CHAPTER 3

DIFFERENTIATION ANALYTIC FUNCTIONS

             13. DERIVATIVES. RULES FOR DIFFERENTIATING COMPLEX FUNCTIONS

Let f(z) be a function of a complex variable which is defined and single-valued on a set E, and let z₀ be any point of E which is a limit point of E. Then the difference quotient

is a function of z defined for any point z ≠ z₀ of the set E. The limit of (3.1) as z → z₀, z ∈ E, provided it exists, is called the derivative of the function f(z) at the point z₀ relative to the set E, denoted by f′_E(z₀) or simply f′(z₀), and the function f(z) itself is said to be differentiable at the point z₀ relative to the set E.

In the special case where E is an interval (finite or infinite) of the real axis, f(z)is a function of the real variable z = x and takes values which are in general complex:

If ψ(x) ≡ 0,if the values of f(z) are real, our definitions of a derivative and of differentiability reduce to the usual definitions given in elementary calculus. However, if ψ(x) 0, then, writing

we conclude that f′(x₀) exists if and only if the derivatives φ′(x₀) and ψ′(x₀) exist, and that

(see Sec. 10). For example, if

(where x is any real number), then

Introducing the notation Δz = z – z₀ for the increment of the independent variable and Δ_Ef(z) = f(z) – f(z₀) for the increment of the function f(z) (relative to the set E), we have the following necessary and sufficient condition for differentiability:

THEOREM 3.1.The function f(z) is differentiable at the point z₀ ∈ E (relative to the set E) if and only if f(z) can be written in the form

where ε(z, z₀) → 0 as Δz→0 (z ∈ E) and A is a constant independent of Δz and ε.

Proof. If f(z) has a derivative f′_E(z₀) at z₀, then, by definition

where ε(z, z₀) → 0 as Δz → 0. Multiplying (3.3) by Δz, we find that Δ_Ef (z) can be written in the form (3.2) with A = f′_E (z₀). Conversely, if Δ_Ef(z) can be written in the form (3.2), then dividing by Δz and taking the limit as Δz → 0, we find that f′_E(z₀) exists and equals A.

Remark. It is an immediate consequence of (3.2) that if f(z) is differentiable at the point z₀ ∈ E, it is also continuous at z₀.¹

By the differential of a function f(z) which is differentiable at the point z₀ ∈ E, we mean the principal linear part of the increment Δ_Ef(z), i.e., the quantity

where, as in elementary calculus, we set dz = Δz (see Prob. 1). We can then write

For simplicity, we usually omit the subscript E in expressions like (3.2) through (3.5), unless we want to emphasize the role of the set E.

Example. To clarify the role of the set E relative to which the derivative is taken, let E be the real axis and consider the function

Obviously, the derivative f′_E(z) exists for all z = x ∈ E, since

We now extend the function f(z) to the whole complex plane , by writing (3.6) for all z = x + iy ∈ . Then f(z) is continuous for any z ∈ and coincides with the original function when z ∈ E. However, the difference quotient is now

which has no limit as z → z₀ ∈ . In fact, (3.7) approaches 0 if z → z₀ along the line x = x₀, whereas (3.7) approaches 1 if z → z₀ along the line y = y₀. In other words, f′(z₀) does not exist for any z ∈ . Thus it is already apparent from this simple example that the requirement that f(z) be differentiable regarded as a function of a complex variable z = x + iy is much more stringent than the requirement that f(z) be differentiable only for real values of z.

It follows from our definition of a derivative and the properties of limits (see Sec. 10) that the basic differentiation rules familiar from elementary calculus can be extended to the case of functions of a complex variable. We now list some of these rules. In each case, f(z), f₁(z), f₂(z), etc., is assumed to be differentiable at a given point z ∈ E (relative to E).

Rule 1. If f(z) ≡ z, then

whereas if f(z) = const, then

Loosely speaking, the derivative of the independent variable is 1, and the derivative of any constant is 0.

