Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx

Now we compute the derivative.
d
dz
log z =
e
−ıθ
∂
∂r
(Log r + ıθ)
=
e
−ıθ
1
r
=
1
z
Solution 8.14
The complex derivative in the coordinate directions is
d
dz
=
e
−ıθ
∂
∂r
= −
ı
r
e

−ıθ
∂
∂θ
.
We substitute f = u + ıv into this identity to obtain the Cauchy-Riemann equation in polar co ordinates.
e
−ıθ
∂f
∂r
= −
ı
r
e
−ıθ
∂f
∂θ
∂f
∂r
= −
ı
r
∂f
∂θ
u
r
+ ıv
r
= −
ı
r

(u
θ
+ ıv
θ
)
We equate the real and imaginary parts.
u
r
=
1
r
v
θ
, v
r
= −
1
r
u
θ
u
r
=
1
r
v
θ
, u
θ
= −rv

r
Solution 8.15
Since w is analytic, u and v satisfy the Cauchy-Riemann equations,
u
x
= v
y
and u
y
= −v
x
.
414
Using the chain rule we can write the derivatives with respect to x and y in terms of u and v.
∂
∂x
= u
x
∂
∂u
+ v
x
∂
∂v
∂
∂y
= u
y
∂
∂u

+ v
y
∂
∂v
Now we examine φ
x
− ıφ
y
.
φ
x
− ıφ
y
= u
x
Φ
u
+ v
x
Φ
v
− ı (u
y
Φ
u
+ v
y
Φ
v
)

φ
x
− ıφ
y
= (u
x
− ıu
y
) Φ
u
+ (v
x
− ıv
y
) Φ
v
φ
x
− ıφ
y
= (u
x
− ıu
y
) Φ
u
− ı (v
y
+ ıv
x

) Φ
v
We use the Cauchy-Riemann equations to write u
y
and v
y
in terms of u
x
and v
x
.
φ
x
− ıφ
y
= (u
x
+ ıv
x
) Φ
u
− ı (u
x
+ ıv
x
) Φ
v
Recall that w

= u

x
+ ıv
x
= v
y
− ıu
y
.
φ
x
− ıφ
y
=
dw
dz
(Φ
u
− ıΦ
v
)
Thus we see that,
∂Φ
∂u
− ı
∂Φ
∂v
=

dw
dz


−1

∂φ
∂x
− ı
∂φ
∂y

.
We write this in operator notation.
∂
∂u
− ı
∂
∂v
=

dw
dz

−1

∂
∂x
− ı
∂
∂y

415

The complex conjugate of this relation is
∂
∂u
+ ı
∂
∂v
=

dw
dz

−1

∂
∂x
+ ı
∂
∂y

Now we apply both these operators to Φ = φ.

∂
∂u
+ ı
∂
∂v

∂
∂u
− ı

∂
∂v

Φ =

dw
dz

−1

∂
∂x
+ ı
∂
∂y

dw
dz

−1

∂
∂x
− ı
∂
∂y

φ

∂

2
∂u
2
+ ı
∂
2
∂u∂v
− ı
∂
2
∂v∂u
+
∂
2
∂v
2

Φ
=

dw
dz

−1


∂
∂x
+ ı
∂

∂y

dw
dz

−1


∂
∂x
− ı
∂
∂y

+

dw
dz

−1

∂
∂x
+ ı
∂
∂y

∂
∂x
− ı

∂
∂y


φ
(w

)
−1
is an analytic function. Recall that for analytic functions f, f

= f
x
= −ıf
y
. So that f
x
+ ıf
y
= 0.
∂
2
Φ
∂u
2
+
∂
2
Φ
∂v

2
=

dw
dz

−1


dw
dz

−1

∂
2
∂x
2
+
∂
2
∂y
2


φ
∂
2
Φ
∂u

2
+
∂
2
Φ
∂v
2
=




dw
dz




−2

∂
2
φ
∂x
2
+
∂
2
φ
∂y

2

Solution 8.16
1. We consider
f(z) = log |z| + ı arg(z) = log r + ıθ.
The Cauchy-Riemann equations in polar coordinates are
u
r
=
1
r
v
θ
, u
θ
= −rv
r
.
416
We calculate the derivatives.
u
r
=
1
r
,
1
r
v
θ

