2 Analytic Functions 2.1 Functions and Mappings Let S be a set of complex numbers. Definition: A function f on S is a rule that assigns to each value in z ∈ S a complex number w, denoted f (z )= w. • The number w is called the image of f at z . The set S is called the domain of the function. • If T ⊆ S then the set of images of z ∈ T is called the image of T . • The image of S is called the range of f . • The set of all points z that have w as their image is called the inverse image of w. • Suppose f (x + iy)= u + iv then u and v both depend on x and y, so we say f (x + iy)= u(x, ty)+ iv(x, y). If v(x, y) = 0 then f is a real-valued function. • When we use f to associated z =(x, y) and w =(u, v) we call f a mapping. • If n ∈ Z + and a 0 , ..., a n ∈ C with a n 6= 0, then P (z )= a 0 + a 1 z + ... + a n z n is a polynomial of degree n. • Quotients P (z )/Q(z ) of polynomials are called rational functions. • Using polar coordinates, we have f (re iθ )= u + iv then we may write f (z )= u(r, θ)+ iv(r, θ) Examples: 1. Let S = C \ 0 and define f by w =1/z . 1
18
Embed
2 Analytic Functions - Duquesne Universityhaensch/304Materials/Chapter... · 2018. 5. 24. · 2 Analytic Functions 2.1 Functions and Mappings Let Sbe a set of complex numbers. De
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
2 Analytic Functions
2.1 Functions and Mappings
Let S be a set of complex numbers.
Definition: A function f on S is a rule that assigns to each value in z ∈ S a complex numberw, denoted f(z) = w.
• The number w is called the image of f at z. The set S is called the domain of the function.
• If T ⊆ S then the set of images of z ∈ T is called the image of T .
• The image of S is called the range of f .
• The set of all points z that have w as their image is called the inverse image of w.
• Supposef(x+ iy) = u+ iv
then u and v both depend on x and y, so we say
f(x+ iy) = u(x, ty) + iv(x, y).
If v(x, y) = 0 then f is a real-valued function.
• When we use f to associated z = (x, y) and w = (u, v) we call f a mapping.
• If n ∈ Z+ and a0, ..., an ∈ C with an 6= 0, then
P (z) = a0 + a1z + ...+ anzn
is a polynomial of degree n.
• Quotients P (z)/Q(z) of polynomials are called rational functions.
• Using polar coordinates, we have
f(reiθ) = u+ iv
then we may writef(z) = u(r, θ) + iv(r, θ)
Examples:
1. Let S = C \ 0 and define f by w = 1/z.
1
2. Suppose f(z) = z2, then
f(x+ iy) = (x+ iy)2+ = x2 − y2 + 2ixy
and sou(x, y) = x2 − y2 and v(x, y) = 2xy
3. Consider the real-valued function
f(z) =| z |2= x2 + y2 + i0.
4. Suppose w = z2 and z = reiθ. Then
w =(reiθ
)2= r2e2iθ = r2 cos(2θ) + ir2 sin(2θ)
and sou(r, θ) = r2 cos(2θ) and v(r, θ) = r2 sin(2θ)
5. Let z 6= 0, then r =| z | and Θ = Arg(z), and
z12 =√re
iΘ+2πk2
where k = 0, 1 and hence
z12 = ±
√re
iΘ2
Choosing only the positive value, we get
f(z) =√re
iΘ2 r > 0,−π < Θ ≤ π.
This is a well-defined function on C \ 0. Define f(0) = 0, then f is defined on C.
6. We talk about some geometric characteristics of a mapping. The mapping
w = z + 1 = (x+ 1) + iy
is a translation of every point one unit to the right. The mapping
w = iz = rei(θ+π/2)
where z = reiθ is a rotation of every point counter clockwise a quarter turn, i.e. π/2radians. The mapping
w = z = x− iy
is a reflection of every point over the real axis.
2
7. Let’s study the mapping w = z2 as a transformation from the xy-plane into the uv-plane.How does that mapping look? From Example 2, we know that
u(x, y) = x2 − y2 and v(x, y) = 2xy.
Now consider the real-valued hyperbola
x2 − y2 = c1 (c1 > 0)
and consider how this can be mapped onto the line u = c1. If (x, y) is a point on theright-hand branch (RHB) or LHB, then we already know that
c1 = x2 − y2
sou = c1
From here it follows thatx = ±
√y2 − c1
sov = ±2y
√y2 − c1.