Rule 2 (Differentiation of sums, products and powers). If c is a constant, then

Moreover we have

In particular,

and

Rule 3 (Differentiation of a quotient). If f₂(z) ≠ 0, then

Rule 4 (Differentiation of a composite function). Suppose the function f(z) with domain E and range is differentiable at the point z₀ ∈ E, and suppose is an infinite set with w₀ = f(z₀) as a limit point. Moreover, suppose the function φ(w) has domain and is differentiable at the point w₀ ∈ . Then the Composite function

is differentiable at the point z₀, and

The proof is trivial if there exists a neighborhood (z₀) of the point z₀ such that f(z)= w ≠ w₀ for all z ∈ (z₀) ∩ E, for then

Since w → w₀ as z → z₀ (see the remark on p. 44), taking the limit of (3.9) as z → z₀ we immediately obtain (3.8).

Now suppose every neighborhood of z₀ contains a point z ∈ E, z ≠ z₀ such that f(z) = f(z₀) = w₀. Then there exists a sequence {z_n} of distinct points in E converging to z₀ such that

For this sequence

and hence the derivative

(which by hypothesis exists) must vanish. Therefore the right-hand side of (3.8) vanishes, and to prove (3.8) we have to show that its left-hand side also vanishes. Since the difference quotient

is obviously zero for every z ≠ z₀ such that f(z) = f(z₀), the proof reduces to showing that

vanishes for every sequence {z*_n} converging to z₀ such that f(z*_n) = w*_n ≠ w₀ = f(z₀). However, for such a sequence, (3.10) equals

[cf. (3.9)], which approaches

as n → ∞. But as just shown, d_Ef(z₀)/dz vanishes, and hence (3.8) holds.

Rule 5 (Differentiation of an inverse function). Suppose w = f(z) is a one-to-one function defined on E, and suppose the inverse function z = f^–1(w) = φ(w) is continuous on , the range of f(z). Then if f(z) is differentiable at the point z₀ ∈ E and if f′_E(z₀) ≠ 0, the function φ(w) is differentiable at the point w₀ = f(z₀) ∈ , and

The proof goes as follows: Since the function w = f(z) is one-to-one, w ≠ w₀ implies z ≠ z₀, and hence the difference quotient of the function φ(w) can be written in the form

Moreover, since φ(w) is continuous on , φ(w) → φ(w₀) as w → w₀, i.e., z → z₀ as w → w₀. Therefore

as asserted.

Problem 1. Justify the formula dz = Δz (“the differential and the increment of the independent variable are equal”) and hence formula (3.4).

Problem 2. Show that the function f(z) → is not differentiable at any point z₀ relative to any neighborhood (z₀).

14. THE CAUCHY-RIEMANN EQUATIONS. ANALYTIC FUNCTIONS

From now on, we shall be concerned mainly with functions defined on some domain E = G, and we shall drop the subscript E in f′_E(z) and d_Ef(z)/dz. Suppose u(x, y) is a real function of two real variables x and y defined on G. Then u(x, y) is said to be differentiable at the point (x₀, y₀) ∈ G if u(x, y) can be written in the form

where

as (x, y) → (x₀, y₀). The coefficients A(x₀, y₀)and B(x₀, y₀) in the right-hand side of (3.11) are just the partial derivatives of the function u(x, y) at the point (x₀, y₀):

THEOREM 3.2 (Cauchy-Riemann equations). Let

be a function of a complex variable defined on a domain G. Then a necessary and sufficient condition for f(z) to be differentiable (as a function of a complex variable) at the point z₀ = x₀ + iy₀ ∈ G is that the functions u(x, y) and v(x, y) be differentiable (as functions of the two real variables x and y) at the point (x₀, y₀) and satisfy the Cauchy-Riemann equations

at (x₀, y₀). If these conditions are satisfied, f′(z₀) can be represented in any of the forms

where the partial derivatives are all evaluated at (x₀, y₀).

Proof. First we prove that the conditions are necessary. Consider the increments

of the independent variable z and of the function f(z). According to Theorem 3.1, if f(z) is differentiable at z₀, then

where ε → 0 as Δz → 0. Writing

and taking the real and imaginary parts of (3.14), we find that

where ε₁, ε₂ → 0 as Δx, Δy → 0, since

It follows that the functions u(x, y) and v(x, y) are differentiable at (x₀, y₀) and

But (3.15) immediately implies (3.12) and (3.13).

