Complex Functions

In this lecture, first we shall introduce a complex-valued function of a complex variable, and then for such a function define the concept of limit and continuity at a point.

LetSbe a set of complex numbers. Acomplex function(complex-valued of a complex variable)f defined onSis a rule that assigns to eachz=x+iy in S a unique complex numberw=u+iv and written as f :S →C.The numberwis called the value off atzand is denoted byf(z); i.e.,w=f(z).

The set S is called the domain of f, the set W = {f(z) : z ∈ S}, often denoted asf(S),is called the rangeorimage off,andf is said to mapS ontoW.The functionw=f(z) is said to be fromS intoW if the range of S underf is a subset ofW.When a function is given by a formula and the domain is not specified, the domain is taken to be the largest set on which the formula is defined. A function f is calledone-to-one (orunivalent, or injective) on a setSif the equationf(z1) =f(z2),wherez1andz2are inS, implies thatz1 =z2. The functionf(z) = iz is one-to-one, butf(z) =z² is not one-to-one sincef(i) =f(−i) =−1.A one-to-one and onto function is called bijective. We shall also consider multi-valued functions: amulti- valued function is a rule that assigns a finite or infinite non-empty subset ofC for each element of its domain S.In Lecture 2, we have already seen that the functionf(z) = argz is multi-valued.

As every complex number z is characterized by a pair of real numbers xand y, a complex functionf of the complex variablez can be specified by two real functionsu=u(x, y) andv=v(x, y). It is customary to write w=f(z) =u(x, y)+iv(x, y).The functionsuandv,respectively, are called the real and imaginary parts off. The common domain of the functionsu andv corresponds to the domain of the functionf.

Example 5.1.

For the functionw=f(z) = 3z²+ 7z,we have f(x+iy) = 3(x+iy)²+ 7(x+iy) = (3x²−3y²+ 7x) +i(6xy+ 7y), and henceu= 3x²−3y²+ 7xandv= 6xy+ 7y.Similarly, for the function w=f(z) =|z|²,we find

f(x+iy) = |x+iy|² = x²+y²,

and henceu=x²+y² andv= 0.Thus, this function is a real-valued func- tion of a complex variable. Clearly, the domain of both of these functions is

R.P. Agarwal et al., An Introduction to Complex Analysis,

DOI 10.1007/978-1-4614-0195-7_5, © Springer Science+Business Media, LLC 2011 28

Complex Functions 29 C. For the functionw=f(z) =z/|z|,the domain is C\{0},and its range is|z|= 1.

Example 5.2.

Thecomplex exponential function f(z) =e^z is defined by the formula (3.1). Clearly, for this function,u=e^xcosyandv =e^xsiny, which are defined for all (x, y)∈IR². Thus, for the functione^zthe domain isC.The exponential function provides a basic tool for the application of complex variables to electrical circuits, control systems, wave propagation, and time-invariant physical systems.

Recall that a vector-valued function of two real variables F(x, y) = (P(x, y), Q(x, y)) is also called a two-dimensional vector filed. Using the standard orthogonal unit basis vectors iand j,we can express this vector field as F(x, y) =P(x, y)i+Q(x, y)j. There is a natural way to represent this vector field with a complex function f(z). In fact, we can use the functions P andQas the real and imaginary parts of f,in which case we say that the complex function f(z) = P(x, y) +iQ(x, y) is the complex representationof the vector fieldF(x, y) =P(x, y)i+Q(x, y)j.Conversely, any complex function f(z) = u(x, y) +iv(x, y) has an associated vector field F(x, y) =u(x, y)i+v(x, y)j.From this point of view, both F(x, y) = P(x, y)i+Q(x, y)jandf(z) =u(x, y) +iv(x, y) can be called vector fields.

This interpretation is often used to study various applications of complex functions in applied mathematical problems.

Letfbe a function defined in some neighborhood ofz0,with the possible exception of the pointz0itself. We say that thelimitoff(z) aszapproaches z0(independent of the path) is the numberw0if|f(z)−w0| →0 as|z−z0| → 0 and we write limz→z₀f(z) =w₀.Hence,f(z) can be made arbitrarily close tow0if we choosez sufficiently close toz0.Equivalently, we say thatw0 is the limit off aszapproachesz₀if, for any given >0,there exists aδ >0 such that

0 < |z−z₀| < δ =⇒ |f(z)−w₀| < .