So upward movement along RHB maps upward along u = c1 and downward movementalong LHB moves upward along u = c1.
We can also consider the hyperbola (dotted line)
2xy = c2 (c2 > 0).
If (x, y) is a point on either branch, then
2xy = c2
sov = c2.
3
For the RHB we havey =
c2
2x(0 < x <∞)
and so
u = x2 −( c2
2x
)2= x2 − c2
2
4x2
and sincelimx→0+
u = −∞ and limx→∞
=∞
we see that moving down along RHB is the same as moving left to right along v = c2.
For the LHB, we see that
x =c2
2y(−∞ < y < 0)
and so
u =
(c2
2y
)2
− y2 =c2
2
4y2− y2
and sincelimy→0+
u =∞ and limy→∞
= −∞
we see that moving up along LHB is the same as moving left to right along v = c2.
8. Consider the domainD := {(x, y) : x > 0, y > 0, xy < 1}
consisting of all points lying on the upper branch of the hyperbolas
2xy = c
where 0 < c < 2. We know:
• Moving down along 2xy corresponds to moving left-to-right along v = c.
• For any c between 0 and 2 we get a corresponding unique horizontal line. So D mapsonto the horizontal strip 0 < v < 2.
9. Using polar coordinates,w = z2 = r2ei2θ
when z = reiθ. So to get the image of any point z we need to:
• square the modulus
• double the argument
Therefore, points on a circle, are transformed to a larger circle, moving counterclockwise.So
quarter circle→ half circle
andhalf circle→ circle
4
The upper half plane is mapped to the entire plane. But then the mapping is not one toone, since
positive real axis→ positive real axis
andnegative real axis→ positive real axis
but you already knew that!
2.2 Limits
Let f be a function defined at all points z in some deleted neighborhood of z0, that is,
0 <| z − z0 |< c.
Definition: We say the limit of flimz→z0
f(z) = w0
if w = f(z) can be made arbitrarily close to w0 if we choose z close to z0. Technically, thismeans for every ε > 0, there is a δ > 0 such that
| f(z)− w0 |< ε whenver 0 <| z − z0 |< δ.
This means
5
• For every ε-nbhd| w − w0 |< ε
of w0 there is a deleted nbhd0 <| z − z0 |< δ
for which every point z has an image in the ε-nbhd.
• The imagine of the deleted nbhd isn’t necessarily the ε-nbhd.
• Once δ has been found, it can be replaced with any δ′ < δ.
Theorem 1. When a limit of a function f(z) exists at a point z0, it is unique.
Proof. Suppose thatlimz→z0
f(z) = w0 and limz→z0
f(z) = w1.
Then for every ε > 0 there are δ0 > 0 and δ1 > 0 such that
by the triangle inequality. And so from the statements above,
0 <| z − z0 |< δ =⇒| f(z)− w0 |<ε
2+ε
2= ε.
The other direction can be proved similarly.
7
Theorem 3. Suppose that
limz→z0
f(z) = w0 and limz→z0
F (z) = W0
thenlimz→z0
[f(z) + F (z)] = w0 +W0
andlimz→z0
[f(z)F (z)] = w0W0
and if W0 6= 0,
limz→z0
f(z)
F (z)=w0
W0
Now let’s consider limits at infinity. We can consider C ∪∞ by the following diagram.
So a neighborhood of infinity can just be considered as a neighborhood of the north pole.
Theorem 4. If z0 and w0 are points in the z and w planes respectively, then
limz→z0
1
f(z)= 0 =⇒ lim
z→z0f(z) =∞
and
limz→∞
f
(1
z
)= w0 =⇒ lim
z→0f(z) = w0.
Moreover,
limz→0
1
f(z)=∞ =⇒ lim
z→0
1
f (1/z)= 0.
Examples:
1. limz→−1z+1iz+3 = 0 =⇒ limz→−1
iz+3z+1 =∞
2. limz→0(2/z)+i(1/z)+1 = limz→0
2+iz1+z = 2 =⇒ limz→∞
2z+iz+1 = 2
3. limz→0(1/z2)+1(2/z3)−1
= limz→0z+z3
2−z3 = 0 =⇒ limz→∞2z3−1z2+1
8
2.3 Continuity
Definition: A function f is continuous at a point z0 if all three of the following are satisfied:
1. limz→z0 f(z) exists
2. f(z0) exists.