To prove that the conditions of the theorem are sufficient, we reverse the preceding argument. Thus, suppose u(x, y) and v(x, y) are differentiable at the point (x₀, y₀), and suppose (3.12) holds. Then

where α_l, α₂, β₁, β₂ → 0 as Δx, Δy → 0 and we have written

Substituting (3.16) into the formula Δf(z) = Δu + i Δv, we find that

or

where

Since α₁, α₂, β₁, β₂ → 0 as Δx, Δy → 0, it follows that ε → 0 as Δz → 0, and hence, according to (3.18) and Theorem 3.1, f(z) is differentiable at z₀, with derivative

The various representations (3.13) of f′(z₀) are now immediate consequences of (3.19) and (3.17).

A function f(z) which is differentiable on a domain G, i.e., at every point of G, is said to be analytic (synonymously, holomorphic or regular) on G. If f(z) is analytic in a neighborhood of z₀, f(z) is said to be analytic at z₀.

Remark. As we know from calculus,2 a sufficient condition for the differentiability of the functions u(x, y) and v(x, y) on a domain G is that the partial derivatives

exist and be continuous on G. Therefore a sufficient condition for the function f(z) = u + iv to be analytic on G is that the partial derivatives (3.20) exist, be continuous and satisfy the equations (3.12) on G.

Example 1. For the function

defined in the whole plane, we have

with continuous partial derivatives

Therefore the Cauchy-Riemann equations (3.12) are satisfied, and the function (3.21) is analytic in the whole plane, with derivative

Example 2. For the function f(z) = x, considered in the example on p. 45, we have

and

Since

the Cauchy-Riemann equations are not satisfied, and hence this function is not differentiable anywhere in the plane.

In many cases, it is important to express the differentiability conditions for a function f(z) = u + iv at a point z ≠ 0 in terms of the polar coordinates

The appropriate necessary and sufficient conditions for differentiability are that u(r, Φ) and v(r, Φ) be differentiable (as functions of the two real variables r and Φ) and satisfy the polar form of the Cauchy-Riemann equations, i.e.,

(at a given nonzero point P). To verify these conditions, we must prove that 1) u and v are differentiable as functions of r and Φ at P if and only if they are differentiable as functions of x and y at P, and 2) under these conditions, the equations (3.22) are equivalent to the equations (3.12). The first assertion follows from the familiar fact that a differentiable function (u = u(x, y), say) of differentiable functions (x = r cos Φ, y = r sin Φ, say) is also differentiable (with respect to the new variables r and Φ).³ The second assertion can be verified by direct calculation. For example, if u and v are differentiable functions of x and y, and if the equations (3.12) hold, then

Writing (3.23) in the form

and solving for ∂u/∂x and ∂u/∂y, we obtain

and hence

This formula is convenient for calculating f′(z) with the help of polar coordinates. Using (3.22), we can write f′(z) in the form

Example. Consider the function

where m and n > 0 are integers (see Sec. 4). This function is defined on the domain G consisting of all nonzero points of the complex plane, and is multiple-valued (unless m/n is an integer), since Φ = Arg z is multiple-valued. Before we can talk about the derivative of z^m/n, we must first make z^m/n single-valued in the following sense: Let z₀ ∈ G, and choose a neighborhood (z₀) which does not contain the origin. Fix one of the values of Φ₀ = Arg z₀, and for the argument of any other point z ∈(z₀) choose the unique value Φ satisfying the condition

(see Figure 3.1). Using this value of Φ in (3.25), we obtain a single-valued function defined on (z₀), which we call a single-valued branch of the function (3.25), denoted by z^m,n as before. Thus at any point z ∈ (z₀), including z₀ itself, we can now write

FIGURE 3.1

and

i.e., f(z) satisfies the equations (3.22), and is therefore differentiable on (z₀). According to formula (3.24),

so that the rule for differentiating a fractional power z^mln is formally the same as the rule for differentiating the corresponding function x^m/n of a real variable. It should be kept in mind that our calculation is subject to the condition z ≠ 0, which can only be dropped if m/n is a nonnegative integer.