=
1
r
u
θ
= 0, −rv
r
= 0
Since the Cauchy-Ri emann equations are satisfied and the partial derivatives are continuous, f(z) is analytic in
|z| > 0, |arg(z)| < π. The complex derivative in terms of polar coordinates is
d
dz
=
e
−ıθ
∂
∂r
= −
ı
r
e
−ıθ
∂
∂θ
.
We use this to differentiate f(z).
df
dz
=
e

−ıθ
∂
∂r
[log r + ıθ] =
e
−ıθ
1
r
=
1
z
2. Next we consider
f(z) =

|z|
e
ı arg(z)/2
=
√
r
e
ıθ/2
.
The Cauchy-Riemann equations for polar coordinates and the polar form f(z) = R(r, θ)
e
ıΘ(r,θ)
are
R
r
=

R
r
Θ
θ
,
1
r
R
θ
= −RΘ
r
.
We calculate the derivatives for R =
√
r, Θ = θ/2.
R
r
=
1
2
√
r
,
R
r
Θ
θ
=
1
2

√
r
1
r
R
θ
= 0, −RΘ
r
= 0
Since the Cauchy-Ri emann equations are satisfied and the partial derivatives are continuous, f(z) is analytic in
|z| > 0, |arg(z)| < π. The complex derivative in terms of polar coordinates is
d
dz
=
e
−ıθ
∂
∂r
= −
ı
r
e
−ıθ
∂
∂θ
.
417
We use this to differentiate f(z).
df
dz

=
e
−ıθ
∂
∂r
[
√
r
e
ıθ/2
] =
1
2
e
ıθ/2
√
r
=
1
2
√
z
Solution 8.17
1. We consider the function
u = x Log r −y arctan(x, y) = r cos θ Log r − rθ sin θ
We compute the Laplacian.
∆u =
1
r
∂

∂r

r
∂u
∂r

+
1
r
2
∂
2
u
∂θ
2
=
1
r
∂
∂r
(cos θ(r + r Log r) − θ sin θ) +
1
r
2
(r(θ sin θ − 2 cos θ) − r cos θ Log r)
=
1
r
(2 cos θ + cos θ Log r − θ sin θ) +
1

r
(θ sin θ − 2 cos θ − cos θ Log r)
= 0
The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations.
v
r
= −
1
r
u
θ
, v
θ
= ru
r
v
r
= sin θ(1 + Log r) + θ cos θ, v
θ
= r (cos θ(1 + Log r) −θ sin θ)
We integrate the first equation with respect to r to determine v to within the constant of integration g(θ).
v = r(sin θ Log r + θ cos θ) + g(θ)
We differentiate this expression with respect to θ.
v
θ
= r (cos θ(1 + Log r) −θ sin θ) + g

(θ)
418
We compare this to the second Cauchy-Riemann equation to see that g


(θ) = 0. Thus g(θ) = c. We have
determined the harmonic conjugate.
v = r(sin θ Log r + θ cos θ) + c
The corresponding analytic function is
f(z) = r cos θ Log r − rθ sin θ + ı(r sin θ Log r + rθ cos θ + c).
On the positive real axis, (θ = 0), the function has the value
f(z = r) = r Log r + ıc.
We use analytic continuation to determine the function in the complex plane.
f(z) = z log z + ıc
2. We consider the function
u = Arg(z) = θ.
We compute the Laplacian.
∆u =
1
r
∂
∂r

r
∂u
∂r

+
1
r
2
∂
2
u

∂θ
2
= 0
The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations.
v
r
= −
1
r
u
θ
, v
θ
= ru
r
v
r
= −
1
r
, v
θ
= 0
We integrate the first equation with respect to r to determine v to within the constant of integration g(θ).
v = −Log r + g(θ)
419
We differentiate this expression with respect to θ.
v
θ
= g


(θ)
We compare this to the second Cauchy-Riemann equation to see that g

(θ) = 0. Thus g(θ) = c. We have
determined the harmonic conjugate.
v = −Log r + c
The corresponding analytic function is
f(z) = θ − ı Log r + ıc
On the positive real axis, (θ = 0), the function has the value
f(z = r) = −ı Log r + ıc
We use analytic continuation to determine the function in the complex plane.
f(z) = −ı log z + ıc
3. We consider the function
u = r
n
cos(nθ)
We compute the Laplacian.
∆u =
1
r
∂
∂r

r
∂u
∂r

+
1

r
2
∂
2
u
∂θ
2
=
1
r
∂
∂r
(nr
n
cos(nθ)) − n
2
r
n−2
cos(nθ)
= n
2
r
n−2
cos(nθ) − n
2
r
n−2
cos(nθ)
= 0
420