3. limz→z0 f(z) = f(z0)
If f and g are continuous at z0, then
• f ± g is continuous at z0
• fg is continuous at z0
• fg is continuous at z0 provided that g(z0) 6= 0.
Theorem 5. The composition of continuous functions is continuous.
Theorem 6. If a function f(z) is continuous and non-zero at a point z0 then f(z) 6= 0 throughsome neighborhood of that point.
Proof. Assuming f(z) is continuous, for all ε > 0 there is some δ > 0 so that
| z − z0 |< δ =⇒| f(z)− f(z0) |< ε.
So let ε = |f(z0)|2 . Then
| z − z0 |< δ =⇒| f(z)− f(z0) |< | f(z0) |2
.
But suppose there is some z in the neighborhood | z − z0 |< δ for which f(z) = 0, then
| f(z0) |< | f(z0) |2
a contradiction.
9
Theorem 7. If the component functions u and v are continuous at a point z0 = (x0, y0), thenso is f . Conversely, if f is continuous at z0 then u and v are continuous at z0.
Proof. Follows immediately from limit rules.
Recall the following:
• A region is closed =⇒ it contains all of its boundary points.
• A region is bounded =⇒ it lies inside a circle centered at the origin.
Theorem 8. If a function f is continuous throughout a region R that is closed and bounded,there exists a nonnegative real number M such that
| f(z) |≤M
for all points z ∈ R, and equality holds for at least one z.
Proof. Since f(x, y) = u(x, y) + iv(x, y) is continuous it follows that√u(x, y)2 + v(x, y)2
is continuous on R, and therefore must reach a maximum value in R, call it M .
2.4 Derivatives
Definition: The derivative of f at z0 is the limit
f ′(z0) = limz→z0
f(z)− f(z0)
z − z0
and f is differentiable at z0 if f ′(z0) exists.
Letting∆z = z − z0 (z 6= z0)
we get
f ′(z0) = limz→z0
f(z0 + ∆z)− f(z0)
∆z
and
f defined on nbhd around z0 =⇒ f(z0 + ∆z) defined for | ∆z | small.
Sometimes we setw = f(z + ∆z)− f(z)
and so
f ′(z) = lim∆z→0
∆w
∆z=dw
dz.
Examples:
10
1. Suppose that f(z) = 1/z. For any nonzero points,
Now ∆z must approach the point 0, if it approaches along the real axis, then
∆z = (x− x0) + i0 =⇒ ∆z = ∆z.
and if it approaches along the imaginary axis, then
∆z = 0 + i(y − y0) =⇒ ∆z = −∆z.
Therefore in each case we have
f ′(z) = 1 or f ′(z) = −1
which don’t agree. So no limit exists.
3. Let f(z) =| z |2. Then
∆w
∆z=| z + ∆z |2 − | z |2
∆z=
(z + ∆z)(z + ∆z)− zz∆z
= z + ∆z + z∆z
∆z. (3)
As in example 2, we have∆z = ∆z or ∆z = −∆z
depending how we approach. So we have
∆w
∆z= z + ∆z + z or
∆w
∆z= z + ∆z − z,
respectively. If limz→z0 exists, then the uniqueness of limits tells us
z + z = z − z,
in other words, z = 0. To show that this limit exists, observe that when z = 0, then (3)becomes
∆w
∆z= ∆z
which of course exists.
11
Some interesting facts now emerge:
• f can be differentiable at a point, but not in any nbhd of the point.
• When f(z) =| z |2= (x+iy)(x−iy) = x2+y2 its real part u(x, y) = x2+y2 is differentiableeverywhere, and it’s complex part v(x, y) = 0 is differentiable everywhere, but f is not!
• The component functions u and v of f(z) =| z |2 are continuous everywhere, this meanscontinuity does not imply differentiability.
Mostly the rules for differentiation are the same as in calculus. For c ∈ C,
• ddz c = 0
• ddz c · f(z) = c · f ′(z)
• ddz z
n = nzn−1 for n ∈ Z+.
