PHYSICS 116A

Homework 3 Solutions

1. Boas, 2.14–9. Evaluate (−1)i in x+ iy form.

We write −1 = ei(π+2πn) (n = 0 , ±1 , ±2 , . . . , ) so

(−1)i = ei i(π+2πn) = e−π−2πn ,

which, rather surprisingly, is a set of real numbers. The principal value of (−1)i = e−π corre-sponding to the choice of n = 0 in the above equation.

2. Boas, 2.15–7. Write arctan(i√2) in x+ iy form.

Start with

z = tanw =sinw

cosw=

1

i

(

eiw − e−iw

eiw + e−iw

)

=1

i

(

e2iw − 1

e2iw + 1

)

.

Solving for e2iw,

iz =e2iw − 1

e2iw + 1=⇒ e2iw =

1 + iz

1− iz.

Hence,

w = arctan z =1

2iln

(

1 + iz

1− iz

)

. (1)

Setting z = i√2, we obtain:

arctan(i√2) =

1

2iln

(

1−√2

1 +√2

)

=1

2iln[

−(1−√2)2]

. (2)

after multiplying the numerator and denominator above by 1−√2 and noticing that (1−

√2)(1+√

2) = 1− 2 = −1. We shall make use of the definition of the multi-valued complex logarithm

ln z = Ln|z|+ i arg z = Ln|z|+ i(Argz + 2πn) , n = 0,±1,±2 . . . , (3)

with z = −(1−√2)2. In particular, we have Ln|z| = 2Ln(

√2−1) and Arg z = π. Hence, Eqs. (3)

and (2) yield:

arctan(i√2) =

(

n+ 12

)

π − iLn(√2− 1) , n = 0 , ±1 , ±2 , . . . .

3. Boas, 2.16–11. Prove that

cos θ + cos 3θ + cos 5θ + · · ·+ cos(2n− 1)θ =sin 2nθ

2 sin θ, (4)

sin θ + sin 3θ + sin 5θ + · · ·+ sin(2n− 1)θ =sin2 nθ

sin θ. (5)

Consider the geometric series:

S ≡ eiθ + e3iθ + e5iθ + · · ·+ e(2n−1)θ . (6)

1

Using the results of eq. (1.4) on p. 2 of Boas,

a+ ar + ar2 + · · ·+ arN−1 =a(1− rN )

1− r,

we identify a = eiθ, r = e2iθ and n = N . Hence,

S ≡ eiθ + e3iθ + e5iθ + · · ·+ e(2n−1)θ =eiθ(1− e2inθ)

1− e2iθ.

This result can be simplified by multiply both the numerator and denominator by −e−iθ. Theend result is:

S =e2inθ − 1

eiθ − e−iθ=

einθ(einθ − e−inθ)

eiθ − e−iθ=

einθ sinnθ

sin θ,

where we have used

sin θ ≡ eiθ − e−iθ

2i

to obtain the final result. In particular, using Re(einθ) = cosnθ and Im(einθ) = sinnθ, it followsthat:

Re S =cosnθ sinnθ

sin θ=

sin 2nθ

2 sin θ,

Im S =sin2 nθ

sin θ,

after employing the trigonometric identity sin 2nθ = 2 sinnθ cosnθ. Equating the real and imagi-nary parts of Eq. (6), we conclude that

Re S = cos θ + cos 3θ + cos 5θ + · · ·+ cos(2n− 1)θ =sin 2nθ

2 sin θ,

Im S = sin θ + sin 3θ + sin 5θ + · · ·+ sin(2n− 1)θ =sin2 nθ

sin θ,

which reproduces the results of Eqs. (4) and (5).

4. Boas Ch. 2, §17, Qu. 22. Show that tanh−1 z = 12 ln

(

1+z1−z

)

.

We write

z = tanhw =ew − e−w

ew + e−w=

e2w − 1

e2w + 1,

soz(

e2w + 1)

= e2w − 1 ,

which has solution

e2w =1 + z

1− z.

Taking the log gives

w ≡ tanh−1 z =1

2ln

(

1 + z

1− z

)

.

