2 Lines and circles on the complex plane

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September 24, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 1 Complex numbers

In this section, we will survey the algebraic and geometric properties of the complex number field, which is denoted by C throughout this course. Properties of the real number field (denoted by R) have been introduced in the course of advanced calculus and are assumed here.

Definition 1.1. The complex number field C is the field extension of degree 2 of the real number field R. There is a basis{1, i} for C which satisfies i2=−1. An element in C is called a complex number.

In the course of modern algebra, one can show that there is only one finite field extension of R, and it is of degree 2. We find an element i in that field such that i2=−1, and then show that {1, i} is a basis for this field, which can be regarded as a vector space over R. Every complex number can then be uniquely expressed as

z = x + iy, x, y∈ R.

Under this expression, the numbers

x = Re z, y = Im z

are called the real part and the imaginary part of the complex number z respectively. Sometimes it is convenient to represent a complex number z = x + iy by the point (x, y) with rectangular coordinates on the complex plane.

The sum and the product of two complex numbers are given by:

(x1+ iy1) + (x2+ iy2) = (x1+ x2) + i(y1+ y2),

(x1+ iy1)· (x2+ iy2) = (x1y1− x2y2) + i(x1y2+ x2y1).

The multiplicative inverse of a non-zero complex number z = x + iy is z−1= 1

z = x

x2+ y2 − i y

x2+ y2. (Check it: z· z−1= 1.) The (complex) conjugate of a complex number z = x + iy is defined as

¯

z = x− iy;

and the modulus, or the absolute value of z is

|z| =p

x2+ y2.

(The square root is conventionally defined in R+.) It is straightforward to check that |z| = |¯z| and z¯z = |z|2. Furthermore, (C,| · |) becomes a normed space, that is, it satisfies the following conditions:

1. |z| ≥ 0 for all z ∈ C, and |z| = 0 if and only if z = 0.

2. |λz| = |λ| · |z| for z ∈ C and λ ∈ R.

3. |z1+ z2| ≤ |z1| + |z2| (triangle inequality.) (Proof of the triangle inequality:

|z1+ z2|2= (z1+ z2)(¯z1+ ¯z2)

=|z1|2+ 2 Re(z12) +|z2|2

≤ |z1|2+ 2|z1z2| + |z2|2

= (|z1| + |z2|)2.

Since both bases are non-negative, we take the square roots and the inequality is proved.)

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The space C is then equipped with the topology endowed by the norm structure. Indeed, it is the same as that of R2under the Euclidean metric. One might check that

|z1z2| = |z1| · |z2|, z1, z2∈ C.

We may define a bilinear formh·, ·i on C by

hz1, z2i = z12, z1, z2∈ C.

Clearly hz, zi = |z|2. It follows that h·, ·i defines an inner product structure on C. With the completeness of R, C becomes a Hilbert space, that is, a complete inner product space.

There is another way to express a complex number, called the polar coordinate. Take r =|z|, x = r cos θ, y = r sin θ,

then z is written as

z = r(cos θ + i sin θ).

When z = 0, the coordinate θ is undefined.

As we identify complex numbers with vectors on the complex plane, the number r is the distance between the origin and the point z, while θ is the oriented angle between the positive real axis (x-axis) and the radial vector representing z. Technically speaking, there are infinitely many choices for θ, for we may add any multiple of 2π to the oriented angle; each choice of θ is called an argument of z, and the set of all arguments of z is denoted by arg z.

Among these values, there is exactly one Θ such that−π < Θ ≤ π; it is called the principal argument of z and denoted by Arg z. Note that

arg z = Arg z + 2nπ, n∈ Z.

As a consequence, two non-zero complex numbers z1= r1(cos θ1+ i sin θ1) and z2= r2(cos θ2+ i sin θ2) are equal if and only if

r1= r2 and θ1= θ2+ 2nπ for some integer n.

Using this expression, the multiplication of complex numbers are easy to write down. If zj= rj(cos θj+ i sin θj), j = 1, 2, then

z1z2= r1r2 cos(θ1+ θ2) + i sin(θ1+ θ2).

This leads to de Moivre’s formula by mathematical induction and inversion:

zn= rn(cos nθ + i sin nθ), n∈ Z.

If n is a positive integer and z0= r0(cos θ0+ i sin θ0) is a non-zero complex number, the equation zn = z0has n distinct solutions, which are called the nth roots of z0. Using de Moivre’s formula, the solutions are

z = √n r0

h

cosθ0+ 2kπ n



+ i sinθ0+ 2kπ n

i

, k = 0, 1, . . . , n− 1.

When n≥ 3, the roots lie at the vertices of a regular polygon of n sides inscribed in the circle, centered at the origin with radius √n

r0.

2 Lines and circles on the complex plane

As in algebraic geometry, we would like to describe some geometric objects like lines and circles by equations. Keep in mind that some variables we will be considering are complex-valued.

The easiest equation is for a circle, because we have a notion of distance in hand. Coming from the definition, the circle centered at a point z0 with radius r > 0 is defined by the equation

|z − z0| = r. (2.1)

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Squaring and expanding Equation (2.1), we reach an equivalent form z ¯z− ¯z0z− z0z + (¯ |z0|2− r2) = 0.

Similarly one can try to write down the equations for other conic curves. For instance, an equivalent definition for an ellipse is that the sum of the distances between a point to two fixed ones is constant; translating this sentence into an equation of a complex variable yields

|z − z1| + |z − z2| = 2a.

For straight lines, we use the following strategy: a line on the xy-plane is defined by the equation

ax + by = c, a, b, c∈ R. (2.2)

(Some technical assumptions are omitted.) For x (resp. y) being the real part (resp. imaginary part) of the complex variable z, we have the identifications

x = 1

2(z + ¯z), y = 1

2i(z− ¯z).

So plug in Equation (2.2) and we see that

a 2 −ib

2

 z +a

2 +ib 2



¯

z = c. (2.3)

If we let w0= (a + ib)/2, then Equation (2.3) becomes

¯

w0z + w0z = c.¯ (2.4)

Exercise. What happens if c in Equation (2.4) is a complex number but not a real one?

3 Topological jargons

We now describe the topology in C, defined by the norm| · |. An ε neighborhood of a point z0∈ C is the set B(z0, ε) :={z ∈ C | |z − z0| < ε}.