• ddz [f(z) + g(z)] = f ′(z) + g′(z)
• ddzf(z)g(z) = f ′(z)g(z) + f(z)g′(z).
• ddz
[f(z)g(z)
]= g(z)f ′(z)−f(z)g′(z)
[g(z)]2when g(z) 6= 0.
• For F (z) = g[f(z)] with f differentiable at z0 and g differentiable at f(z0), we have
F ′(z0) = g′[f(z0)]f ′(z0).
Examples:
1. Let’s find the derivative of f(z) = (1 − 4z2)3. First, write w = (1 − 4z2) and W = w3.Then
These equations in (4) are called the Cauchy Riemann equations.
Theorem 9. Suppose thatf(z) = u(x, y) + iv(x, y)
and that f ′(z) exists at a point z0 = x0 + iy0. Then the first order partial derivatives of u andv exist and satisfy
ux = vy and uu = −vx.
Also,f ′(z0) = ux(x0, y0) + ivx(x0, y0).
13
Examples:
1. The function f(z) = z2 is differentiable everywhere. To verify C-R are satisfied, recall
f(z) = z2 = x2 − y2 + 2ixy
sou(x, y) = x2 − y2 and v(x, y) = 2xy
Therefore,ux(x, y) = 2x and vx(x, y) = 2y
anduy(x, y) = −2y and vy(x, y) = 2x
soux = vy and uu = −vx,
andf ′(z) = 2x+ 2iy = 2z,
as expected.
2. When f(z) =| z |2 we know this is only differentiable at z = 0. And we have
f(z) =| z |2= x2 + y2
sou(x, y) = x2 + y2 and v(x, y) = 0
thereforeux(x, y) = 2x and uy(x, y) = 2y
so C-R can only be satisfied when x = y = 0.
3. Let f(z) = u(x, y) + iv(x, y) defined by
f(z) =
{z2/z when z 6= 0
0 when z = 0
then its real and imaginary parts are
u(x, y) =x3 − 3xy2
x2 + y2and v(x, y) =
y3 − 3x2y
x2 + y2
when (x, y) 6= (0, 0) and u(x, y) = v(x, y) = 0 when (x, y) = (0, 0). Because
ux(0, 0) = lim∆x→0
u(0 + ∆x, 0)− u(0, 0)
∆x= lim
∆x→0
∆x
∆x= 1
and
uy(0, 0) = lim∆y→0
u(0, 0 + ∆y)− u(0, 0)
∆y= lim
∆y→0
∆y
∆y= 1
14
so ux = vy and similarly show that uy = −vx. But look, f ′(0) does not exist if we consider
∆w
∆z=f(z + ∆z)− f(z)
∆z
and when z = 0 this becomes
∆w
∆z=f(∆z)
∆z=
(∆z)2/∆z
∆z=
(∆z)2
(∆z)2
if we approach along (∆x, 0) this is 1, if we approach along (0,∆y) then this is 1, Supposewe approach along the line (x, x), then ∆z = (∆x,∆x) and
∆w
∆z=
(∆x− i∆x)2
(∆x+ i∆x)2= −1.
So these conditions are not sufficient!
Theorem 10. Let the function f(z) = u(x, y) + iv(x, y) be defined throughout some ε-nbhd ofa point z0 = x0 + iy0 and suppose that
1. the first-order partial derivatives of the functions u and v with respect to x and y existeverywhere in the neighborhood;
2. the partial derivates are continuous at (x0, y0) and satisfy the C-R equations,
ux = vy and uy = −vx,
then f ′(z0) exists andf ′(z0) = ux + ivx
evaluated at (x0, y0).
Examples:
1. Considerf(z) = exeiy = ex cos(y) + iex sin(y)
thenu(x, y) = ex cos(y) and v(x, y) = ex sin(y).
These exist and are continuous everywhere, moreover, ux = vy and uy = −vx, so
f ′(z) = ex cos(y) + iex sin(y) = exeiy = f(z).
2. The function f(z) =| z |2 with
u(x, y) = x2 + y2 and v(x, y) = 0
has a derivative at 0, and f ′(0) = 0 + i0 = 0.
3. Be careful, before we use the expression for f ′(z) in the theorem we must be sure thatf ′(z) at z0 exists. For example, in the example above, we could easily say
f ′(x+ iy) = 2x
for any z, this would not be true.
