PHZ 3113 Fall 2010
Homework #6, Due Friday, October 22
i
i
1. Compute ln[ (ii ) ]. Compute ln[ i(i ) ].
π
Use ab = eb ln a [(14.1) on p.73]. Write i = ei 2 . Then,
π
ii = (ei 2 )i = ei
2π
2
π
= e− 2 = −i,
i
π
π
(ii ) = (e− 2 )i = e−i 2 ,
i
π
i(i ) = (ei 2 )e
−π
2
π −π
2
= ei 2 e
Log is inverse of exponentiation, and so
i
ln[ (ii ) ] = −i π2 ,
π
i
ln[ i(i ) ] = i π2 e− 2 .
To either of these you can add 2πni for any integer n.
2. Find all z such that cos z = i.
We need to find solutions to
1 iz
−iz
= i,
(eiz )2 − 2i eiz + 1 = 0,
e +e
2
so
√
eiz = i(1 ± 2).
(this is a complex number,
√ which can be written as a magnitude and a phase, and the phase
is different for the ‘1 − 2’ choice). Taking the log gives
√
√
π
π
iz = Ln(1 + 2) + i( + 2πn),
iz = Ln( 1 − 1) + i(− + 2πn).
2
2
and solutions z for the inverse cosine of i are
√
π
z = − i Ln( 2 + 1) + 2πn,
2
z=−
√
π
− i Ln( 2 − 1) + 2πn.
2
∞
X
1
(iπ − 1)n .
3. Compute
n!
n=0
The exponential series converges for any argument,
(iπ − 1)n+1 /(n + 1)! iπ − 1 =
lim n + 1 → 0,
n→∞ (iπ − 1)n /n!
1
eiπ−1 = eiπ e−1 = − .
e
4. In optics, the intensity of the interference between 2n + 1 sources arrayed along the y-axis
separated by distance d one from the next and radiating in phase is proportional to the
square of the amplitude
n
X
ψ=
eij∆φ ,
j=−n
where the phase difference between sources is ∆φ = kd sin θ (k = 2π/λ is the wavenumber
of the radiation, θ is angle from the x-axis). Compute the sum and determine |ψ|2 . Identify
values of θ where the amplitude has peaks.
Let z = ei∆φ ; then
ψ=
n
X
z j = z −n + · · · + z n = z −n (1 + · · · + z 2n ) =
j=−n
z −n − z n+1
1−z
1
1
sin[(n + 12 )∆φ]
ei(n+1)∆φ − e−in∆φ
ei(n+ 2 )∆φ − e−i(n+ 2 )∆φ
=
=
=
1
1
ei∆φ − 1
sin( 12 ∆φ)
ei 2 ∆φ − e−i 2 ∆φ
1
(for the first step in the final line, multiply both numerator and denominator by e−i 2 ∆φ ),
and
sin2 [(n + 12 )∆φ]
|ψ|2 =
.
sin2 ( 12 ∆φ)
The numerator oscillates rapidly, n or 2n or so times for each phase of a slower modulation
by the denominator. Peaks appear where the denominator vanishes, sin 12 ∆φ = 0,
1
1 2π
∆φ =
d sin θ = nπ,
2
2 λ
nλ = d sin θ.
The function is not infinite at these points; as ∆φ → 0, the numerator goes to [(n + 12 ) ∆φ]2
and the denominator to ( 12 ∆φ)2 ; and so |ψ|2 → 4(n + 12 )2 . The numerator goes to zero at
(n + 12 ) ∆φ = ±π, and so the width of a peak is ∆φ = ±π/(n + 12 ). Peaks become taller and
narrower with n.
The plot (next page) shows |ψ|2 for n = 10.
5. Use complex numbers to sum the series
∞
X
1
sin[(2k + 1)x] e−(2k+1)y
2k + 1
k=0
for y ≥ 0. Sketch the behavior of your solution as a function of x for y → 0+ .
Every step is straightforward, but there are many steps. First, note that the sum can be
written as a function of the complex variable z = x + iy, and in particular as powers of the
function Z = eiz = eix−y = (cos x + i sin x) e−y ,
Z 2k+1 = (eiz )2k+1 = e(2k+1)(ix−y) = (cos[(2k + 1)x] + i sin[(2k + 1)x]) e−(2k+1)y .
Thus, the series is
∞
X
k=0
∞
X
1
1
Im[Z k ] = Im
Zk.
(2k + 1)
(2k + 1)
k=0
The series for ln(1 ± Z) have coefficients that go as 1/n,
ln(1 + Z) =
∞
X
1
(−1)n−1 Z n ,
n=1
n
∞
X
1 n
Z ;
− ln(1 − Z) =
n
n=1
and adding the two series leaves only odd terms,
∞
X
X 1
1
1
n
[ln(1 + Z) − ln(1 − Z)] =
Z =
Z 2k+1.
2
n
(2k + 1)
k=0
odd n
So, the sum we want is
#
"
1 1 + Z 1 1 + eix e−y Im
.
ln
ln
= Im
2
1−Z
2
1 − eix e−y
You may also invoke [cf. (13.7) on p.30]
tan−1 Z =
Z Z
Z ZX
∞
∞
X
dt
1
2k
Z 2k+1,
=
t dt =
2
2k + 1
0 1+t
0 k=0
k=0
but you then need to apply something like the procedure of Problem 2 to find the arctan of
a complex number, which leads to the same expression.
Next, write the argument of the log as a simple complex number (with a real denominator),
1 + eix e−y
(1 + eix e−y )(1 − e−ix e−y )
1 − e−2y + e−y (eix − e−ix )
1 − e−2y + 2i e−y sin x
=
=
=
1 + e−2y + 2 e−y cos x
1 − eix e−y
(1 − eix e−y )(1 − e−ix e−y )
1 + e−2y + e−y (eix + e−ix )
As a complex number reiθ , this has a magnitude, which (fortunately) we do not need to
know, and a phase, given by
tan θ =
2 sin x e−y
2 sin x
sin x
.
=
=
ey − e−y
sinh y
1 − e−2y
The sum is then
h1
i
h1
sin x i 1
1
Im ln(reiθ ) = Im (ln r + iθ) = θ = tan−1
.
2
2
2
2
sinh y
As y → 0, the denominator becomes small, and the argument of the arctan becomes very
large; its value is essentially + π2 for positive sin x and − π2 for negative sin x. The plot shows
the result as a function of x for y = 1, 0.1, 0.01, and 0.001 (or so). You can add any 2πn
to the arctan once, and all other values are shifted by the same value (all are connected
continuously).