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Physics 451 Fall 2004 Homework Assignment #9 — Solutions
Textbook problems: Ch. 6: 6.1.3, 6.1.7, 6.2.5, 6.2.6, 6.3.3, 6.4.3, 6.4.
Chapter 6
6.1.3 Prove algebraically that
|z 1 | − |z 2 | ≤ |z 1 + z 2 | ≤ |z 1 | + |z 2 |
Interpret this result in terms of vectors. Prove that
|z − 1 | < |
z^2 − 1 | < |z + 1|, for <(z) > 0
We start by evaluating |z 1 + z 2 |^2
|z 1 + z 2 |^2 = (z 1 + z 2 )(z 1 ∗ + z∗ 2 ) = |z 1 |^2 + |z 2 |^2 + z 1 z∗ 2 + z∗ 1 z 2 = |z 1 |^2 + |z 2 |^2 + (z 1 z∗ 2 ) + (z 1 z 2 ∗ )∗^ = |z 1 |^2 + |z 2 |^2 + 2<(z 1 z∗ 2 )
We now put a bound on the real part of z 1 z∗ 2. First note that, for any complex quantity ζ, we have |ζ|^2 = (<ζ)^2 + (=ζ)^2 ≥ (<ζ)^2. Taking a square root gives |ζ| ≥ |<ζ| or −|ζ| ≤ <ζ ≤ |ζ|. For the present case (where ζ = z 1 z∗ 2 ) this gives −|z 1 ||z 2 | ≤ <(z 1 z∗ 2 ) ≤ |z 1 ||z 2 |. Using this inequality in (1), we obtain
|z 1 |^2 + |z 2 |^2 − 2 |z 1 ||z 2 | ≤ |z 1 + z 2 |^2 ≤ |z 1 |^2 + |z 2 |^2 + 2|z 1 ||z 2 |
or (|z 1 | − |z 2 |)^2 ≤ |z 1 + z 2 |^2 ≤ (|z 1 | + |z 2 |)^2 Taking the square root then proves the triangle inequality. The reason this is called the triangle inequality is that, in terms of vectors, we can think of z 1 , z 2 and z 1 + z 2 as the three sides of a triangle
1 2
z 1
z
Then the third side (|z 1 + z 2 |) of a triangle can be no longer than the sum of the lengths of the other two sides (|z 1 | + |z 2 |) nor shorter than the difference of lengths (|z 1 | − |z 2 |).
Finally, for the second inequality, we start by proving that
|z + 1|^2 = |z|^2 + 1 + 2<z = (|z|^2 + 1 − 2 <z) + 4<z = |z − 1 |^2 + 4 |z − 1 |^2
for 0. This implies that |z + 1| > |z − 1 | for 0. The picture here is that if z is on the right half of the complex plane then it is closer to the point 1 than the point −1.
z +^1 z^ −^1
z
z
Given this result, it is simple to see that
|z − 1 |^2 < |z − 1 ||z + 1| < |z + 1|^2
or, by taking a square root
|z − 1 | < |
(z − 1)(z + 1)| < |z + 1|
which is what we set out to prove.
6.1.7 Prove that
a)
N∑ − 1
n=
cos nx = sin(N x/2) sin x/ 2 cos(N − 1) x 2
b)
N∑ − 1
n=
sin nx = sin(N x/2) sin x/ 2 sin(N − 1) x 2
We may solve parts a) and b) simultaneously by taking the complex combination
S =
N∑ − 1
n=
cos nx + i
N∑ − 1
n=
sin nx =
N∑ − 1
n=
(cos nx + i sin nx) =
N∑ − 1
n=
einx
The real part of S gives part a) and the imaginary part of S gives part b). When written in this fashion, we see that S is a terminating geometric series with ratio r = eix. Thus
S =
N∑ − 1
n=
rn^ = 1 − rN 1 − r
1 − eN ix 1 − eix^
e 12 N ix(e 12 N ix^ − e−^12 N ix) e 12 ix(e 12 ix^ − e−^12 ix)
6.2.6 If there is some common region in which w 1 = u(x, y) + iv(x, y) and w 2 = w 1 ∗ = u(x, y) − iv(x, y) are both analytic, prove that u(x, y) and v(x, y) are constants.
If u + iv and u − iv are both analytic, then they must both satisfy the Cauchy- Riemann equations. This corresponds to
∂u ∂x
∂v ∂y
∂u ∂y
∂v ∂x
(from u + iv) and ∂u ∂x
∂v ∂y
∂u ∂y
∂v ∂x (from u − iv). Clearly this indicates that
∂u ∂x
∂u ∂y
∂v ∂x
∂v ∂y
Since all partial derivatives vanish, u and v can only be constants.
6.3.3 Verify that (^) ∫ 1+i 0
z∗dz
depends on the path by evaluating the integral for the two paths shown in Fig. 6.10. 2
x
1 + i y (^) z
2
1
1
We perform this integral as a two-dimensional line integral ∫ z∗dz =
(x − iy)(dx + idy)
For path 1, we first integrate along the x-axis (y = 0; dy = 0) and then along the y-axis (x = 1; dx = 0) ∫ (^) 1+i
0
z∗dz =
0
(x − iy)
y=
dx +
0
(x − iy)
x=
idy
0
xdx +
0
(i + y)dy = 12 x^2
1 0
1 0 = 1 + i
Similarly, for path 2, we find ∫ (^) 1+i
0
z∗dz =
0
(x − iy)
x= idy +
0
(x − iy)
y= dx
0
ydy +
0
(x − i)dx = 12 y^2
∣∣^1
0
∣∣^1
0 = 1 − i
So we see explicitly that the integral depends on the path taken (1 + i 6 = 1 − i).
6.4.3 Solve Exercise 6.3.4 [
C dz/(z (^2) +z) where C is a circle defined by |z| > 1] by separating the integrand into partial fractions and then applying Cauchy’s integral theorem for multiply connected regions.
Note that, by applying Cauchy’s integral formula to a constant function f (z) = 1, we may derive the useful expression ∮
C
dz z − z 0 = 2πi (2)
provided point z 0 is contained inside the contour C (it is zero otherwise). Then, using partial fractions, we see that ∮
C
dz z^2 + z
C
dz z(z + 1)
C
z
z + 1
dz =
C
dz z
dz z + 1 Since C is a circle of radius greater than one, it encompasses both points z 0 = 0 and z 0 = −1. Thus, using (2), we find ∮
C
dz z^2 + z = 2πi − 2 πi = 0
Note that, if the radius of C is less than one, we would have encircled only the pole at z 0 = 0. The result would then have been 2πi instead of zero.
6.4.4 Evaluate (^) ∮
C
dz z^2 − 1 where C is the circle |z| = 2.
Again, we use partial fractions and (2) ∮
C
dz z^2 − 1
C
dz (z + 1)(z − 1)
C
z − 1
z + 1
dz
C
dz z − 1
C
dz z + 1 = πi − πi = 0
Here it is important that the contour of radius 2 encircles both points z 0 = − 1 and z 0 = 1.
