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.

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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.

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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.

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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

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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

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• 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

| f(z)− w0 |< ε whenver 0 <| z − z0 |< δ0.

and| f(z)− w1 |< ε whenver 0 <| z − z0 |< δ1.

Sincew1 − w0 = [f(z)− w0]− [f(z)− w1]

from the triangle inequality we have

| w1 − w0 |=| [f(z)− w0]− [f(z)− w1] |≤| f(z)− w0 | + | f(z)− w1 | .

Now pick δ < δ0, δ1, then when 0 <| z − z0 |< δ we have

| w1 − w0 |≤ ε+ ε = 2ε.

But | w1 − w0 | is a non-negative constant, but ε can be chosen to be arbitrarily small, so

w1 − w0 = 0 that is w1 = w0

Examples:

1. Let f(z) = iz/2 with domain the open disk | z |< 1, then

limz→1

f(z) =i

2.

For any z is the disk (note that z 6= 1), we have∣∣∣∣f(z)− i

2

∣∣∣∣ =

∣∣∣∣ iz − i2

∣∣∣∣ =| z − 1 |

2,

so for any ε > 0 we have ∣∣∣∣f(z)− i

2

∣∣∣∣ < ε =⇒| z − 1 |< 2ε.

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2. Let f(z) = zz , then

limz→0

f(z)

does not exist. Watch what happens when a number (x, y) approaches 0. If we approachalong the real axis, then

f(z) =x+ i0

x− i0= 1

but if we approach along the imaginary axis, then

f(z) =0 + iy

0− iy= −1

since these don’t agree, no limit can exist. Remember, the limit is unique!

Theorem 2. Suppose that

f(z) = u(x, y) + iv(x, y) (z = x+ iy)

andz0 = x0 + iy0, w0 = u0 + iv0.

If

lim(x,y)→(x0,y0)

u(x, y) = u0 and lim(x,y)→(x0,y0)

v(x, y) = v0 (1)

then

limz→z0

f(z) = w0, (2)

and, conversely, if (2) is true, then so is statement (1).

Proof. We will suppose (1) holds, and we will show that (2) holds. If (1) holds, this means thatfor every ε > 0 there are δ1, δ2 > 0 such that

0 <| z − z0 |< δ1 =⇒| u− u0 |<ε

2

and0 <| z − z0 |< δ2 =⇒| v − v0 |<

ε

2.

Now pick δ < δ1, δ2, then

| f(z)− w0 |=| (u− iv)− (u0 − iv0) |=| (u− u0) + i(v − v0) |≤| u− u0 | + | v − v0 |

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.

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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

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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.

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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:

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1. Suppose that f(z) = 1/z. For any nonzero points,

lim∆z→0

∆w

∆z= lim

∆z→0

1z+∆z −

1z

∆z= lim

∆z→0

−1

z(z + ∆z)

therefore

f ′(z) =−1

z2

2. Suppose that f(z) = z, then

lim∆z→0

∆w

∆z= lim

∆z→0

z + ∆z − z∆z

= lim∆z→0

z + ∆z − z∆z

= lim∆z→0

∆z

∆z.

Recall that∆z = z − z0 = x+ iy − (x0 + iy0) = (x− x0) + i(y − y0).

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.

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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

d

dz(1− 4z2)3 = 3w2 · (−8z) = −24z(1− 4z2)2

2.5 Cauchy-Riemann Equations

Letf(z) = u(x, y) + iv(x, y)

assume that f ′(z0) exits, and write

z0 = x0 + iy0 and ∆z = ∆x+ i∆y,

and∆w = f(z0 + ∆z)− f(z0).

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Then

∆w

∆z=

f(z0 + ∆z)− f(z0)

∆x+ i∆y

=u(x0 + ∆x, y0 + ∆y) + iv(x0 + ∆x, y0 + ∆y)− u(x0, y0)− iv(x0, y0)

∆x+ i∆y

=u(x0 + ∆x, y0 + ∆y)− u(x0, y0) + iv(x0 + ∆x, y0 + ∆y)− iv(x0, y0)

∆x+ i∆y

=u(x0 + ∆x, y0 + ∆y)− u(x0, y0))

∆x+ i∆y+ i

v(x0 + ∆x, y0 + ∆y)− v(x0, y0)

∆x+ i∆y.

Since f ′(z0) exits, this expression is valid as ∆z approaches 0 in any manner!

• Horizontal Approach: Let ∆z = (∆x, 0) approach (0, 0), then

f ′(z0) = lim∆x→0

u(x0 + ∆x, y0)− u(x0, y0))

∆x+ i

v(x0 + ∆x, y0)− v(x0, y0)

∆x= ux(x0, y0) + ivx(x0, y0)

• Vertical Approach: Let ∆z = (0,∆y) approach (0, 0), then

f ′(z0) = lim∆y→0

u(x0, y0 + ∆y)− u(x0, y0))

i∆y+ i

v(x0, y0 + ∆y)− v(x0, y0)

i∆y

= lim∆y→0

v(x0, y0 + ∆y)− v(x0, y0)

∆y− iu(x0, y0 + ∆y)− u(x0, y0))

∆y

= vy(x0, y0)− iuy(x0, y0)

From we see that the existence of f(z0) requires that

ux(x0, y0) + ivx(x0, y0) = vy(x0, y0)− iuy(x0, y0)

or in other words

ux(x0, y0) = vy(x0, y0) and vx(x0, y0) = −uy(x0, y0). (4)

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).

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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

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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.

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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)

Similarly,vr = vx cos(θ) + vy sin(θ) and vθ = −vxr sin(θ) + vyr cos(θ)

andux = vy and uy = −vy =⇒ rur = vθ and uθ = −rvr

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.

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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

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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

f(z)f(z) = c2 6= 0 =⇒ f(z) 6= 0 in D

=⇒ f(z) =c2

f(z)

=⇒ f(z) is analytic everywhere

=⇒ f(z) is constant, by previous example

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