It consists of all points z lying inside but not on a circle centered at z0 with the positive radius ε. When the value of ε is understood or is immaterial in the discussion, we often refer the set as just a neighborhood. On the complex plane, a neighborhood looks like a circular disk without the circle that bounds it. Occasionally, it is convenient to speak of a deleted neighborhood

B0(z0, ε) :={z ∈ C | 0 < |z − z0| < ε}

consisting of all points z in an ε neighborhood of z0 except for the point z0 itself.

Definition 3.1. Let S be a subset of C. A point z0 is said to be an interior point of S if there is some neighborhood of z0 lying completely in S; it is called an exterior point of S if there is some neighborhood of z0 lying completely outside of S. If z0 is neither an interior point nor an exterior point of S, it is called a boundary point of S. The collection of all boundary points of S, denoted by ∂S, is called the boundary of S.

Exercise. A point z0 is a boundary point of S if and only if every neighborhood of z0 contains a point in S and another point not in S.

Exercise. Let A be the set{x ∈ R | 0 ≤ x ≤ 1}. What is the boundary ∂A of A in C? (Be careful!)

Definition 3.2. A set is open when it contains none of its boundary points. A set is closed when it contains all of its boundary points. The closure ¯S of a set S is the union of the set S itself with its boundary ∂S, hence the closure is always closed.

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Remark. Under this definition, the empty set∅ is both open and closed!

Exercise. A set is open if and only if all of its points are interior points.

Exercise. A set S is open if and only if its complement C\ S is closed. (Caution: ∂(C \ S) needs to be identified.) Definition 3.3. An open set S is connected if it cannot be written as a disjoint union of two non-empty open subsets.

A connected open set is called a domain. A domain together with some, none, or all of its boundary points is referred to as a region.

Remark. A set S is path connected if any pair of points in S can be joined by a path completely lying in S. Hence when a set is open in C, it is connected if and only if it is path connected. A detailed discussion about this notion will be more suited in a course of point set topology.

Definition 3.4. A set S is bounded if every point of S lies inside some circle |z| = R; otherwise S is called unbounded.

Definition 3.5. A point z0 is said to be an accumulation point of a set S if each deleted neighborhood of z0 contains at least one point of S. If a point z0∈ S is not an accumulation point of S, it is called an isolated point of S.

Exercise. What is the set of all accumulation points of the set{1/n | n ∈ N} in C?

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October 8, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 4 Functions of a complex variable

In this section, we will define functions whose variables could assume complex values. It should be noted first that not all functions that were encountered before can be done in this way. One has to be very careful on those functions.

We shall emphasize on the concept of domains of functions of a complex variable. The domain of a function is the set where the function is defined. When the domain of definition is not mentioned, we agree that the largest possible set is to be taken. Two functions are considered the same only when they possess the same domain and their values are identical at every point in that domain. This distinction will be important when we discuss analytic continuation later this year.

Suppose w = u + iv is the value of a function f at z = x + iy, that is, u + iv = f (x + iy).

Using the identification C' R2, we can sometimes think of f as a function from R2to R2: (u, v) = f (x, y) = u(x, y), v(x, y),

or equivalently,

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

Example. Take f (z) = z2= (x + iy)2= (x2− y2) + 2ixy. Then the above identification produces two functions u, v of two real variables as

u(x, y) = x2− y2, v(x, y) = 2xy.

At this stage, it is clear that we can consider polynomials in the variables z and ¯z. Furthermore, rational functions, which are quotients of two such polynomials, are defined where the denominator does not vanish. But other elementary functions like trigonometric functions and exponential functions cannot be defined in the conventional way (ask yourself: what is sin i? We don’t have the Euler formula at hand yet!) The answer to this question will be postponed until we have enough tools.

5 Limits and continuity

Definition 5.1. A function f (z) of a complex variable z is said to have the limit A as z tends to z0, denoted by

zlim→z0

f (z) = A, (5.1)

if and only if for every ε > 0 there exists a number δ > 0 such that |f(z) − A| < ε whenever z lies in the deleted δ-neighborhood B0(z0, δ) of z0, that is, 0 <|z − z0| < δ. A function f(z) is said to be continuous at z0 if and only if

zlim→z0

f (z) = f (z0). (5.2)

A continuous function is a function which is continuous at all points where it is defined.

With the complex plane C equipped with the topology induced by the norm| · | introduced in Section 1, it is an easy exercise to prove the following theorem.

Theorem 5.2. Suppose that

f (z) = u(x, y) + iv(x, y), z0= x0+ iy0, w0= u0+ iv0. Then

zlim→z0

f (z) = w0 if and only if

lim

(x,y)→(x0,y0)

u(x, y) = u0 and lim

(x,y)→(x0,y0)

v(x, y) = v0.

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Proof. The topology on C is the same as the product topology on R2.

As one can expect, there are properties concerning limits which are just the same as those in real-valued functions.

Theorem 5.3. Suppose that

zlim→z0

f (z) = w0 and lim

z→z0

F (z) = W0. Then,

zlim→z0

f (z) + F (z) = w0+ W0,

zlim→z0f (z) · F (z) = w0· W0, and if W06= 0,

zlim→z0

f (z) F (z) = w0

W0.

The concept of continuity is the same as that in advanced calculus. We will simply state the definition.

Definition 5.4. A function f of a complex variable is continuous at a point z = a when

zlim→af (z) = f (a).

Let Ω be an open subset of C. We say that f is continuous in Ω if and only if it is continuous at every point in Ω.

There is a concept of infinity in the course of complex variables. In the space of real variables, we have two such notions, +∞ and −∞, denoting the infinities for the positive and negative directions, respectively. On the contrary, there is only one infinity∞ in the extended complex plane, which is obtained by the so-called stereographic projection. Consider the unit sphere ∂B(0, 1) in R3 and think of the complex plane as the plane passing through the equator of the sphere. Every point z in the complex plane can be joined with the north pole of the sphere by a straight line, which intersects at another point on the sphere. The correspondence can be reversed from any point on the sphere other than the north pole to a point in the complex plane. The exception, which is the north pole, then corresponds to what we think of the point at infinity∞ on the extended complex plane. Since the space of the unit sphere is homogeneous, the topology near∞ is no other than that near any other points.

Definition 5.5. We say that a complex-valued function f has the limit ∞ near the point z = a, denoted by

zlim→af (z) =∞, if, for any positive number R > 0, there is a δ > 0 such that

0 <|z − a| < δ =⇒ |f(z)| > R.