Figure 5.1 x

y

z

·

₀

·

0

z

u v

0

f(z) w₀

·

Example 5.3.

By definition, we shall show that (i) limz→1−i(z²−2) =

−2 + 2iand (ii) limz→1−i|z²−2|=√ 8.

(i). Given any >0,we have

|z²−2−(−2 + 2i)| = |z²−2i| = |z²+ 2i| = |z²+ 2i|

= |z−(1−i)||z+ (1−i)|

≤ |z−(1−i)|(|z−(1−i)|+ 2|1−i|)

≤ |z−(1−i)|(1 + 2√

2) if |z−(1−i)|<1

< if |z−(1−i)|<min

1,

1 + 2√ 2

. (ii). Given any >0,from (i) we have

||z²−2| −√

8| = ||z²−2| − | −2 + 2i||

≤ |z²−2−(−2 + 2i)|

< if |z−(1−i)|<min

1,

1 + 2√ 2

.

Example 5.4.

(i). Clearly, limz→z₀z=z₀. (ii). From the inequalities

|Re(z−z₀)| ≤ [(Re(z−z₀))²+ (Im(z−z₀))²]^1/2 = |z−z₀|,

|Im(z−z0)| ≤ |z−z0|,

it follows that limz→z₀Rez= Rez0, limz→z₀Imz= Imz0.

Example 5.5.

lim_z→0(z/z) does not exist. Indeed, we have lim

z→0

alongx-axis

z

z = lim

x→0

x+i0 x−i0 = 1,

lim z→0

alongy-axis

z

z = lim

y→0

0 +iy

0−iy = −1.

The following result relates real limits of u(x, y) and v(x, y) with the complex limit off(z) =u(x, y) +iv(x, y).

Theorem 5.1.

Letf(z) =u(x, y) +iv(x, y), z0 =x0+iy0, andw0 = u0+iv0.Then, limz→z₀f(z) =w0if and only if limx→x₀, y→y₀u(x, y) =u0

and limx→x₀, y→y₀v(x, y) =v0.

In view of Theorem 5.1 and the standard results in calculus, the follow- ing theorem is immediate.

Theorem 5.2.

If lim_z→z0f(z) = A and lim_z→z0g(z) = B, then (i) lim_z→z0(f(z)±g(z)) = A±B, (ii) lim_z→z0f(z)g(z) = AB, and (iii) limz→z₀

f(z) g(z) = A

B ifB= 0.

Complex Functions 31 For thecomposition of two functions f and g denoted and defined as (f◦g)(z) =f(g(z)),we have the following result.

Theorem 5.3.

If limz→z₀g(z) =w₀ and limw→w₀f(w) =A,then

zlim→z₀f(g(z)) = A = f

zlim→z₀g(z)

.

Now we shall define limits that involve∞.For this, we note thatz→ ∞ means|z| → ∞,and similarly,f(z)→ ∞means|f(z)| → ∞.

The statement limz→z₀f(z) =∞means that for any M >0 there is a δ >0 such that 0 <|z−z0| < δ implies|f(z)| > M and is equivalent to limz→z₀1/f(z) = 0.

The statement lim_z→∞f(z) = w₀ means that for any > 0 there is an R >0 such that|z| > R implies |f(z)−w0|< , and is equivalent to lim_z→0f(1/z) =w₀.

The statement lim_z→∞f(z) =∞means that for anyM >0 there is an R >0 such that|z|> Rimplies|f(z)|> M.

Example 5.6.

Since

2z+ 3

3z+ 2 = 2 + 3/z 3 + 2/z,

limz→∞(2z+ 3)/(3z+ 2) = 2/3.Similarly, limz→∞(2z+ 3)/(3z²+ 2) = 0 and limz→∞(2z²+ 3)/(3z+ 2) =∞.

Letf be a function defined in a neighborhood ofz0.Then,f is contin- uousatz0if limz→z₀f(z) =f(z0).Equivalently,f iscontinuousatz0if for any given >0,there exists aδ >0 such that

|z−z₀| < δ =⇒ |f(z)−f(z₀)| < .