15
2.6 Polar Coordinates
For z0 6= 0 we usex = r cos(θ) and y = r sin(θ)
then when w = f(z) = u(r, θ) + iv(r, θ) then using the chain rule for multivariable real-valuedfunction
∂u
∂r=∂u
∂x
∂x
∂r+∂u
∂y
∂y
∂r
and∂u
∂θ=∂u
∂x
∂x
∂θ+∂u
∂y
∂y
∂θ
so we can write
ur = ux cos(θ) + uy sin(θ) and uθ = −uxr sin(θ) + uyr cos(θ)
Theorem 11. Let the function f(z) = u(r, θ) + iv(r, θ) be defined throughout some ε-nbhd of apoint z0 = r0e
iθ0 and suppose that
1. the first-order partial derivatives of the functions u and v with respect to r and θ existeverywhere in the neighborhood;
2. the partial derivates are continuous at (r0, θ0) and satisfy the C-R equations,
rur = vθ and uθ = −rvr,
then f ′(z0) exists andf ′(z0) = e−iθ(ur + ivr)
evaluated at (r0, θ0).
Examples:
1. If
f(z) =1
z2=
1
(reiθ)2 =
1
r2e−2iθ =
1
r2(cos(2θ)− i sin(2θ))
so
u(r, θ) =1
r2cos(2θ) and v(r, θ) = − 1
r2sin(2θ).
Therefore,
rur =−2
r2cos(2θ) = vθ and uθ = − 2
r2sin(2θ) = −rvr.
16
Since partials exists, are continuous, and polar C-R are satisfied at every non-zero point,the derivative exists when z 6= 0 and
f ′(z) = e−iθ(−2
r3cos(2θ) + i
2
r3sin(2θ)
)=−2e−iθ
r3(cos(2θ)− i sin(2θ))
=−2e−iθ
r3(cos(−2θ) + i sin(−2θ))
=−2e−iθ
r3· e−2iθ
=−2
r3e3iθ
=−2
z3
2. Now we will show that every branch of the square root function has a derivate in itsdomain of definition. Recall that
f(z) =√z =√reiθ/2 =
√r cos (θ/2) + i
√r sin (θ/2)
so
rur =
√r
2cos (θ/2) = vθ
and
uθ =−√r
2sin (θ/2) = −rvr
exists and are continuous on their domain of definition, therefore,
f ′(z) = e−iθ(
1
2√r
cos (θ/2) + i1
2√r
sin (θ/2)
)=e−iθ
2√reiθ/2 =
1
2√reiθ/2
=1
2f(z)
2.7 Analytic Functions
Definition: A complex function f is
• analytic in an open set S if it has a derivative at every point in S;
• analytic at z0 if it is analytic in some nbhd of z0;
• entire if it analytic for every point in C.
• If a function is analytic at all but one point, we call that point a singularity.
Examples:
1. f(z) = 1/z is analytic for all z 6= 0
17
2. f(z) =| z |2 is not analytic anywhere since der. only exists at z = 0.
3. All polynomials are entire.
Some comments:
• Analytic in D =⇒ continuous in D
• Analytic in D =⇒ C-R satisfied in D
• sum/product/quotient/composition of analytic =⇒ analytic.
Theorem 12. If f ′(z) = 0 everywhere in a domain D, then f(z) must be constant throughoutD.
Examples:
1. The quotient
f(z) =z2 + 3
(z + 1)(z2 + 5)
is analytic except at singular points −1 and i√
5.
2. Suppose f(z) = u(x, y) + iv(x, y) and f(z) are both analytic throughout their domain,D. We will show that f(z) must constant throughout D. Consider,
f(z) = U(x, y) + iV (x, y)
where U = u and V = −v. Because f is analytic, and because of C-R, we know
ux = vy and uy = −vx
are true in D. Moreover,Ux = Vy and Uy = −Vx
implyingux = −vy and uy = vx.
Therefore ux = 0 and vx = 0 so f ′(z) = 0 and therefore f(z) is constant.
3. Suppose f(z) is analytic throughout the domain D, and suppose that the modulus usconstant, so | f(z) |= c for all z ∈ D. We will show that f(z) is constant throughout D.If c = 0, then this is obvious, since f(z) = 0 everywhere. Suppose c 6= 0, then zz =| z |2implies that