforward to show
the integral must also be zero.
You have to be careful with this problem or you can come away with the wrong impression. It is
important to realize that this particular integral is not zero because of the Cauchy integral theorem (after
all, it violates the assumptions of that theorem for any contour, C, that encloses any nonanlytic point of the
integrand, in this case z = 0, −1) but in spite of it.
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