A functionf is said to be continuous on a setS if it is continuous at each point ofS.

Example 5.7.

The functions f(z) = Re (z) and g(z) = Im (z) are continuous for allz.

Example 5.8.

The functionf(z) =|z|is continuous for allz.For this, letz0be given. Then

zlim→z₀|z| = lim

z→z₀

(Rez)²+ (Imz)² =

(Rez₀)²+ (Imz₀)² = |z₀|. Hence,f(z) is continuous atz₀.Sincez₀is arbitrary, we conclude thatf(z) is continuous for allz.

It follows from Theorem 5.1 that a functionf(z) = u(x, y) +iv(x, y) of a complex variable is continuous at a pointz₀ =x₀+iy₀ if and only if u(x, y) andv(x, y) are continuous at (x₀, y₀).

Example 5.9.

The exponential function f(z) = e^z is continuous on the whole complex plane sincee^xcosyande^xsiny both are continuous for all (x, y)∈IR².

The following result is an immediate consequence of Theorem 5.2.

Theorem 5.4.

If f(z) and g(z) are continuous at z0, then so are (i)f(z)±g(z),(ii)f(z)g(z),and (iii) f(z)/g(z) providedg(z0)= 0.

Now letf :S→W, S1 ⊂S,andW₁⊂W.Theinverse image denoted as f⁻¹(W₁) consists of all z ∈ S such that f(z) ∈ W₁. It follows that f(f⁻¹(W₁))⊂W₁ andf⁻¹(f(S₁))⊃S₁.By definition, in terms of inverse image continuous functions can be characterized as follows: A function is continuous if and only if the inverse image of every open set is open.

Similarly, a function is continuous if and only if the inverse image of every closed set is closed.

For continuous functions we also have the following result.

Theorem 5.5.

Letf :S→Cbe continuous. Then,

(i). a compact set ofS is mapped onto a compact set inf(S),and (ii). a connected set ofS is mapped onto a connected set off(S).

It is easy to see that the constant function and the function f(z) =z are continuous on the whole plane. Thus, from Theorem 5.4, we deduce that thepolynomial functions; i.e., functions of the form

P(z) = a₀+a₁z+a₂z²+· · ·+anzⁿ, (5.1) whereai, 0≤i≤nare constants, are also continuous on the whole plane.

Rational functionsin z,which are defined as quotients of polynomials; i.e., P(z)

Q(z) = a₀+a₁z+· · ·+anzⁿ

b₀+b₁z+· · ·+b_mz^m, (5.2) are therefore continuous at each point where the denominator does not vanish.

Example 5.10.

We shall find the limits asz→2iof the functions f1(z) = z²−2z+ 1, f2(z) = z+ 2i

z , f3(z) = z²+ 4 z(z−2i). Since f₁(z) and f₂(z) are continuous at z = 2i, we have lim_z→2if₁(z) = f₁(2i) =−3−4i,lim_z→2if₂(z) =f₂(2i) = 2.Since f₃(z) is not defined at

Complex Functions 33 z= 2i,it is not continuous. However, for z= 2iandz= 0,we have

f₃(z) = (z+ 2i)(z−2i)

z(z−2i) = z+ 2i

z = f₂(z)

and so lim_z→2if₃(z) = lim_z→2if₂(z) = 2. Thus, the discontinuity off₃(z) atz= 2ican be removed by settingf2(2i) = 2.The function f3(z) is said to have aremovable discontinuityatz= 2i.

Problems

5.1. For each of the following functions, describe the domain of defini- tion that is understood:

(a).f(z) = z

z²+ 3, (b).f(z) = z

z+z, (c). f(z) = 1 1− |z|².

5.2. (a). Write the function f(z) = z³+ 2z+ 1 in the formf(z) = u(x, y) +iv(x, y).

(b). Suppose thatf(z) =x²−y²−2y+i(2x−2xy).Expressf(z) in terms ofz.

5.3. Show that when a limit of a function f(z) exists at a point z0,it is unique.

5.4. Use the definition of limit to prove that:

(a). lim

z→z₀(z²+ 5) =z₀²+ 5, (b). lim

z→1−iz²= (1 +i)², (c). lim

z→z₀z=z0, (d). lim

z→2−i(2z+ 1) = 5−2i.