Problem 1. Show that the function f(z) = z Re z is differentiable only at the point z = 0, and find f′(0).

Comment. Thus f(z) = z Re z is differentiable but not analytic at z = 0.

Problem 2. Show that the function

has partial derivatives ∂u/∂x and ∂u/∂y at the origin, but is not differentiable there.

Problem 3. Show that the function f(z) = satisfies the Cauchy-Riemann equations at the point z = 0, but is not differentiable there.

Problem 4. Establish the following generalization of the Cauchy-Riemann equations: If f(z) = u + iv is differentiable at a point z₀ = x₀ + iy₀ of a domain G, then

at (x₀, y₀) where ∂/∂s and ∂/∂n denote directional differentiation in any two orthogonal directions s and n at (x₀, y₀) such that n is obtained from s by making a counterclockwise rotation.

Problem 5. Use the preceding problem to deduce the equations (3.22).

15. GEOMETRIC INTERPRETATION OF Arg f′(z) AND |f′(z)|. CONFORMAL MAPPING

Let l be a continuous curve with equation z = λ(t), t ∈ [a, b], and suppose λ(t) is differentiable at a point t₀ ∈ [a, b] (relative to the set [a, b]). Let {t_n} be an arbitrary sequence of points in [a, b] converging to t₀ (t_n ≠ t₀), and consider the difference quotient

Obviously r_n → r₀ = λ′(t₀) as n → ∞.

DEFINITION. The curve l is said to have a tangent at the point z₀ = λ(t₀) if the limit

exists, and then the tangent is said to have inclination θ. Geometrically, the tangent to l at z₀ is represented by the ray τ emanating from z₀ which makes the angle θ with the positive real axis.⁴

Remark 1. Clearly, if λ′(t₀) ≠ 0, l has a tangent at z₀, since

(see Sec. 6, Prob. 3), and then the inclination of the tangent is just the argument of the complex number λ′(t₀). On the other hand, if λ′(t₀) = 0, l may or may not have a tangent at z₀, since the fact that r_n → 0 implies only that |r_n| → 0 and says nothing about the behavior of Arg r_n. However, if l has a tangent at z₀, then λ(t_n) ≠ λ(t₀) for all t_n sufficiently close to t₀, since Arg 0 is meaningless.⁵

Remark 2. As just defined, the tangent is a ray, not a vector. If λ′(t₀) ≠ 0, we can also introduce a tangent vector to l at z₀, defined as the vector of length |λ′(t₀)| which makes the angle λ with the positive real axis.

THEOREM 3.3. Let G be a domain, and let f(z) be a continuous function of a complex variable defined on G. Suppose f(z) has a nonzero derivative f′(z₀) at a point z₀ ∈ G, and let l be a curve which passes through z₀ and has a tangent τ at z₀. Then w = f(z) maps l into a curve L in the w-plane which passes through the point w₀ = f(z₀) and has a tangent T at w₀. Moreover, the inclination of T exceeds the inclination of τ by the angle Arg f′(z₀).

Proof. Suppose l has the equation z = λ(t), t ∈ [a, b], and let z₀ = λ(t₀). By hypothesis,

exists, where r_n is given by (3.26). The function w = f(z) maps l into a curve L in the w-plane with equation

where w₀ = f(z₀) = ∧(t₀). Let {t_n} be an arbitrary sequence of points in [a, b] converging to t₀, and let

Then the tangent to L at w₀ has inclination

provided this limit exists. Clearly we have

where the first factor in the right-hand side is well defined, since λ(t_n) ≠ λ(t₀) for all t_n sufficiently close to t₀ (l is assumed to have a tangent at z₀). Therefore

where w_n = ∧(t_n), z_n = λ(t_n) and θ is the inclination of τ at z₀.6 It follows from (3.29) that exists and that

as asserted. We note that things are particularly simple in the case where λ′(t₀) ≠ 0, since then

by the rule for differentiating the composite function (3.28).