The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations.
v
r
= −
1
r
u
θ
, v
θ
= ru
r
v
r
= nr
n−1
sin(nθ), v
θ
= nr
n
cos(nθ)
We integrate the first equation with respect to r to determine v to within the constant of integration g(θ).
v = r
n
sin(nθ) + g(θ)
We differentiate this expression with respect to θ.
v
θ
= nr
n

cos(nθ) + g

(θ)
We compare this to the second Cauchy-Riemann equation to see that g

(θ) = 0. Thus g(θ) = c. We have
determined the harmonic conjugate.
v = r
n
sin(nθ) + c
The corresponding analytic function is
f(z) = r
n
cos(nθ) + ır
n
sin(nθ) + ıc
On the positive real axis, (θ = 0), the function has the value
f(z = r) = r
n
+ ıc
We use analytic continuation to determine the function in the complex plane.
f(z) = z
n
4. We consider the function
u =
y
r
2
=
sin θ

r
421
We compute the Laplacian.
∆u =
1
r
∂
∂r

r
∂u
∂r

+
1
r
2
∂
2
u
∂θ
2
=
1
r
∂
∂r

−
sin θ

r

−
sin θ
r
3
=
sin θ
r
3
−
sin θ
r
3
= 0
The function u is harmonic. We find the harmonic conjugate v by solving the Cauchy-Riemann equations.
v
r
= −
1
r
u
θ
, v
θ
= ru
r
v
r
= −

cos θ
r
2
, v
θ
= −
sin θ
r
We integrate the first equation with respect to r to determine v to within the constant of integration g(θ).
v =
cos θ
r
+ g(θ)
We differentiate this expression with respect to θ.
v
θ
= −
sin θ
r
+ g

(θ)
We compare this to the second Cauchy-Riemann equation to see that g

(θ) = 0. Thus g(θ) = c. We have
determined the harmonic conjugate.
v =
cos θ
r
+ c

The corresponding analytic function is
f(z) =
sin θ
r
+ ı
cos θ
r
+ ıc
422
On the positive real axis, (θ = 0), the function has the value
f(z = r) =
ı
r
+ ıc.
We use analytic continuation to determine the function in the complex plane.
f(z) =
ı
z
+ ıc
Solution 8.18
1. We calculate the first partial derivatives of u = (x −y)
2
and v = 2(x + y).
u
x
= 2(x −y)
u
y
= 2(y − x)
v

x
= 2
v
y
= 2
We substitute these expressions into the Cauchy-Riemann equations.
u
x
= v
y
, u
y
= −v
x
2(x −y) = 2, 2(y − x) = −2
x −y = 1, y − x = −1
y = x −1
Since the Cauchy-Riemann equation are satisfied along the line y = x−1 and the partial derivatives are continuous,
the function f(z) is differentiable there. Since the function is not differentiable in a neighborhood of any point,
it is nowhere analytic.
423
2. We calculate the first partial derivatives of u and v.
u
x
= 2
e
x
2
−y
2

(x cos(2xy) − y sin(2xy))
u
y
= −2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
v
x
= 2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
v
y
= 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy))
Since the Cauchy-Riemann equations, u
x

= v
y
and u
y
= −v
x
, are satisfied everywhere and the partial derivatives
are continuous, f(z) is everywhere differentiable. Since f(z) is differentiable in a neighborhood of every point, it
is analytic in the complex plane. (f(z) is entire.)
Now to evaluate the derivative. The complex derivative is the derivative in any direction. We choose the x
direction.
f

(z) = u
x
+ ıv
x
f

(z) = 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy)) + ı2
e
x
2
−y

2
(y cos(2xy) + x sin(2xy))
f

(z) = 2
e
x
2
−y
2
((x + ıy) cos(2xy) + (−y + ıx) sin(2xy))
Finding the derivative is easier if we first write f(z) in terms of the complex variable z and use complex differen-
tiation.
f(z) =
e
x
2
−y
2
(cos(2x, y) + ı sin(2xy))
f(z) =
e
x
2
−y
2
e
ı2xy
f(z) =
e