5.5. Find each of the following limits:

(a). lim

z→2+3i(z−5i)², (b). lim

z→2

z²+ 3

iz , (c). lim

z→3i

z²+ 9 z−3i, (d). lim

z→i

z²+ 1

z⁴−1, (e). lim

z→∞

z²+ 1

z²+z+ 1−i, (f). lim

z→∞

z³+ 3iz²+ 7 z²−i . 5.6. Prove that:

(a). lim

z→0

z z

2

does not exist, (b). lim

z→0

z² z = 0.

5.7. Show that if lim

z→z₀f(z) = 0 and there exists a positive num- ber M such that |g(z)| ≤ M for all z in some neighborhood of z0, then

zlim→z₀f(z)g(z) = 0.Use this result to show that limz→0ze^i/|z|= 0.

5.8. Show that if lim

z→z₀f(z) =w₀,then lim

z→z₀|f(z)|=|w₀|.

5.9. Suppose thatf is continuous atz₀ and g is continuous at w₀ = f(z0). Prove that the composite functiong◦f is continuous atz0.

5.10.Discuss the continuity of the function f(z) =

⎧⎨

⎩ z³−1

z−1 , |z| = 1 3, |z|= 1 at the points 1, −1, i,and−i.

5.11.Prove that the function f(z) = Arg(z) is discontinuous at each point on the nonpositive real axis.

5.12 (Cauchy’s Criterion).Show that limz→z₀f(z) =w0if and only if for a given > 0 there exists a δ > 0 such that for any z, z satisfying

|z−z0|< δ, |z−z0|< δ,the inequality|f(z)−f(z)|< holds.

5.13.Prove Theorem 5.5.

5.14.The functionf :S→Cis said to beuniformly continuousonSif for every given >0 there exists aδ=δ()>0 such that|f(z₁)−f(z₂)|<

for all z₁, z₂ ∈ S with |z₁−z₂| < δ. Show that on a compact set every continuous function is uniformly continuous.

Answers or Hints

5.1. (a). z² =−3 ⇐⇒ z =±√

3i, (b). z+z= 0 ⇐⇒ z is not purely imaginary; i.e., Re(z)= 0,(c). |z|²= 1 ⇐⇒ |z| = 1.

5.2. (a). (x+iy)³+ 2(x+iy) + 1 = (x³−3xy²+ 2x+ 1) +i(3x²y−y³+ 2y).

(b). Usex= (z+z)/2, y= (z−z)/2ito obtainf(z) =z²+ 2iz.

5.3. Suppose that limz→z₀f(z) = w₀ and limz→z₀f(z) = w₁. Then, for any positive number , there are positive numbers δ₀ and δ₁ such that

|f(z)−w₀|< whenever 0<|z−z₀| < δ₀ and |f(z)−w₁|< whenever 0 <|z−z₀|< δ₁.So, if 0 <|z−z₀|< δ= min{δ₀, δ₁}, then|w₀−w₁|=

| −(f(z)−w₀) + (f(z)−w₁)| ≤ |f(z)−w₀|+|f(z)−w₁| < 2; i.e.,

|w₀−w₁|<2.But,can be chosen arbitrarily small. Hence,w₀−w₁= 0, or w₀=w₁.

5.4. (a).|z²+ 5−(z₀²+ 5)|=|z−z₀||z+z₀| ≤ |z−z₀|(|z−z₀|+ 2|z₀|)

≤(1 + 2|z₀|)|z−z₀| if |z−z₀|<1

< if 0<|z−z0|<min

1+2|z₀|,1

,

(b).|z²−(1 +i)²|=|z²−(1−i)²| ≤ |z−(1−i)||z+ (1−i)|

≤ |z−(1−i)|(|z−(1−i)|+2|1−i|)<5|z−(1−i)| if |z−(1−i)|<1

Complex Functions 35

< if |z−(1−i)|<min{1, /5}, (c).|z−z₀|=|z−z₀|< if|z−z₀|< ,

(d).|2z+1−(5−2i)|=|2z−(4−2i)|= 2|z−(2−i)|< if|z−(2−i)|< /2.

