Returning to Eq. (1), we now impose the boundary condition
u
(
r
=
a, θ
) =
u
0
, and solve
for the coefficients by inverting the Fourier sine series; we do the inversion in the usual way,
by multiplying both sides by a trig function with index
m
, integrating over angles, and using
orthogonality on the RHS to collapse the sum to
n
=
m
ˆ
β
0
u
(
r
=
a, θ
) sin
mπ
β
θ
dθ
=
X
n
B
n
a
k
n
ˆ
β
0
sin
nπ
β
θ
sin
mπ
β
θ
dθ
=
B
m
a
k
m
×
β
2
The LHS in the above equation works out as follows
LHS
=
u
0
ˆ
β
0
sin
mπ
β
θ
dθ
=
u
0
β
mπ

cos
mπ
β
θ
β
0
=
2
u
0
β
mπ
,
m
=
odd
=
0
,
m
=
even
So putting together the last two equations, we at last have an expression for the coefficients
B
n
a
k
n
=
4
u
0
nπ
,
n
=
odd
4
Problem II
(a)
Use the CauchyRiemann equations to show that the function
sin
z
is analytic through
out the finite complex plane.
(b)
Show that if a function
f
(
z
)
is analytic in a region

z

< R,
then the function
f
(1
/z
)
is analytic in the region

z

>
1
/R
.
(c)
Find the Laurent series expansion for the function
f
(
z
) =
z
2
sin
1
z
(d)
Use the residue theorem to evaluate the integral
I
=
‰
C
z
2
sin
1
z
dz
where
C
is a counterclockwise oriented closed contour that encloses the origin.
Solution (a)
We can use the standard sine identity to find the real and imaginary parts of the function
f
(
z
)
=
sin (
x
+
iy
) = sin
x
cos
iy
+ cos
x
sin
iy
=
sin
x
cosh
y
+
i
cos
x
sinh
y
=
u
(
x, y
) +
iv
(
x, y
)
So now let’s check that the CauchyRiemann equations are satisfied. First,
∂u
∂x
= cos
x
cosh
y
?
=
∂v
∂y
= cos
x
cosh
y
which is true throughout the finite complex plane (the point at infinity is excluded, since
the hyperbolic trig functions blow up). Second
∂u
∂y
= cos
x
sinh
y
?
=

∂v
∂x
=

cos
x
sinh
y
which is also true throughout the finite complex plane.
Solution (b)
We want to prove that the expression
d
dz
f
1
z
exists in the region

z

>
1
/R
. This is almost trivial, given that
f
(
z
)
is analytic in the region

z

< R
. However, a complete proof requires a little extra finesse. To do this carefully, define
w
=
1
z
5
and use the chain rule to write
d
dz
f
1
z
=
dw
dz
×
df
(
w
)
dw
=

w
2
df
(
w
)
dw
Now the derivative on the RHS exists for

w

< R
, and since
w
2
is finite in this domain, the
derivative on the LHS also exists in this domain. Thus we have established that
f
(1
/z
)
is
analytic for

z

>
1
/R
.
Solution (c)
Thanks to the previous proof, we can simply use the Taylor series expansion for
sin
z
to
work this out
z
2
sin
1
z
=
z
2
∞
X
n
=0
1
(2
n
+ 1)!
1
z
2
n
+1
=
z
+
∞
X
n
=1
c

n
z
1

2
n
where I split off the regular and principal parts of the series, with
c

n
= 1
/
(2
n
+ 1)!
Note
about convergence: since
sin
z
is analytic for all finite
z
,
sin(1
/z
)
is anaytical for all
z >
0
.
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 Winter '20
 Physics, Trigonometry, Derivative, Taylor Series, Euler's formula, θ dθ