كومبلكس

608

C H A P T E R

1 3

Complex Numbers 
and Functions. Complex
Differentiation

The transition from “real calculus” to “complex calculus” starts with a discussion of
complex numbers and their geometric representation in the complex plane. We then
progress to analytic functions in Sec. 13.3. We desire functions to be analytic because
these are the “useful functions” in the sense that they are differentiable in some domain
and operations of complex analysis can be applied to them. The most important equations
are therefore the Cauchy–Riemann equations in Sec. 13.4 because they allow a test of
analyticity of such functions. Moreover, we show how the Cauchy–Riemann equations
are related to the important Laplace equation.

The remaining sections of the chapter are devoted to elementary complex functions

(exponential, trigonometric, hyperbolic, and logarithmic functions). These generalize the
familiar real functions of calculus. Detailed knowledge of them is an absolute necessity
in practical work, just as that of their real counterparts is in calculus.

Prerequisite: Elementary calculus.
References and Answers to Problems: App. 1 Part D, App. 2.

13.1

Complex Numbers and 
Their Geometric Representation

The material in this section will most likely be familiar to the student and serve as a
review.

Equations without real solutions, such as 

or 

were

observed early in history and led to the introduction of complex numbers.

1 

By definition,

a complex number z is an ordered pair (x, y) of real numbers x and y, written

z

⫽ (x, y).

x

2

⫺ 10x ⫹ 40 ⫽ 0,

x

2

⫽ ⫺1

1

First to use complex numbers for this purpose was the Italian mathematician GIROLAMO CARDANO

(1501–1576), who found the formula for solving cubic equations. The term “complex number” was introduced
by CARL FRIEDRICH GAUSS (see the footnote in Sec. 5.4), who also paved the way for a general use of
complex numbers.

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x is called the real part and y the imaginary part of z, written

By definition, two complex numbers are equal if and only if their real parts are equal

and their imaginary parts are equal.

(0, 1) is called the imaginary unit and is denoted by i,

(1)

Addition, Multiplication. Notation 

Addition of two complex numbers 

and 

is defined by

(2)

Multiplication is defined by

(3)

These two definitions imply that

and

as for real numbers 

Hence the complex numbers “extend ” the real numbers. We

can thus write

because by (1), and the definition of multiplication, we have

Together we have, by addition, 

In practice, complex numbers 

are written

(4)

or e.g., 

(instead 

of 

i4).

Electrical engineers often write j instead of i because they need i for the current.
If 

then 

and is called pure imaginary. Also, (1) and (3) give

(5)

because, by the definition of multiplication, i

2

⫽ ii ⫽ (0, 1)(0, 1) ⫽ (⫺1, 0) ⫽ ⫺1.

i

2

⫽ ⫺1

z

⫽ iy

x

⫽ 0,

17

⫹ 4i

z

⫽ x ⫹ yi,

z

⫽ x ⫹ iy

z

ⴝ (x, y)

(x, y)

⫽ (x, 0) ⫹ (0, y) ⫽ x ⫹ iy.

iy

⫽ (0, 1)y ⫽ (0, 1)(

 

y, 0)

⫽ (0

#

y

⫺ 1

#

0,

 

0

#

0

⫹ 1

#

y)

⫽ (0, y).

(x, 0)

⫽ x.

     

Similarly,

     

(0, y)

⫽ iy

x

1

, x

2

.

(x

1

, 0)(x

2

, 0)

⫽ (x

1

x

2

, 0)

(x

1

, 0)

⫹ (x

2

, 0)

⫽ (x

1

⫹ x

2

, 0)

z

1

z

2

⫽ (x

1

, y

1

)(x

2

, y

2

)

⫽ (x

1

x

2

⫺ y

1

y

2

,

 

x

1

y

2

⫹ x

2

y

1

).

z

1

⫹ z

2

⫽ (x

1

, y

1

)

⫹ (x

2

, y

2

)

⫽ (x

1

⫹ x

2

,

 

y

1

⫹ y

2

).

z

2

⫽ (x

2

, y

2

)

z

1

⫽ (x

1

, y

1

)

z

⫽ x ⫹ iy

i

⫽ (0, 1).

x

⫽ Re z,

     

y

⫽ Im z.

SEC. 13.1

Complex Numbers and Their Geometric Representation

609

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For addition the standard notation (4) gives [see (2)]

For multiplication the standard notation gives the following very simple recipe. Multiply
each term by each other term and use 

when it occurs [see (3)]:

This agrees with (3). And it shows that 

is a more practical notation for complex

numbers than (x, y).

If you know vectors, you see that (2) is vector addition, whereas the multiplication (3)

has no counterpart in the usual vector algebra.

E X A M P L E   1

Real Part, Imaginary Part, Sum and Product of Complex Numbers

Let and . 