(x+ıy)
2
f(z) =
e
z
2
f

(z) = 2z
e
z
2
424
Solution 8.19
1. Assume that the Cauchy-Riemann equations in Cartesian coordinates
u
x
= v
y
, u
y
= −v
x
are satisfied and these partial derivatives are continuous at a point z. We write the derivatives in polar coordinates
in terms of derivatives in Cartesian coordinates to verify the Cauchy-Riemann equations in polar coordinates. First
we calculate the derivatives.
x = r cos θ, y = r sin θ
w
r
=

∂x
∂r
w
x
+
∂y
∂r
w
y
= cos θw
x
+ sin θw
y
w
θ
=
∂x
∂θ
w
x
+
∂y
∂θ
w
y
= −r sin θw
x
+ r cos θw
y
Then we verify the Cauchy-Riemann equations in polar coordinates.

u
r
= cos θu
x
+ sin θu
y
= cos θv
y
− sin θv
x
=
1
r
v
θ
1
r
u
θ
= −sin θu
x
+ cos θu
y
= −sin θv
y
− cos θv
x
= −v
r
This proves that the Cauchy-Riemann equations in Cartesian coordinates hold only if the Cauchy-Riemann equa-

tions in polar coordinates hold. (Given that the partial derivatives are continuous.) Next we prove the converse.
Assume that the Cauchy-Riemann equations in polar coordinates
u
r
=
1
r
v
θ
,
1
r
u
θ
= −v
r
425
are satisfied and these partial derivatives are continuous at a p oint z. We write the derivatives in Cartesian
coordinates in terms of derivatives in polar coordinates to verify the Cauchy-Riemann equations in Cartesian
coordinates. First we calculate the derivatives.
r =

x
2
+ y
2
, θ = arctan(x, y)
w
x
=

∂r
∂x
w
r
+
∂θ
∂x
w
θ
=
x
r
w
r
−
y
r
2
w
θ
w
y
=
∂r
∂y
w
r
+
∂θ
∂y

w
θ
=
y
r
w
r
+
x
r
2
w
θ
Then we verify the Cauchy-Riemann equations in Cartesian coordinates.
u
x
=
x
r
u
r
−
y
r
2
u
θ
=
x
r

2
v
θ
+
y
r
v
r
= u
y
u
y
=
y
r
u
r
+
x
r
2
u
θ
=
y
r
2
v
θ
−

x
r
v
r
= −u
x
This proves that the Cauchy-Riemann equations in polar coordinates hold only if the Cauchy-Riemann equations
in Cartesian coordinates hold. We have demonstrated the equivalence of the two forms.
2. We verify that log z is analytic for r > 0 and −π < θ < π using the polar form of the Cauchy-Riemann equations.
Log z = ln r + ıθ
u
r
=
1
r
v
θ
,
1
r
u
θ
= −v
r
1
r
=
1
r
1,

1
r
0 = −0
426
Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous for r > 0, log z is
analytic there. We calculate the value of the derivative using the polar differentiation formulas.
d
dz
Log z =
e
−ıθ
∂
∂r
(ln r + ıθ) =
e
−ıθ
1
r
=
1
z
d
dz
Log z =
−ı
z
∂
∂θ
(ln r + ıθ) =
−ı

z
ı =
1
z
3. Let {x
i
} denote rectangular coordinates in two dimensions and let {ξ
i
} be an orthogonal coordinate system .
The distance metric coefficients h
i
are defined
h
i
=


∂x
1
∂ξ
i

2
+

∂x
2
∂ξ
i


2
.
The Laplacian is
∇
2
u =
1
h
1
h
2

∂
∂ξ
1

h
2
h
1
∂u
∂ξ
1

+
∂
∂ξ
2

h

1
h
2
∂u
∂ξ
2

.
First we calculate the distance metric coefficients in polar coordinates.
h
r
=