2

5. The series for the principal value of the complex-valued logarithm,

Ln(1− z) = −∞∑

n=1

zn

n, (7)

converges for all |z| ≤ 1, z 6= 1. In particular, consider the conditionally convergent series,

S ≡∞∑

n=1

einθ

n, where 0 < θ < 2π . (8)

Using Eq. (7) with z = eiθ, it follows that

S = −Ln(1− eiθ) . (9)

The principal value of the complex logarithm is given by:

Ln(1− eiθ) = Ln|1− eiθ|+ iArg(1− eiθ) . (10)

To evaluate this expression we write the complex number z = 1− eiθ in polar form, i.e.

z = 1− eiθ = ReiΘ ,

soS = − [LnR+ iΘ] , (−π < Θ ≤ π) . (11)

A simple way to do this is:

1− eiθ = ReiΘ = eiθ/2[

e−iθ/2 − eiθ/2]

= −2ieiθ/2 sin(θ/2) = 2ei(θ−π)/2 sin(θ/2) , (12)

after using sin(θ/2) = 12i(e

iθ/2 − e−iθ/2) and −i = e−iπ/2. Since sin(θ/2) > 0 for 0 < θ < 2π, itfollows that

R ≡ |1− eiθ| = 2 sin(θ/2),

and we can then identify the principal value of the argument of z as

Θ ≡ Arg(1− eiθ) = 12(θ − π) . (13)

Note that when 0 < θ < 2π, it follows that Θ lies between −π and π, which is inside the range ofthe principal value of the argument function as defined in class. Hence, Eq. (11) simplifies to:

S = −Ln(1− eiθ) = −Ln

(

2 sinθ

2

)

+ 12 i(π − θ) , for 0 < θ < 2π . (14)

(a) By taking the real part of Eq. (8), evaluate

∞∑

n=1

cosnθ

n, where 0 < θ < 2π ,

3

as a function of θ. Check that your answer has the right limit for θ = π.

Noting that cosnθ = Re einθ, it follows from Eqs. (8) and (14) that

∞∑

n=1

cosnθ

n= ReS = −Ln

(

2 sinθ

2

)

, for 0 < θ < 2π . (15)

We can check Eq. (15) in the special case of θ = π. Using the results, sin(π/2) = 1, cosnπ = (−1)n

and −(−1)n = (−1)n+1, it follows that:

Ln 2 =∞∑

n=1

(−1)n+1

n= 1− 1

2 + 13 − 1

4 + · · · ,

a well-known result.

(b) By taking the imaginary part of Eq. (8), prove that

∞∑

n=1

sinnθ

n= 1

2(π − θ) , where 0 < θ < 2π .

Noting that sinnθ = Im einθ, it follows from Eqs. (8) and (14) that

∞∑

n=1

sinnθ

n= ImS = 1

2(π − θ) , for 0 < θ < 2π . (16)

Again, we can check the special case of θ = π. Since sinnπ = 0, Eq. (16) reduces in this limit tothe trivial equation 0 = 0 .

6. Evaluate the integral

In =

∫ ∞

0tne−kt2 dt ,

for k > 0 and n > −1.

Let us define a new variable, u = kt2 or equivalently t =√

u/k. Then,

du = 2ktdt =⇒ dt =du

2kt=

du

2k

(

k

u

)1/2

=du

2k1/2u1/2.

Hence,

In =1

kn/2

∫ ∞

0un/2e−u du

2k1/2u1/2du =

1

2k(n+1)/2

∫ ∞

0u(n−1)/2e−u du

4

Using the definition of the Gamma function,

Γ(x) =

∫ ∞

0tx−1e−t dt ,

we identify x = 12(n+ 1). Thus,

In =1

2k(n+1)/2Γ

(

n+ 1

2

)

As a check, let us set k = 1 and n = 0. Then, we obtain I0 =12Γ(

12) =

12

√π as expected. The factor

of k(n+1)/2 can be deduced using simple dimensional analysis. If t has dimensions of time, then kmust have dimensions of t−2, since the argument of the exponential must be dimensionless.∗ Sincedt also has dimensions of time, the entire integral has dimensions of tn+1, which must be respectedby the final result for In. Indeed k−(n+1)/2 has dimensions of tn+1, and so the dimensions of ourfinal result for In are consistent. In particular, it is possible to first carry out the computationwith k = 1, and then determine the k dependence of In strictly on dimensional grounds!

7. Boas, Ch. 11 §3, Qu. 5. Simplify Γ(12)Γ(4)/Γ(92).

Using xΓ(x) = Γ(x + 1) repeatedly, one obtains Γ(92) = 72 Γ(

72) = 7

2 · 52 Γ(

52), etc. until finally

obtaining Γ(92) =72 · 5

2 · 32 · 1

2 Γ(12) . Hence, using Γ(4) = 3! = 6, it follows that

Γ(12)Γ(4)

Γ(29)= 2

7 · 25 · 2

3 · 2 · 6 =32

35.