Definition 5.6. We say that a complex-valued function f has the limit L near the point of infinity z =∞, denoted by

zlim→∞f (z) = L, if, for any ε > 0, there is an R > 0 such that

|z| < R =⇒ |f(z) − L| < ε.

Exercise. Formulate the precise definition of the following notion lim

z→∞f (z) =∞.

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October 15, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 6 Derivatives

The derivative of a function of a complex variable is defined in the usual way, but there is some tricky part as we investigate the limit.

Definition 6.1. The derivative of a function f (z) of a complex variable at the point z = a is defined as f0(a) := lim

z→a

f (z)− f(a)

z− a , (6.1)

provided that the limit exists, and we say that f is differentiable at the point a. When Ω is an open subset of C and f is differentiable at every point in Ω, we will say that f is analytic on Ω.

Example. The derivative of a monomial f (z) = zn for n∈ N is f0(z) = nzn−1.

In the case of a real variable, a point x can approach some fixed point a on the real line only from two directions, left and right. However this is not true anymore in the case of a complex variable. To illustrate what really happens, we make a change of variable to reach an equivalent definition of Equation (6.1):

f0(a) := lim

h→0

f (a + h)− f(a)

h . (6.2)

We also write the real and the imaginary parts of f as f = u + iv since the operation of taking limits is linear over C. If h is real, then

lim

h→0

f (x + h, y)− f(x, y)

h =∂u

∂x + i∂v

∂x. (6.3)

However, if ik is purely imaginary, then lim

k→0

f (x, y + k)− f(x, y)

ik =−i

 lim

k→0

f (x, y + k)− f(x, y) k



=−i∂u

∂y +∂v

∂y. (6.4)

Comparing the real and the imaginary parts of Equations (6.3) and (6.4), we see that a necessary condition for a function f to be differentiable is

∂u

∂x =∂v

∂y, ∂u

∂y =−∂v

∂x. (6.5)

These equations are called the Cauchy-Riemann equations, abbreviated as the CR equations.

Example. The monomial g(z) = ¯zn can only be differentiated at z = 0 when n≥ 2. The function h(z) = ¯z cannot be differentiated anywhere.

On the other hand, with a stronger assumption we can now prove its converse.

Proposition 6.2. If u(x, y) and v(x, y) have continuous first-order partial derivatives which satisfy Equation (6.5), then f (z) = u(z) + iv(z) is analytic with continuous derivative f0(z).

Proof. For sufficiently small h, k > 0, we can write

u(x + h, y + k) = ∂u

∂xh +∂u

∂yk + ε1 v(x + h, y + k) = ∂v

∂xh +∂v

∂yk + ε2,

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where the remainders ε1, ε2 tend to zero more rapidly than h + ik in the sense that lim

h+ik→0

ε1

√h2+ k2 = 0 = lim

h+ik→0

ε2

√h2+ k2.

With the notion f (z) = u(x, y) + iv(x, y), we obtain by virtue of Equation (6.5) f (z + h + ik)− f(z) =∂u

∂x + i∂v

∂x



(h + ik) + (ε1+ iε2) and hence

lim

h+ik→0

f (z + h + ik)− f(z)

h + ik = ∂u

∂x+ i∂v

∂x. We then conclude f (z) is analytic.

It should be noted that the hypothesis in Proposition 6.2 can be weakened, but it will not be pursued here.

We shall prove later that the derivative of an analytic function is itself analytic, that is, it can be differentiated again. Hence u and v will have continuous partial derivatives of all orders, and in particular the mixed derivatives are equal. If we let ∆ = ∂xx+ ∂yy denote the Laplace operator in R2, we obtain by the Cauchy-Riemann equations (6.5):

∆u = ∂

∂x

 ∂u

∂x

 + ∂

∂y

 ∂u

∂y



= ∂

∂x

 ∂v

∂y

 + ∂

∂y



−∂v

∂x



= 0.

Similarly ∆v = 0 as well. A function u that satisfies the Laplace’s equation ∆u = 0 is said to be harmonic. Hence the previous argument shows that the real and the imaginary parts of an analytic function are harmonic. If two harmonic functions u and v satisfy the Cauchy-Riemann equations (6.5), then v is said to be a harmonic conjugate of u.

Exercise. What happens if v is a harmonic conjugate of u and also u is a harmonic conjugate of v?

Recall that if z = x + iy, then

x = z + ¯z

2 , y =z− ¯z 2i . Motivated by these expressions, we define the differential operator

¯z= ∂

∂ ¯z := 1 2

 ∂

∂x + i ∂

∂y

.

One then checks that a function f (z) = u(x, y) + iv(x, y) satisfies the Cauchy-Riemann equations if and only if f lies in the kernel of the operator ∂z¯, that is,

z¯f = ∂f

∂ ¯z = 1 2 h

(ux− vy) + i(vx+ uy)i

= 0.

Exercise. Define another differential operator

z= ∂

∂z := 1 2

 ∂

∂x − i ∂

∂y

 . Show by direct computation that the Laplace operator ∆ can be written as

∆f = ∂2

∂x2 + ∂2

∂y2



f = 4 ∂2

∂z∂ ¯zf when f is a function with continuous derivatives up to order 2.

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October 22, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables

7 The exponential function, the trigonometric functions and the hy- perbolic functions

Consider the function f (x) = exp(x) = ex for the real variable x. This function is smooth and satisfies the property:

f0(x) = f (x), f (0) = 1.

By the existence and uniqueness in the theory of ODE, we know that this is the unique function with the above property. Motivated from the real-variable case, we seek a function in a complex variable with the same property:

f0(z) = f (z) for all z, f (0) = 1. (7.1)

Definition 7.1. Define the exponential function

f (z) = f (x + iy) = ex(cos y + i sin y). (7.2) Then f (z) is the unique function which satisfies (7.1).

In short, the exponential function will be denoted by exp z, or simply ez. Note that the exponent function has a period of 2πi, that is, exp(z) = exp(z + 2nπi) for every integer n∈ Z.

Obviously f (z) satisfies the initial condition f (0) = 1. Now we check the differentiability: set u(x, y) = excos y and v(x, y) = exsin y, we obtain

ux= vy = excos y, uy=−vx=−exsin y.

Since all of first-order partial derivatives are continuous, we conclude that f is analytic everywhere by Proposition 6.2.