∂x
∂r

2
+

∂y
∂r

2
=

cos
2
θ + sin
2

θ = 1
h
θ
=


∂x
∂θ

2
+

∂y
∂θ

2
=

r
2
sin
2
θ + r
2
cos
2
θ = r
Then we find the Laplacian.
∇
2

φ =
1
r

∂
∂r
(rφ
r
) +
∂
∂θ

1
r
φ
θ

In polar coordinates, Laplace’s equation is
φ
rr
+
1
r
φ
r
+
1
r
2
φ

θθ
= 0.
427
Solution 8.20
1. We compute the Laplacian of u(x, y) = x
3
− y
3
.
∇
2
u = 6x −6y
Since u is not harmonic, it is not the real part of on analytic function.
2. We compute the Laplacian of u(x, y) = sinh x cos y + x.
∇
2
u = sinh x cos y −sinh x cos y = 0
Since u is harmonic, it is the real part of on analytic function. We determine v by solving the Cauchy-Riemann
equations.
v
x
= −u
y
, v
y
= u
x
v
x
= sinh x sin y, v

y
= cosh x cos y + 1
We integrate the first equation to determine v up to an arbitrary additive function of y.
v = cosh x sin y + g(y)
We substitute this into the second Cauchy-Riemann equation. This will determine v up to an additive constant.
v
y
= cosh x cos y + 1
cosh x cos y + g

(y) = cosh x cos y + 1
g

(y) = 1
g(y) = y + a
v = cosh x sin y + y + a
f(z) = sinh x cos y + x + ı(cosh x sin y + y + a)
Here a is a real constant. We write the function in terms of z.
f(z) = sinh z + z + ıa
428
3. We compute the Laplacian of u(r, θ) = r
n
cos(nθ).
∇
2
u = n(n −1)r
n−2
cos(nθ) + nr
n−2
cos(nθ) − n

2
r
n−2
cos(nθ) = 0
Since u is harmonic, it is the real part of on analytic function. We determine v by solving the Cauchy-Riemann
equations.
v
r
= −
1
r
u
θ
, v
θ
= ru
r
v
r
= nr
n−1
sin(nθ), v
θ
= nr
n
cos(nθ)
We integrate the first equation to determine v up to an arbitrary additive function of θ.
v = r
n
sin(nθ) + g(θ)

We substitute this into the second Cauchy-Riemann equation. This will determine v up to an additive constant.
v
θ
= nr
n
cos(nθ)
nr
n
cos(nθ) + g

(θ) = nr
n
cos(nθ)
g

(θ) = 0
g(θ) = a
v = r
n
sin(nθ) + a
f(z) = r
n
cos(nθ) + ı(r
n
sin(nθ) + a)
Here a is a real constant. We write the function in terms of z.
f(z) = z
n
+ ıa
Solution 8.21

1. We find the velocity potential φ and stream function ψ.
Φ(z) = log z + ı log z
Φ(z) = ln r + ıθ + ı(ln r + ıθ)
φ = ln r −θ, ψ = ln r + θ
429
Figure 8.7: The velocity potential φ and stream function ψ for Φ(z) = log z + ı log z.
A branch of these are plotted in Figure 8.7.
Next we find the stream lines, ψ = c.
ln r + θ = c
r =
e
c−θ
These are spirals which go counter-clockwise as we follow them to the origin. See Figure 8.8. Next we find the
velocity field.
v = ∇φ
v = φ
r
ˆ
r +
φ
θ
r
ˆ
θ
v =
ˆ
r
r
−
ˆ

θ
r
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Figure 8.8: Streamlines for ψ = ln r + θ.
The velocity field is shown in the first plot of Figure 8.9. We see that the fluid flows out from the origin along
the spiral paths of the streamlines. The second plot shows the direction of the velocity field.
2. We find the velocity potential φ and stream function ψ.
Φ(z) = log(z −1) + log (z + 1)
Φ(z) = ln |z − 1| + ı arg(z − 1) + ln |z + 1|+ ı arg(z + 1)
φ = ln |z
2
− 1|, ψ = arg(z − 1) + arg(z + 1)
The velocity potential and a branch of the stream function are plotted in Figure 8.10.
The stream lines, arg(z − 1) + arg(z + 1) = c, are plotted in Figure 8.11.
Next we find the velocity field.
v = ∇φ
v =
2x(x
2
+ y
2
− 1)
x
4
+ 2x
2
(y
2
− 1) + (y
2

+ 1)
2
ˆ
x +
2y(x
2
+ y
2
+ 1)
x
4
+ 2x
2
(y
2
− 1) + (y
2
+ 1)
2
ˆ
y
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Figure 8.9: Velocity field and velocity direction field for φ = ln r −θ.
The velo ci ty field is shown in the first plot of Figure 8.12. The fluid is flowing out of sources at z = ±1. The
second plot shows the direction of the velocity field.
Solution 8.22
1. (a) We factor the denominator to see that there are first order poles at z = ±ı.
z
z
2