8. Boas, Ch. 11 §3, Qu. 13. Express as a Γ function

∫ 1

0x2(

ln1

x

)3

dx .

Introduce a new variable x = e−u. Then dx = −e−udu and ln(1/x) = ln eu = u. Noting thatx = 0 =⇒ u = ∞ and x = 1 =⇒ u = 0, it follows that

∫ 1

0x2(

ln1

x

)3

dx =

∫ ∞

0u3e−3u du =

1

34

∫ ∞

0v3e−v dv =

Γ(4)

81=

2

27.

In the second step above, I used the overall minus sign to interchange the lower and upper limits ofintegration. At the third step, I changed the integration variable once more to v = 3u (the limitsof integration are unchanged). Finally, I used the definition of the Gamma function [eq. (3.1) onp. 538 of Boas] followed by Γ(4) = 3! = 6 to obtain the final result.

∗The simplest way to see that the argument of any exponential must be dimensionless is to consider the power seriesexpansion, ez = 1 + z + z2/2! + z3/3! + · · · . Suppose z had dimensions of length, for example. Then ez would be thesum of a dimensionless number plus a number with units of length plus a number with units of squared-length, and soon. But, one cannot consistently sum two quantities that possess different length dimensions. The only way to avoid thisinconsistency is to conclude that z is dimensionless.

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9. Boas, problem 11.5–3. Show that the binomial coefficient(

pn

)

can be written in terms of Gammafunctions as:

(

p

n

)

=Γ(p+ 1)

n! Γ(p− n+ 1). (17)

The binomial coefficient is defined in eq. (13.6) on p. 28 of Boas as:

(

p

n

)

=p(p− 1)(p− 2) · · · (p− n+ 1)

n!.

To prove that it can be written as in Eq. (17), we first multiply numerator and denominator byΓ(p− n+ 1),

(

p

n

)

=p(p− 1)(p− 2) · · · (p− n+ 1)Γ(p− n+ 1)

n! Γ(p− n+ 1). (18)

By repeatedly employing xΓ(x) = Γ(x+ 1), the numerator of Eq. (18) can be written as:

p(p− 1)(p− 2) · · · (p− n+ 3)(p− n+ 2)(p− n+ 1)Γ(p− n+ 1)

= p(p− 1)(p− 2) · · · (p− n+ 3)(p− n+ 2)Γ(p− n+ 2)

= p(p− 1)(p− 2) · · · (p− n+ 3)Γ(p− n+ 3)

= · · ·

= p(p− 1)(p− 2)Γ(p− 2) = p(p− 1)Γ(p− 1) = pΓ(p) = Γ(p+ 1) .

Inserting this last result back into Eq. (18) yields

(

p

n

)

=Γ(p+ 1)

n! Γ(p− n+ 1)

as requested.

10. Boas, Ch. 11 §5, Qu. 5, part (a).

Use eq. (5.4) on p. 541 of Boas to show that

Γ(

12 − n

)

Γ(

12 + n

)

= (−1)n π , (19)

assuming that n is an integer.†

We start from the reflection formula [eq. (5.4) on p. 541 of Boas]:

Γ(p)Γ(1− p) =π

sinπp. (20)

†Boas restricts n to be a positive integer. However, Eq. (19) holds for negative integers as well. In fact, noting that(−1)n = (−1)−n for any integer n, it follows that Eq. (19) is unmodified under the interchange of n and −n. Note thatfor n = 0, Eq. (19) also yields a correct result, [Γ( 1

2)]2 = π.

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If we set p = 12 − n, then

1− p = 1− (12 − n) = n+ 12 .

Inserting p = 12 − n into Eq. (20) yields:

Γ(

12 − n

)

Γ(

12 + n

)

=π

sin[

π(

12 − n

)] . (21)

Finally, we make use of the trigonometric identity,

sin[

π(

12 − n

)]

= sin(

12π − nπ

)

= sin(

12π)

cos(nπ)− cos(

12π)

sin(nπ) = cos(nπ) ,

after using sin(12π) = 1 and cos(12π) = 0. Using the fact that cos(nπ) = (−1)n for any integer n,it follows that

sin[

π(

12 − n

)]

= (−1)n , for any integer n .

Inserting this result back into Eq. (21) and using (−1)−n = (−1)n for any integer n, we obtain:

Γ(

12 − n

)

Γ(

12 + n

)

= (−1)n π , for any integer n ,

which is the desired result.

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