And its derivative is

f0(z) = ux+ ivx= excos y + i exsin y = f (z).

Suppose there is another function g(z) satisfying (7.1). Then the derivative of f (z)− g(z) is zero, hence f(z) − g(z) must be a constant by the mean value theorem. Using the initial condition at z = 0 we conclude that g(z) must be identical to f (z), which proves the uniqueness.

The expression has the advantage of expressing the multiplication in polar coordinates. As suggested, we have ez+w= ez· ew.

One can easily verify this identity under the polar coordinates.

It is now not surprising to reach the following definition.

Definition 7.2 (Euler’s formula). For a real number θ, the value e is defined as e= cos θ + i sin θ.

Now it is time to define the trigonometric functions for a complex variable. Using some elementary algebra, we have the following definition.

Definition 7.3. The trigonometric functions are defined by cos z = eiz+ e−iz

2 , sin z = eiz− e−iz 2i .

It is then an easy exercise to find the derivatives of these functions. We state here a proposition as a reference.

Proposition 7.4. The derivatives of cos z and sin z are

D(cos z) =− sin z, D(sin z) = cos z. (7.3)

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The other trigonometric functions tan z, cot z, sec z and csc z are defined in the same way as before, and their derivatives can be found likewise.

Exercise. Use Equation (7.3) to show that sin2z + cos2z = 1 for all z∈ C.

Definition 7.5. The hyperbolic functions are defined as cosh z = ez+ e−z

2 , sinh z = ez− e−z

2 .

We list a few identities whose verifications are left as exercise.

Exercise.

(1) D(cosh z) = sinh z; D(sinh z) = cosh z.

(2) cosh2z− sinh2z = 1.

(3) sinh z = sinh x cos y + i cosh x sin y; cosh z = cosh x cos y + i sinh x sin y.

(4) | sinh z|2= sinh2x + sin2y; | cosh z|2= sinh2x + cos2y.

Also the other hyperbolic functions are defined as usual, which will be omitted here.

8 The logarithmic function

The logarithmic function is supposed to be the inverse function of the exponential function. However from the discussion earlier, we realize that the equation ew= z has infinitely many complex solutions w for any non-zero z.

Since we do not have any preference from one to another, we define the logarithmic “function” as follows.

Definition 8.1. The logarithmic function is defined as the set

log z :={w ∈ C | ew= z} for any nonzero complex number z.

Indeed, this is not a function under the traditional sense. We call such things multi-valued functions, and the study of multi-valued functions in complex analysis will lead to the concept of Riemann surfaces. Basically, one of the motivations for Riemann surfaces is to find the right domains and ranges such that a multi-valued holomorphic function becomes single-valued over them.

To solve the equation ew= z, we first write the non-zero complex number z under the polar coordinates:

z = re. Then we see that the solution w can be expressed as

w = ln r + i Θ + 2nπ

for any n∈ Z

= ln|z| + i arg z.

Therefore we write

log z = ln|z| + i arg z. (8.1)

The principal value of log z is obtained from replacing arg z by Arg z in Equation (8.1), hence Log z = ln|z| + i Arg z.

From this we know that 0≤ Im Log z < 2π. Furthermore, the map

Log : C\ {0} → {w ∈ C | 0 ≤ Im w < 2π}

becomes a single-valued function. Sometimes we call this function the principal branch of the logarithmic function.

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Now we ask whether the logarithmic function is differentiable or not. Since differentiability is a local property, we can restrict the range of log z to any branch: α < Im log z < α + 2π. First we call the Cauchy-Riemann equations under the polar coordinates.

Exercise. If f (z) = f (re) = u(r, θ) + iv(r, θ), then the Cauchy-Riemann equations are given by ur= 1

rvθ, 1

ruθ=−vr. (8.2)

It is then an easy verification that u(r, θ) = ln r, v(r, θ) = θ for log z satisfy the Cauchy-Riemann equations (8.2).

We now use the chain rule to compute the derivative of log z:

exp(log z) = z

exp(log z)· D(log z) = 1 (take D on both sides) D(log z) = 1

exp(log z) = 1

z (|z| > 0, α < arg z < α + 2π).

9 Complex exponents

When z6= 0 and c is any complex number, the function zc is defined by means of the equation

zc:= exp(c log z). (9.1)

Note that zc is now a multi-valued function, for we have many choices for log z.

Example. We compute

i−2i= exp(−2i log i) = exp −2i(2n +1

2)πi = exp (4n + 1)π

for all n∈ Z. If we specify a branch for zc, then its derivative can be found as d

dz(zc) = d

dz exp(c log z) = exp(c log z) ·c

z = c·exp(c log z)

exp(log z) = c exp (c− 1) log z = czc−1.

Note that everything appearing above is a single-valued function once we choose a branch once and for all. Likewise, the principal branch of zc is obtained by choosing the principal branch Log z in Equation (9.1). One should also note that the nth root of a non-zero complex number z is defined in this way, that is,

n

z = zn1 = exp 1 nlog z.

It can be verified that √n

z is an n-to-1 function.

According to Equation (9.1), the exponential function with base c, where c is any non-zero complex constant, is written as

cz:= exp(z log c). (9.2)

Again czis a multi-valued function. If we specify a branch for log c, the derivative of czwith respect to z is computed

as d

dz(cz) = d

dz exp(z log c) = exp(z log c) · log c = czlog c.

10 Inverse trigonometric and hyperbolic functions

Inverses of the trigonometric and hyperbolic functions are described in terms of logarithms. We will illustrate them by an example.

Example. We want to find sin−1z. By definition for the inverse function, sin−1z are those complex numbers w satisfying sin w = z. Hence

eiw− e−iw

2i = z. (10.1)

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The equation (10.1) is in fact quadratic in eiw, since multiplying eiw on the both sides of the equation yields (eiw)2− 2izeiw− 1 = 0.

Solving for eiwby using the quadratic formula, we see that eiw= iz +p

−z2+ 1 sin−1z = w =−i log iz +p

1− z2.

Here, of course, both√

1− z2and the logarithm are multi-valued functions.

Other inverse trigonometric and hyperbolic functions are found in similar ways, also one can work out their derivatives once a branch has been chosen. We will leave them for interested readers.

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October 22, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables, Appendix A

We make up the proof for the uniquess of the exponential function here. Suppose g is a function satisfying g0(z) = g(z) and g(0) = 1. Define another function

h(z) = exp(−z) · g(z).