+ 1
=
z
(z − ı)(z + ı)
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Figure 8.10: The velocity potential φ and stream function ψ for Φ(z) = log(z − 1) + log(z + 1).
Since the function behaves like 1/z at infinity, it is analytic there.
(b) The denominator of 1/ sin z has first order zeros at z = nπ, n ∈ Z. Thus the function has first order poles
at these locations. Now we examine the point at infinity with the change of variables z = 1/ζ.
1
sin z
=
1
sin(1/ζ)
=
ı2
e
ı/ζ
−
e
−ı/ζ
We see that the point at infinity is a singularity of the function. Since the denominator grows exponentially,
there is no multiplicative factor of ζ
n
that will make the function analytic at ζ = 0. We conclude that the
point at infinity is an essential singularity. Since there is no deleted neighborhood of the point at infinity
that does contain first order poles at the locations z = nπ, the point at infinity is a non-isolated singularity.
(c)
log


1 + z
2

= log(z + ı) + log(z − ı)
There are branch points at z = ±ı. Since the argument of the l ogarithm is unbounded as z → ∞ there is
a branch point at infinity as well. Branch points are non-isolated singularities.
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Figure 8.11: Streamlines for ψ = arg(z − 1) + arg(z + 1).
(d)
z sin(1/z) =
1
2
z

e
ı/z
+
e
ı/z

The point z = 0 is a sin gularity. Since the function grows exponentially at z = 0. There is no multiplicative
factor of z
n
that will make the function analytic. Thus z = 0 is an essential singularity.
There are no other singularities in the finite complex plane. We examine the point at infinity.
z sin

1
z


=
1
ζ
sin ζ
The point at infinity is a singularity. We take the limit ζ → 0 to demonstrate that it is a removable
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Figure 8.12: Velocity field and velocity direction field for φ = ln |z
2
− 1|.
singularity.
lim
ζ→0
sin ζ
ζ
= lim
ζ→0
cos ζ
1
= 1
(e)
tan
−1
(z)
z sinh
2
(πz)
=
ı log

ı+z

ı−z

2z sinh
2
(πz)
435
There are branch points at z = ±ı due to the logarithm. These are non-isolated singularities. Note that
sinh(z) has first order zeros at z = ınπ, n ∈ Z. The arctangent has a first order zero at z = 0. Thus there
is a second order pole at z = 0. There are second order poles at z = ın, n ∈ Z \{0} due to the hyperbolic
sine. Since the hyperbolic sine has an essential singularity at infinity, the function has an essential singularity
at i nfini ty as well. The point at infinity is a non-isolated si ngularity because there is no neighborhood of
infinity that does not contain second order poles.
2. (a) (z − ı)
e
1/(z−1)
has a simple zero at z = ı and an isolated essential singularity at z = 1.
(b)
sin(z − 3)
(z − 3)(z + ı)
6
has a removable singularity at z = 3, a pole of order 6 at z = −ı and an essential singularity at z
∞
.
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Chapter 9
Analytic Continuation
For every complex problem, there is a solution that is simple, neat, and wrong.
- H. L. Mencken
9.1 Analytic Continuation
Suppose there is a function, f

1
(z) that is analytic in the domain D
1
and another analytic function, f
2
(z) that is
analytic in the domain D
2
. (See Figure 9.1.)
If the two domains overlap and f
1
(z) = f
2
(z) in the overlap region D
1
∩ D
2
, then f
2
(z) is called an analytic
continuation of f
1
(z). This is an appropriate name since f
2
(z) continues the definition of f
1
(z) outside of its original
domain of definition D
1
. We can define a function f(z) that is analytic in the union of the domains D

1
∪D
2
. On the
domain D
1
we have f(z) = f
1
(z) and f(z) = f
2
(z) on D
2
. f
1
(z) and f
2
(z) are called function elements. There is an
analytic continuation even if the two domains only share an arc and not a two dimensional region.
With more overlapping domains D
3
, D
4
, . . . we could perhaps extend f
1
(z) to more of the complex plane. Sometimes
it is i mpossible to extend a function beyond the boundary of a domain. This is known as a natural boundary. If a
437