Now let l₁ and l₂ be two curves with a common initial point z₀, which have tangents τ₁ and τ₂ at z₀, and suppose the angle between τ₁ and measured from τ₁ to τ₂. Suppose l₁ and l₂ have images L₁ and L₂ under f(z). Then, according to Theorem 3.3, if f′(z₀) ≠ 0, L₁ and L₂ have tangents T₁ and T₂ at the point w₀ = f(z₀), where T₁ and T₂ are obtained by rotating τ₁ and τ₂ through the same angle Arg f′(z₀). Therefore the angle between L₁ and L₂ equals the angle between l₁ and l₂, and is measured in the same direction, i.e., from L₁ to L₂. In other words, a continuous function w = f(z) with a nonzero derivative f′(z₀) maps all curves in the z-plane which pass through z₀ and have tangents at z₀ into curves in the w-plane which pass through w₀ = f(z₀) and have tangents at w₀, and moreover, the mapping preserves angles between curves. A mapping by a continuous function which preserves angles between curves passing through a given point z₀ is said to be conformal at z₀. If a conformal mapping preserves the directions in which angles are measured (as well as their magnitudes), it is called a conformal mapping of the first kind, but if it reverses the directions in which angles are measured, it is called a conformal mapping of the second kind. Thus Theorem 3.3 has the following consequence:

THEOREM 3.4. Let G be a domain, and let f(z) be an analytic function on G. Thenf(z) is a conformal mapping of the first kind at every point of G where f′(z) ≠ 0.

Example 1. Reflection in the real axis, i.e., the transformation w = , is a conformal mapping of the second kind. A more general example is the complex conjugate

of an analytic function f(z), where f′(z) ≠ 0.

Example 2. At a point where the derivative vanishes, angles may or may not be preserved, as can be seen by comparing the mappings

at the point z = 0.

As we have just seen, Arg f′(z₀) represents the rotation undergone by the tangent to a curve l at the point z₀ ∈ l when transforming to the new curve L = f(l) and the new point w₀ = f(z₀). In particular, if f′(z₀) is a positive real number, the tangents to l at z₀ and to L at w₀ are parallel and point in the same direction.

To explain the geometric meaning of the quantity |f′(z₀)|, i.e., the absolute value of the derivative at z₀, we note that

The numbers |z – z₀| and |f(z) – f(z₀)| are the distance between the points z and z₀ in the z-plane, and the distance between their images f(z) and f(z₀) in the w-plane, respectively. Thus, interpreting

as the linear magnification ratio (or simply the magnification) of the vector z – z₀ under the mapping w = f(z),7 we can regard | f′(z₀)| as the magnification at the point z₀ under w = f (z).

Remark. The size of the magnification at the point z₀ does not depend on the choice of the finite vector z — z₀ drawn from z₀, since |f′(z₀)| is not the actual magnification of any such vector, but rather the limiting magnification as z → z₀.

Problem 1. With the same notation as on p. 55, a curve l is said to have a left-hand tangent (of inclination θ) at the point z₀ = λ(t₀) if the limit (3.27) exists, subject to the extra condition that every point of the sequence {t_n} converging to t₀ be less than t₀. The right-hand tangent is defined similarly by requiring that t_n > t₀ for every n. Give an example of a (continuous) curve l which has a left-hand tangent but no right-hand tangent (and hence no tangent) at a point z₀ ∈ l.

Problem 2. Verify that the function

used in Example 2 above is differentiable at z = 0.

Problem 3. Find the angle through which a curve drawn from the point z₀ is rotated under the mapping w = z² if

Also find the corresponding values of the magnification.

Problem 4. Carry out the same calculations as in the preceding problem, this time applied to the function w = z³.