The initial condition is h(0) = exp(0)· g(0) = 1. Since h is a product of two differentiable functions, h itself is differentiable as well. We find

h0(z) =− exp0(−z) · g(z) + exp(−z) · g0(z) = 0.

Therefore h is a constant function, which is 1 from the initial condition. At last, g(z) = 1

exp(−z) = exp(z).

So the uniqueness is proved.

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October 29, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 11 Contour integrals

In the following sections we will be discussing integrals of complex-valued functions over contours. Before the formal definition of contour integrals, we start with defining integrals of complex-valued functions of a real variable. Suppose w : [a, b]→ C is a complex-valued function. Write

w(t) = u(t) + iv(t), u, v : [a, b]→ R.

Then we define the integral of w(t) over the interval [a, b] as Z b

a

w(t) dt :=

Z b a

u(t) dt + i Z b

a

v(t) dt

provided that both integrals on the right-hand side exist. Indeed, if one performs either Riemann sums or Lebesgue integrals formally to the function w(t), the operation is actually linear on the integrand over C. Therefore this definition is desired.

Example. As an illustration, Z 1

0

(1 + it)2dt = Z 1

0

(1− t2) dt + i Z 1

0

2t dt = 2 3 + i.

The fundamental theorem of calculus, involving antiderivatives, can be extended so as to apply the integrals defined here. Namely, suppose that the functions

w(t) = u(t) + iv(t) and W (t) = U (t) + iV (t)

are continuous on the interval [a, b]. If W0(t) = w(t) when a≤ t ≤ b, then U0(t) = u(t) and V0(t) = v(t) (beware of the differentiation here!) Hence,

Z b a

w(t) dt = Z b

a

u(t) dt + i Z b

a

v(t) dt =U (b) − U(a) + iV (b) − V (a) = W (b) − W (a).

Proposition 11.1. Let w(t) = u(t) + iv(t) be a complex-valued function which is integrable over the interval [a, b].

Then,

Z b a

w(t) dt

≤ Z b

a

|w(t)| dt. (11.1)

Proof. The inequality (11.1) clearly holds when the left-hand side is zero. Thus, in the verification, we may assume thatRb

a w(t) dt = r0e0 for some r0> 0. We rewrite as r0=

Z b a

e−iθ0w(t) dt.

The left-hand side of this equation is a real number, hence so is the right-hand side. Furthermore,

r0= Z b

a

e−iθ0w(t) dt = Re Z b

a

e−iθ0w(t) dt

!

= Z b

a

Re e−iθ0w(t) dt.

Now we use the inequality on the complex number e−iθ0w(t):

Re e−iθ0w(t) ≤

e−iθ0w(t)

=|w(t)|,

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therefore

Z b a

w(t) dt

= r0= Z b

a

Re e−iθ0w(t) dt ≤ Z b

a

|w(t)| dt, and the proposition is proved.

Now we define the concept of a contour on the complex plane.

Definition 11.2. A contour C on the complex plane is a continuous curve on C which may be parametrized by a piecewise C1 function z : [a, b]→ C ' R2.

Notice that a contour is defined with its orientation. If we parametrize the contour C backwards, that is, a parameterization ˜z : [a, b]→ C given by ˜z(t) := z(a + b− t), then the new contour is denoted by −C. Also if ϕ is a continuous and increasing function from [c, d] to [a, b], then z◦ ϕ gives an alternative parametrization to the same contour C.

We write, as always, that z(t) = x(t) + iy(t), a≤ t ≤ b, for a contour C. The derivative of z(t) with respect to t is given by

z0(t) = x0(t) + iy0(t).

It follows from the definition that the real-valued function

|z0(t)| =p

[x0(t)]2+ [y0(t)]2

is integrable over the interval a≤ t ≤ b, for there are only at most finitely many points at which |z0(t)| is discontinuous.

The length L of the contour C is then given by

L = Z b

a

|z0(t)| dt.

Notice that the value L is independent of parametrization of C by the chain rule in the usual calculus.

Except at finitely many points, the unit tangent vector T = z0(t)

|z0(t)| is well-defined with angle of inclination arg z0(t).

Suppose z(t) = x(t) + iy(t), a≤ t ≤ b is a parametrization of a contour C. If z(a) = z(b), we say that the contour C is closed. If a closed contour does not intersect with itself, that is, there are no distinct s, t ∈ (a, b) such that z(s) = z(t), then we call it a simple closed contour. The points on any simple closed contour C are boundary points of two distinct domains, one of which is the interior of C and is bounded. The other, which is the exterior of C, is unbounded. It will be convenient to accept this statement, known as the Jordan curve theorem, as geometrically evident. The proof, a deep result in the course of topology, will not be pursued here.

We now turn to the integrals of complex-valued functions f of a complex variable z. Such an integral is defined in terms of the values f (z) along a given contour C, therefore it is a line integral. The integral is written as

Z

C

f (z) dz or Z z2

z1

f (z) dz,

the latter notation often being used when the value of the integral is dependent only on the endpoints z1, z2 of the contour C.

Definition 11.3. The contour integral of a complex-valued f over a contour C is obtained by Z

C

f (z) dz :=

Z b a

f (z(t)) d[z(t)], (11.2)

where z : [a, b]→ C is a parametrization of the contour C, and d[z(t)] = z0(t) dt.

Again the value of the contour integral is independent of parametrization of C by the discussion above.

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Proposition 11.4. (1) Let−C denote the orientation-reversed contour of a contour C. Then Z

−C

f (z) dz =− Z

C

f (z) dz.

(2) If a contour C consists of a contour C1 from z0 to z1 followed by another contour C2 from z1 to z2, then the contour integral of f over C is the sum of those over C1 and C2, that is,

Z

C

f (z) dz = Z

C1

f (z) dz + Z

C2

f (z) dz.

Proof. (2) is clear. For (1), suppose that z : [a, b]→ C is a parametrization of C. Then −C can be parameterized by w(t) := z(a + b− t) over the interval [a, b]. Hence

Z

−C

f (z) dz = Z b

a

f (w(t)) w0(t) dt =− Z b

a

f (z(a + b− t)) z0(a + b− t) dt = Z a

b

f (z(u)) z0(u) du =− Z

C

f (z) dz.