5.5. (a).−8i,(b).−7i/2,(c). 6i,(d).−1/2,(e). 1,(f). ∞. 5.6. (a). limz→0,z=x(z/z)²= 1, limz→0,y=x(z/z)²=−1.

(b). Let > 0. Choose δ=. Then, 0<|z−0|< δ implies |z²/z−0|=

|z|< .

5.7. Since lim_z→z0f(z) = 0,given any >0,there existsδ >0 such that

|f(z)−0|< /Mwhenever|z−z₀|< δ.Thus,|f(z)g(z)−0|=|f(z)||g(z)| ≤ M|f(z)|< if|z−z₀|< δ.

5.8. Use the fact||f(z)| − |w0|| ≤ |f(z)−w0|.

5.9. Let >0.Sinceg is continuous atw0,there exists aδ1>0 such that

|w−w0|< δ1 implies that |g(w)−g(w0)|< .Now, f is continuous atz0, so there exists aδ2>0 such that|z−z0|< δ2 implies|f(z)−f(z0)|< δ1. Combining these, we find that |z−z0| < δ2 implies |f(z)−f(z0)| < δ1, which in turn implies|(g◦f)(z)−(g◦f)(z0)|=|g[f(z)]−g[f(z0)]|< . 5.10.Continuous at 1,discontinuous at−1, i,−i.

5.11. f is not continuous at z0 if there exists 0 > 0 with the following property: For every δ > 0, there exists zδ such that |zδ −z0| < δ and

|f(zδ)−f(z0)| ≥0. Now letz0=x0<0. Take0= 3π/2.For eachδ >0, let zδ = x₀−i(δ/2). Then, |zδ −z₀| = |iδ/2| =δ/2 < δ, −π < f(zδ) <

−π/2, f(z0) =π,so |f(zδ)−f(z0)|>3π/2 =0, andf is not continuous atz₀.Thus,f is not continuous at every point on the negative real axis. It is also not continuous atz= 0 because it is not defined there.

5.12.If f is continuous at z0, then given > 0 there exists a δ >0 such that |z1−z0| < δ/2 ⇒ |f(z1)−f(z0)| < /2 and |z2−z0| < δ/2 ⇒

|f(z2)−f(z0)| < /2. But then |z1−z2| ≤ |z1−z0|+|z0−z2| < δ ⇒

|f(z1)−f(z2)| ≤ |f(z1)−f(z0)|+|f(z2)−f(z0)|< .For the converse, we assume that 0<|z−z0|< δ, 0<|z−z0|< δ; otherwise, we can takez =z0

and then there is nothing to prove. Letzn →z0, zn =z0,and >0.There is aδ >0 such that 0<|z−z0|< δ, |z−z0|< δimplies|f(z)−f(z)|< , and there is an N such that n ≥N implies 0 <|zn−z0| < δ. Then, for m, n≥N,we have|f(zm)−f(zn)|< .So, w0= limn→∞f(zn) exists. To see that limz→z₀f(z) =w0,take aδ1>0 such that 0<|z−z0|< δ1, 0<

|z−z0| < δ1 implies |f(z)−f(z)| < /2,and an N1 such that n ≥N1

implies 0<|zn−z0|< δ1 and|f(zn)−w0|< /2.Then, 0<|z−z0|< δ1

implies|f(z)−w0| ≤ |f(z)−f(zN₁)|+|f(zN₁)−w0|< .

5.13.(i). Suppose thatf :U →C is continuous andU is compact. Con- sider a covering off(U) to be open sets V.The inverse images f⁻¹(V) are open and form a covering of U. Since U is compact, by Theorem 4.4 we can select a finite subcovering such that U ⊂ f⁻¹(V₁)∪ · · · ∪f⁻¹(V_n). It follows that f(U) ⊂V₁∪ · · · ∪V_n, which in view of Theorem 4.4 implies that f(U) is compact. (ii). Suppose that f :U →C is continuous andU is connected. Iff(U) =A∪B where AandB are open and disjoint, then U =f⁻¹(A)∪f⁻¹(B), which is a union of disjoint and open sets. SinceU

is connected, either f⁻¹(A) =∅orf⁻¹(B) =∅,and hence eitherA=∅ or B=∅.This implies thatf(U) is connected.

5.14.Use Theorem 4.4.

Lecture 6