Problem 5. Which part of the plane is shrunk and which part stretched under the following mappings:

16. THE MAPPING

To illustrate the above considerations, we now examine the fractional linear transformation or Möbius transformation

where a, b, c, d are arbitrary complex numbers (except that c and d are not both zero). First suppose that c = 0. Then L(z) reduces to

and is sometimes called the linear transformation. The transformation (3.31) is defined for all values of z, and if α ≠ 0, its derivative L′(z) is a nonzero constant, so that (3.31) is conformal at every point of the (finite) z-plane. Under this transformation, the tangents to all curves in the z-plane are rotated through the same angle, equal to Arg α, and the magnification at every point equals |α|. If α = 1, then

where k is an integer, and then both the rotation and expansion produce no effect. In this case, the transformation takes the form

which obviously corresponds to displacing the whole plane by the vector β. On the other hand, if α ≠ 1 (and α ≠ 0), the transformation (3.31) can be written in the form

where z₀ is determined from the equation8

Then it is immediately clear that the transformation (3.31) is equivalent to a rotation of the whole plane through the angle Arg α about the point z₀ = β/(l — α), together with a uniform magnification by the factor |α| relative to the point z₀. This magnification is sometimes called a homothetic transformation (or transformation of similitude) with ray center z₀ and ray ratio |α|.

Next suppose that c ≠ 0 in (3.30). Then the derivative

exists, if z ≠ δ, where δ = —d/c. If the determinant

vanishes, then its rows are proportional, i.e.,

where μ is a constant, so that (3.30) reduces to the trivial transformation

If ad – bc ≠ 0, then L′(z) ≠ 0 for all z ≠ δ, and hence the mapping w = L(z) is conformal at all finite points except possibly at z = δ. Under the mapping, the tangents to curves passing through any point z ≠ δ are rotated through an angle equal to

while the magnification at z equals

The angle through which tangents are rotated has the same value for all points with equal values of Arg (z — δ), i.e., along any ray drawn from δ, but otherwise varies from point to point. Similarly, in general the magnification varies with z, but it has the same value for all points with equal values of |z — δ|, i.e., along any circle with center δ. In particular, the magnification is equal to 1 at every point of the circle C with equation

(called the isometric circle of the Möbius transformation), is greater than 1 inside C (approaching ∞ as z → δ), and is less than 1 outside C (approaching 0 as z → ∞). The situation is shown schematically in Figure 3.2.

FIGURE 3.2

Problem 1. As shown on p. 60, the entire linear transformation w = αz + β is equivalent to a rotation and a magnification relative to the fixed point z₀ = β/(1 – α), provided that α ≠ 0. Find the rotation, magnification and (finite) fixed point, if such exists, corresponding to each of the following transformations, and write each in the canonical form w – z₀ = α(z – z₀):

Problem 2. Find the entire linear transformation with fixed point 1 + 2i carrying the point i into the point –i.

Problem 3, Find the entire linear transformation carrying the triangle with vertices at the points 0, 1, i into the similar triangle with vertices at the points 0, 2, 1 + i.

17. CONFORMAL MAPPING OF THE EXTENDED PLANE

As in the preceding section, let c ≠ 0 and ad – bc ≠ 0. Then it is clear that

where δ = —d/c. Suppose we complete the definition of L(z) by setting

Then L(z) is defined in the whole extended plane, and maps the finite point δ into ∞ (the point at infinity) and ∞ into the finite point A. We now show that the mapping w = L(z) is conformal at the points δ and A (it has already been shown that the mapping is conformal everywhere else). First we must suitably define the concept of an angle with its vertex at infinity (for the justification of this definition, see Probs. 3 and 4; also recall the last paragraph of Sec. 8):

DEFINITION.⁹ Two continuous curves γ₁ and γ₂ in the extended plane form an angle of α radians with its vertex at infinity if and only if their images and in the extended plane under the transformation ζ = 1/z form an angle of α radians with its vertex at the origin.

Example. The real and imaginary axes form an angle of π/2 radians with its vertex at infinity. In fact, under the transformation ζ = 1/z, the real and imaginary axes are carried into themselves, and they obviously form an angle of π/2 radians with its vertex at the origin.

