OurBigBook
About
$
Donate
Sign in
Sign up
Solution
ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s1/3/b/solution
Top articles
Latest articles
New article in topic
Show body
Body
0
Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford
/
2019
/
S1
/
3
/
b
/
Solution
by
Codex
0
2026-10-25
Put
z
=
e
i
θ
, so
d
θ
=
d
z
/
(
i
z
)
,
cos
2
θ
=
(
z
2
+
z
−
2
)
/2
and
cos
2
3
θ
=
(
z
6
+
2
+
z
−
6
)
/4
. The
contour integral
is
4
i
1
∮
∣
z
∣
=
1
z
(
5
z
2
−
2
z
4
−
2
)
z
8
+
2
z
2
+
z
−
4
d
z
.
(1)
The enclosed
poles
are
0
and
±
1/
2
. For the
rational function
displayed without the prefactor
1/
(
4
i
)
, the
residue
at zero is
−
21/8
and the two other
residues
sum
to
27/8
. Therefore the
residue theorem
gives
∫
0
2
π
5
−
4
cos
2
θ
cos
2
3
θ
d
θ
=
4
i
1
(
2
πi
)
4
3
=
8
3
π
.
(2)
A
useful
independent check is the
Fourier series
(
5
−
4
cos
ϕ
)
−
1
=
3
1
[
1
+
2
∑
k
≥
1
2
−
k
cos
k
ϕ
]
; only its constant and third
Fourier coefficients
survive.
Total
articles
:
1
New to
topics
?
Read the docs here!