Finally, we assume that for some fixed positive number M ,|f(z)| ≤ M for all points z in the contour C. According to (11.1), we have

Z

C

f (z) dz ≤

Z b a

|f(z(t))z0(t)| dt ≤ M Z b

a

|z0(t)| dt = ML, (11.3)

where L is the length of the contour C (provided it is finite.) This inequality will be of great use when we estimate various integrals later on.

Example. Let C be the contour defined by the semicircle from z =−2i to z = 2i along the circle |z| = 2, going counterclockwise (we will say the orientation of a circle is positive if it is going counterclockwise.) Let us find the value of the contour integral

I = Z

C

¯ z dz.

We first choose a parametrization of C by z = 2eit,−π/2 ≤ t ≤ π/2. Then the contour integral is computed as follows:

I = Z π2

π2

2eitd(2eit) = Z π2

π2

2e−it· 2ieitdt = Z π2

π2

4i dt = 4πi.

Example. Let CR be the semicircular path z = Re, θ∈ [0, π], and z1/2denote the branch of the square root when we fix arg z∈ (−π/2, 3π/2). We want to show that

lim

R→∞

Z

CR

z1/2

z2+ 1dz = 0. (11.4)

For z = Re, |z| = R which we assume to be greater than, say 1029. Then |z1/2| = √

R, and |z2+ 1| ≥ R2/2.

Consequently, at points on CR where the integrand is defined,

z1/2 z2+ 1

√R R2/2 = 2

R3/2 =: MR. Since the length of the contour CR is LR= πR, it follows from (11.3) that

Z

CR

z1/2 z2+ 1dz

≤ MRLR= 2π

√R → 0

as R→ ∞. Hence the limit (11.4) is proved.

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November 5, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 12 Antiderivatives

As the notation suggests, the contour integral

Z z2 z1

f (z) dz

might be independent of the path once the endpoints z1, z2 are fixed. This does not always happen, but it holds under certain circumstances. We will investigate the situations in this section.

Definition 12.1. Let D be a domain in C and f be a function in D. We say that F is an antiderivative of f when F0(z) = f (z) for all z∈ D.

Notice that an antiderivative is not unique if it exists. For if F1 and F2 are antiderivatives of a function f , then the derivative (F1− F2)0(z) = F10(z)− F20(z) = f (z)− f(z) = 0 for all z ∈ D, hence F1(z)− F2(z) is a constant. This property is the same as that in elementary calculus.

Theorem 12.2. Suppose that a function f is continuous on a domain D in C. The following are equivalent:

(a) f has an antiderivative F is D.

(b) The integrals of f (z) along contours lying entirely in D and extending from any fixed point z1 to another fixed point z2 all have the same value.

(c) The integrals of f (z) around closed contours lying entirely in D all have value 0.

Notice that this theorem does not claim that any of these statements is true for a given function f in a given domain D. It only says that if any one of the statements is true, so are the others.

Proof. (a)⇒ (b): We suppose that F0(z) = f (z) for all points z∈ D. If a contour C from z1 to z2 is differentiable, then it can be parametrized by a differentiable function z : [a, b]→ D with z(a) = z1 and z(b) = z2. According to the chain rule, we have

d

dt[F (z(t))] = F0(z(t)) z0(t) = f (z(t)) z0(t).

Therefore the contour integral of f over C is computed as Z

C

f (z) dz = Z b

a

f (z(t)) z0(t) dt = Z b

a

d

dt[F (z(t))] dt

= F (z(t))

b

a = F (z(b))− F (z(a)) = F (z2)− F (z1)

by the fundamental theorem of calculus. Notice that this value is evidently independent of the contour C as long as C connects z1 to z2 and lies entirely in D.

In general, let C be a piecewise differentiable contour from z1 to z2. Then C consists of a finite number of differentiable contours Ck (k = 1, . . . n), each Ck extending from wk−1 to wk (w0= z1, wn= z2). Then

Z

C

f (z) dz =

n

X

k=1

Z

Ck

f (z) dz =

n

X

k=1

F (wk)− F (wk−1) = F (wn)− F (w0) = F (z2)− F (z1).

Again, this value does not depend on how C connects z1 and z2.

(b)⇔ (c): Suppose (b) holds first. We let z1 and z2be any two points on a closed contour C in D and form two paths C1, C2, each with the initial point z1 and the final point z2, such that C = C1− C2. By (b), we have

Z

C1

f (z) dz = Z

C2

f (z) dz

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because they have the same starting and end points. Therefore Z

C

f (z) dz = Z

C1−C2

f (z) dz = Z

C1

f (z) dz− Z

C2

f (z) dz = 0.

On the other hand, if (c) holds, then for any pair of points z1 and z2 in D, we may choose two contours C1and C2

in D connecting z1 to z2. Then C1− C2 is a closed contour that lies entirely in D, henceR

C1−C2f (z) dz = 0, or R

C1f (z) dz =R

C2f (z) dz. Therefore (b) is proved.

(b)⇒ (a): Now suppose (b) is true. We first fix a point z0 in D, and define F (z) :=

Z z z0

f (s) ds, z∈ D.

Notice that F is well-defined because the integral is independent of path by (b). Now we show that F0(z) = f (z).

Suppose z +∆z is any point, distinct from z, lying in some neighborhood of z that is sufficiently small to be contained in D. Then

F (z + ∆z)− F (z) = Z z+∆z

z0

f (s) ds− Z z

z0

f (s) ds = Z z+∆z

z

f (s) ds where the path of integration from z to z + ∆z can be chosen as a line segment. Since

Z z+∆z z

ds = ∆z, it follows that

F (z + ∆z)− F (z)

∆z − f(z) = 1

∆z

Z z+∆z z

[f (s)− f(z)] ds.

By hypothesis f is continuous at the point z. Hence, for each positive number ε, there is a positive number δ such that

|f(s) − f(z)| < ε

whenever|s − z| < δ. Consequently, if the point z + ∆z lies in the δ-neighborhood of z (which can be shrunk to be contained in D if necessary), then

F (z + ∆z)− F (z)

∆z − f(z)

< 1

|∆z| · ε|∆z| = ε.

This is exactly the condition for the limit lim

∆z→0

F (z + ∆z)− F (z)

∆z = f (z),

or, F0(z) = f (z).