Returning to the mapping w = L(z), let γ₁ and γ₂ be two curves forming an angle θ with its vertex at the point δ, and let and be their images in the w-plane. To prove that and form an angle θ with its vertex at infinity, we subject the w-plane to the transformation

Then the curves and go into two curves and , and the point at infinity goes into the origin of coordinates (see Figure 3.3). Obviously, we can go from γ₁ and γ₂ in the z-plane to and in the η-plane by making the Möbius transformation

which is conformal at the point z = δ = —d/c. It follows that the curves and form an angle θ with its vertex at the origin. Therefore the curves and also form an angle 0 with its vertex at infinity. This proves that the mapping w = L(z) is conformal at the point z = δ

The fact that w = L(z) is conformal at ∞ is proved similarly. In fact, if the curves γ₁ and γ₂ go through the point at infinity in the z-plane, their images and in the w-plane go through the point A. Suppose γ₁ and γ₂ form an angle θ with its vertex at infinity. This means that their images γ*₁ and γ*₂ under the transformation ζ = 1/z form an angle θ with its vertex at the origin. But we can obviously go from γ*₁ and γ*₂ to and by making the Möbius transformation

FIGURE 3.3

which is conformal at the point ζ = 0. It follows that and form the same angle θ at the point A = a/c. This proves that the mapping w = L(z) is conformal at ∞. The situation can be summarized by saying that the transformation w = L(z) is a conformal mapping of the extended plane onto itself.

Remark 1. These considerations suggest the following definition: A function f(z) is said to be analytic at z = ∞ if the function f*(ζ) = f(l/ζ) is analytic at ζ = 0. In particular, if f(z) is analytic at z = ∞, the limit

always exists and is finite. We define the derivative of f(z) at z = ∞ to be the quantity

where it should be noted that in general

(see Prob. 2). Then, by the argument given above for the special case of the Möbius transformation, it is easy to see that the mapping w = f(z) is conformal at ∞ if f′(∞) ≠ 0. With this approach, the conformality at ∞ of the Möbius transformation

follows from the fact that

Remark 2. Similarly, if

but if

is analytic at z = a, with derivative φ′(a) ≠ 0, then, just as in the case of the Möbius transformation, the mapping w = f(z) is conformal at z = a.

Problem 1. Prove that the transformation L(z) = αz + β is conformal at infinity if α ≠ 0.

Problem 2. Prove that if f(z) is analytic at infinity, then

Problem 3 (Ml, p. 87). Prove that stereographic projection is conformal, i. e., that under stereographic projection the angle between any two curves on the Riemann sphere (with its vertex at any point except the north pole) equals the angle between the images of the curves in the finite plane.

Problem 4 (Ml, Sec. 25). Prove that two curves γ₁ and γ₂ in the extended plane form an angle of α radians with its vertex at infinity if and only if their images γ*₁ and γ*₂ on the Riemann sphere under stereographic projection form an angle of α radians with its vertex at the north pole.

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¹ For brevity, we shall often omit the phrase “relative to E,” which is tacitly assumed in any context like this.

² See e.g., D. V. Widder, Advanced Calculus, second edition, Prentice-Hall, Inc., Englewood Cliffs, N.J. (1961), p. 17.

³ The reader should verify this assertion, guided by (3.11).

⁴ Formula (3.27) means that given any ε > 0, there is an integer N(ε) > 0 and a sequence {θ_n}, where each θ_n is a value of Arg r_n, such that |θ_n – θ| < ε for all n > N(ε). Clearly, θ is only defined to within a multiple of 2π. The angle θ will always be measured from the positive real axis to the tangent τ (in the counterclockwise direction for a positive value of θ).

⁵ This condition is automatically satisfied if λ′(t₀) ≠ 0, since otherwise r_n = 0 for t arbitrarily close to t₀, which implies that λ′(t₀) = 0, contrary to hypothesis.

⁶ In reversing the order of the operations Arg and we have used the fact that f′(z₀) ≠ 0.

⁷ Here the word magnification is used in a general sense, and can correspond to stretcting if |f′(z₀)| > 1 or shrinking if |f′(z₀)| < 1 [or neither if |f′(z₀)| = 1].

⁸ Obviously, z₀ is invariant under the transformation (3.31), i.e., z₀ is a fixed point of the transformation. If α ≠ 0, the point at infinity is also a fixed point (see Sec. 25).

⁹ The curves γ₁ and γ₂ are unbounded, in the sense of Sec. 12, Prob. 2.