Example. The function 1/z2, which is continuous everywhere except at the origin, has an antiderivative −1/z in the domain|z| > 0. Consequently,

Z z2

z1

dz z2 = −1

z

z2

z1

= 1 z1 − 1

z2 (z16= 0, z26= 0) for any contour from z1to z2that does not pass through the origin. In particular,

Z

C

dz z2 = 0 when C is any circle z = re (r > 0,−π ≤ θ ≤ π) about the origin.

Example. Let f (z) = 1/z on C\ {0} and CR be the closed contour|z| = R with positive orientation. To compute R

CRf (z) dz, we parametrize CR as z(t) = Re, 0≤ θ ≤ 2π. Then the contour integral is Z

CR

f (z) dz = Z

0

R i e R e dθ =

Z 0

i dθ = 2πi.

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In particular this is a nonzero number. Hence the function f does not have an antiderivative on C\ {0}. However, if we remove a ray emitting from the origin, f does have an antiderivative which is a branch of the logarithmic function we have seen in Section 8.

Example. Let us use an antiderivative to evaluate the integral Z

C1

z1/2dz, (12.1)

where the integrand is the branch

z1/2=√

r eiθ/2 (r > 0, 0 < θ < 2π) (12.2) of the square root function and where C1 is any contour from z = −3 to z = 3 that, except for its endpoints, lies above the real axis. Although the integrand is not defined on the ray θ = 0, it is still piecewise continuous on C1, and the integral therefore exists. Indeed, we can use another branch

f1(z) =√

r eiθ/2 (r > 0,−π

2 < θ < 3π 2 ),

which is defined and continuous everywhere on C1. The values of f1(z) at all points on C1 except z = 3 coincide with those of our integrand (12.2); so the integrand can be replaced by f1(z). Since an antiderivative of f1(z) is the function

F1(z) = 2

3r3/2e3iθ/2, (r > 0,−π

2 < θ < 3π 2 ), we can now write

Z

C1

z1/2dz = Z 3

−3

f1(z) dz = F1(z)

3

−3

= 2√

3(1 + i).

Integral (12.1) over any contour C2that extends from z =−3 to z = 3 below the real axis has another value. In this case, we can replace the integrand by the branch

f2(z) =√

r eiθ/2 (r > 0,π

2 < θ < 5π 2 ),

whose values coincide with those of the branch (12.2) in the lower half plane. The analytic function F2(z) = 2

3r3/2e3iθ/2, (r > 0,π

2 < θ < 5π 2 ), is an antiderivative of f2(z). Thus,

Z

C2

z1/2dz = Z 3

−3

f2(z) dz = F2(z)

3

−3

= 2√

3(−1 + i).

Notice how it follows that the integral of the function (12.2) around the closed contour C2− C1 has the value 2√

3(−1 + i) − 2√

3(1 + i) =−4√ 3.

13 Cauchy-Goursat theorem

Let C be a simple closed contour z = z(t) (a≤ t ≤ b), positively oriented (counterclockwise,) and we assume that f is analytic at any point interior to and on C. Then the contour integral is computed as

Z

C

f (z) dz = Z b

a

f [z(t)] z0(t) dt. (13.1)

If we write f (z) = u(x, y) + iv(x, y) and z(t) = x(t) + iy(t), then Equation (13.1) can be written as Z

C

f (z) dz = Z b

a

(ux0− vy0) dt + i Z b

a

(vx0+ uy0) dt. (13.2)

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In terms of line integrals of real-valued functions of two real variables, then Z

C

f (z) dz = Z

C

(u dx− v dy) + i Z

C

(v dx + u dy). (13.3)

Let R be the region bounded by C. If f is analytic in some neighborhood of ¯R and f0 is continuous there, the Green’s theorem implies that

Z

C

f (z) dz = Z Z

R

(−vx− uy) dA + i Z Z

R

(ux− vy) dA.

But, in view of the Cauchy-Riemann equations ux= vy, uy=−vx, Cauchy deduced thatR

Cf (z) dz = 0 in this case.

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November 12, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables

(Continued from the last section)

Goursat was the first to prove that the condition of continuity of f0 can be omitted. Its removal is important and will allow us to show, for example, that the derivative f0 of an analytic function f is again analytic without having to assume the continuity of f0, which follows as a consequence. We now state the revised form of Cauchy’s result:

Theorem 13.1 (Cauchy-Goursat theorem). If a function f is analytic at all points interior to and on a simple closed contour C, then

Z

C

f (z) dz = 0.

Proof. There are several forms of the Cauchy-Goursat theorem, but they differ in their topological rather than in their analytical content. It is natural to begin with with a case in which the topological considerations are trivial.

We assume first that the simple closed contour C is the boundary of a rectangle R = [a, b]× [c, d], positively oriented. For any rectangle S contained in R, define

η(S) = Z

∂S

f (z) dz.

Subdividing R into four identical rectangles R(j), j = 1, 2, 3, 4, we find η(R) = η(R(1)) + η(R(2)) + η(R(3)) + η(R(4)),

for integrals over the common sides cancel each other. It then follows that at least one of the rectangles R(j), j = 1, 2, 3, 4, must satisfy the condition

|η(R(j))| ≥ 1 4|η(R)|.

Denote this rectangle by R1; if several R(j)’s have this property, the choice shall be made according to some definite rule.

This process can be repeated indefinitely, and we obtain a sequence of nested rectangles R⊃ R1 ⊃ R2⊃ · · · ⊃ Rn ⊃ · · · with the property

|η(Rn)| ≥ 1

4|η(Rn−1)| and hence

|η(Rn)| ≥ 1

4n|η(R)|. (13.4)

The rectangles Rn’s converge to a point z∈ R in the sense that Rnwill be contained in a prescribed neighborhood B(z, δ) as soon as n is sufficiently large. First of all, we choose δ so small that f (z) is defined and analytic in B(z, δ).

Secondly, if ε > 0 is given, we can choose δ so that

f (z)− f(z)

z− z − f0(z)

< ε or

|f(z) − f(z)− f0(z)(z− z)| < ε|z − z| (13.5) for all z∈ B0(z, δ) by analyticity. We assume that δ satisfies both conditions and that Rn is contained in B(z, δ).

We make now the observation that

Z

∂Rn

dz = 0 = Z

∂Rn

z dz,

for both functions 1 and z have antiderivatives and Rn is a closed contour. By virtue of these equations we are able to write

η(Rn) = Z

∂Rn

[f (z)− f(z)− f0(z)(z− z)] dz,

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and it follows from Inequality (13.5) that

|η(Rn)| ≤ ε Z

∂Rn

|z − z| · |dz|. (13.6)

In the last integral |z − z| is at most equal to the length dn of the diagonal of Rn. If Ln denotes the length of the perimeter of Rn, the integral is hence at most dnLn. But if d and L are the corresponding quantities for the original rectangle R, it is clear that dn = 2−nd and Ln = 2−nL. By Inequality (13.6) we have hence

|η(Rn)| ≤ 4−nd L ε, and comparison with Inequality (13.4) yields

|η(R)| ≤ d L ε.

Since ε is arbitrary, we must have η(R) = 0.

Now we assume the next scenario that the simple closed contour C is contained in a circular disk ∆ in which f is analytic everywhere. We define a function F (z) by

F (z) = Z

σ

f (z) dz (13.7)

where σ consists of the horizontal line segment from the center (x0, y0) to (x, y0) and the vertical segment from (x, y0) to (x, y); it is immediately seen that ∂F/∂y = if (z). On the other hand, by the previous case for rectangles σ can be replaced by a path consisting of a vertical segment followed by a horizontal segment. This choice defines the same function F (z), and we obtain that ∂F/∂x = f (z). Hence by the Cauchy-Riemann equations F (z) is analytic in ∆ with the derivative f (z), and by Theorem 12.2R

σf (z) dz = 0.

For the most general case, we can choose a triangularization of the region enclosed by the simple closed contour C so that each triangle lies in an open disk in which f is analytic. With the orientations of each triangle assigned consistently, the original contour integral R

Cf (z) dz can be written as a sum of integrals over these triangles, each of which is zero by the argument in the previous paragraph. Hence the original integralR

Cf (z) dz = 0 as well.

Definition 13.2. A domain D is said to be simply connected if every simple closed contour within it enclosed only points in D. A domain which is not simply connected is called multiply connected.

With this definition, there is an immediate corollary for the Cauchy-Goursat theorem:

Corollary 13.3. If a function f is analytic throughout a simply connected domain D, thenR

Cf (z) dz = 0 for every closed contour C lying in D.

The proof is easy if the closed contour C itself is simple or intersects itself a finite number of times. For, if C is simple and lies in D, the function f is analytic at each point interior to and on C; so we apply the Cauchy-Goursat theorem directly. On the other hand, if C is closed but intersects itself a finite number of times, it consists of a finite number of simple closed contours. By apply the Cauchy-Goursat theorem to each of those simple closed contours, we obtain the desired result for C. Subtleties arise if the closed contour has infinitely many self-intersecting points.

Still the corollary is true under this circumstance and we will not elaborate a complete proof here.

Corollary 13.4. A function f which is analytic throughout a simply connected domain D must have an antiderivative in D.

The Cauchy-Goursat theorem can also be extended in a way that involves integrals along the boundary of a multiply connected domain:

Theorem 13.5. Let C and Ck (k = 1, 2, . . . , n) be simple closed contours such that C is described in the coun- terclockwise direction, Ck’s are in the clockwise direction, interior to C and their interior points are disjoint. If a function f is analytic throughout the closed region consisting of all points interior to and on C except for points interior to any of Ck, then

Z

C

f (z) dz +

n

X

k=1

Z

Ck

f (z) dz = 0. (13.8)

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To prove this theorem, we will draw a finite number of auxiliary lines so that the sum of the contours C + C1+ C2+· · · + Cn can be decomposed as the sum of finite simple closed contours. Then we apply the Cauchy-Goursat theorem to each of the simple closed contours and conclude that Equation (13.8) holds.

Example. When C is any positively oriented simple closed contour surrounding the origin, we must have Z

C

dz z = 2πi.

To show this, we need only to construct a positively oriented circle C0 with center at the origin and radius so small that C0 lies entirely inside C. Then the theorem implies that

Z

C

dz z =

Z

C0

dz z = 2πi by earlier computation.

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November 26, 2001 Dr. Y-C. Roger Lin

Lecture note on Complex Variables 14 Cauchy integral formula

Now it is time to establish one of the most fundamental formulae in the complex analysis.

Theorem 14.1 (Cauchy integral formula). Let f be analytic everywhere within and on a simple closed contour C, taken in the positive sense. If z0 is any interior point to C, then

f (z0) = 1 2πi

Z

C

f (z) z− z0

dz. (14.1)

Proof. Since f is continuous at z0, we know that for every ε > 0 there is a positive number δ > 0 such that

|f(z) − f(z0)| < ε whenever |z − z0| < δ.

Let us now choose a positive number ρ that is less than δ and is so small that the positively oriented circle|z −z0| = ρ, denoted by C0, is interior to C.

Since the function f (z) z− z0

is analytic in the closed region consisting of the contours C and C0 and all points between them, we know from Theorem 13.5, the principle of deformation of paths, that

Z

C

f (z) z− z0

dz = Z

C0

f (z) z− z0

dz.

It has been established before that

Z

C0

dz z− z0

= 2πi, therefore,

Z

C

f (z) z− z0

dz− 2πi f(z0) = Z

C0

f (z)− f(z0) z− z0

dz. (14.2)

It remains to estimate the last integral in Equation (14.2). Combining all of our assumptions, we know that

Z

C0

f (z)− f(z0) z− z0

dz

< ε

ρ· 2πρ = 2πε,

because|z − z0| = ρ < δ for all z ∈ ∂B(z0, ρ), and the arc length of C0 is 2πρ. Hence

Z

C

f (z) z− z0

dz− 2πi f(z0)

< 2πε.

Since ε is arbitrary, we conclude that the difference inside the absolute value must be zero, and the theorem is proved.

The Cauchy integral formula tells us that if a function is to be analytic within and on a simple closed contour C, then the values of f interior to C are completely determined by those on C.

Example. Let C be the positively oriented circle|z| = 2. Since the function f (z) = z

9− z2

is analytic within and on C and since the point z =−i is interior to C, the Cauchy integral formula tells us that Z

C

z

(9− z2)(z + i)dz = Z

C

f (z)

z− (−i)dz = 2πi f (−i) = 2πi ·−i 10 = π

5.

An important consequence of the Cauchy integral formula which we are about to prove is that if a function is analytic in a neighborhood of a point, its derivatives of all orders exist around that point and are themselves analytic there. This is done by the following theorem.

Figure

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