Then 

and

Subtraction, Division

Subtraction and division are defined as the inverse operations of addition and multipli-
cation, respectively. Thus the difference

is the complex number z for which

Hence by (2),

(6)

The  quotient

is the complex number z for which 

If we

equate the real and the imaginary parts on both sides of this equation, setting 
we obtain 

The solution is

The practical rule used to get this is by multiplying numerator and denominator of 
by and 

simplifying:

(7)

E X A M P L E   2

Difference and Quotient of Complex Numbers

For and we 

get 

and

Check the division by multiplication to get 

䊏

8

⫹ 3i.

z

1

z

2

⫽

8

⫹ 3i

9

⫺ 2i

⫽

(8

⫹ 3i)(9 ⫹ 2i)

(9

⫺ 2i)(9 ⫹ 2i)

⫽

66

⫹ 43i

81

⫹ 4

⫽

66

85

⫹

43

85

 i.

z

1

⫺ z

2

⫽ (8 ⫹ 3i) ⫺ (9 ⫺ 2i) ⫽ ⫺1 ⫹ 5i

z

2

⫽ 9 ⫺ 2i

z

1

⫽ 8 ⫹ 3i

z

⫽

x

1

⫹ iy

1

x

2

⫹ iy

2

⫽

(x

1

⫹ iy

1

)(x

2

⫺ iy

2

)

(x

2

⫹ iy

2

)(x

2

⫺ iy

2

)

⫽

x

1

x

2

⫹ y

1

 

y

2

x

2

2

⫹ y

2

2

⫹ i 

x

2

 

y

1

⫺ x

1

 

y

2

x

2

2

⫹ y

2

2

 

.

x

2

⫺ iy

2

z

1

>z

2

z

⫽

z

1

z

2

⫽ x ⫹ iy,

   

x

⫽

x

1

x

2

⫹ y

1

 

y

2

x

2

2

⫹ y

2

2

 

,

   

y

⫽

x

2

 

y

1

⫺ x

1

 

y

2

x

2

2

⫹ y

2

2

 

.

(7*)

x

1

⫽ x

2

x

⫺ y

2

y, y

1

⫽ y

2

x

⫹ x

2

y.

z

⫽ x ⫹ iy,

z

1

⫽ zz

2

.

z

⫽ z

1

>z

2

 (z

2

⫽ 0)

z

1

⫺ z

2

⫽ (x

1

⫺ x

2

)

⫹ i

 

(

 

y

1

⫺ y

2

).

z

1

⫽ z ⫹ z

2

.

z

⫽ z

1

⫺ z

2

䊏

z

1

z

2

⫽ (8 ⫹ 3i)(9 ⫺ 2i) ⫽ 72 ⫹ 6 ⫹ i

 

(

⫺16 ⫹ 27) ⫽ 78 ⫹ 11i.

z

1

⫹ z

2

⫽ (8 ⫹ 3i) ⫹ (9 ⫺ 2i) ⫽ 17 ⫹ i,

Re z

1

⫽ 8, Im z

1

⫽ 3, Re z

2

⫽ 9, Im z

2

⫽ ⫺2

z

2

⫽ 9 ⫺ 2i

z

1

⫽ 8 ⫹ 3i

x

⫹ iy

 

⫽ (x

1

x

2

⫺ y

1

 

y

2

)

⫹ i(x

1

 

y

2

⫹ x

2

 

y

1

).

 (x

1

⫹ iy

1

)(x

2

⫹ iy

2

)

⫽ x

1

x

2

⫹ ix

1

 

y

2

⫹ iy

1

x

2

⫹ i

2

y

1

 

y

2

i

2

⫽ ⫺1

(x

1

⫹ iy

1

)

⫹ (x

2

⫹ iy

2

)

⫽ (x

1

⫹ x

2

)

⫹ i(

 

y

1

⫹ y

2

).

610

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

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Complex numbers satisfy the same commutative, associative, and distributive laws as real
numbers (see the problem set).

Complex Plane

So far we discussed the algebraic manipulation of complex numbers. Consider the
geometric representation of complex numbers, which is of great practical importance. We
choose two perpendicular coordinate axes, the horizontal x-axis, called the real axis, and
the vertical y-axis, called the imaginary axis. On both axes we choose the same unit of
length (Fig. 318). This is called a Cartesian coordinate system.

SEC. 13.1

Complex Numbers and Their Geometric Representation

611

y

x

1

1

P

z = x + i y

(Imaginary

axis)

(Real

axis)

Fig. 318.

The complex plane

Fig. 319.

The number 4 

⫺ 3i in

the complex plane

y

x

1

5

–1

–

3

4 

–

 3

i

We now plot a given complex number 

as the point P with coordinates

x, y. The xy-plane in which the complex numbers are represented in this way is called the
complex plane.

2

Figure 319 shows an example.

Instead of saying “the point represented by z in the complex plane” we say briefly and

simply “the point z in the complex plane.” This will cause no misunderstanding.

Addition and subtraction can now be visualized as illustrated in Figs. 320 and 321.

z

⫽ (x, y) ⫽ x ⫹ iy

y

x

z

2

z

1

z

1

 

+ z

2

y

x

z

1

– 

z

2

z

1

z

2

– 

z

2

Fig. 320.

Addition of complex numbers

Fig. 321.

Subtraction of complex numbers

2

Sometimes called the Argand diagram, after the French mathematician JEAN ROBERT ARGAND

(1768–1822), born in Geneva and later librarian in Paris. His paper on the complex plane appeared in 1806,
nine years after a similar memoir by the Norwegian mathematician CASPAR WESSEL (1745–1818), a surveyor
of the Danish Academy of Science. 

c13.qxd  10/30/10  2:14 PM  Page 611

Fig. 322.

Complex conjugate numbers

y

x

5

2

–2

z = x + iy = 5 + 2i

z = x 

–

 iy = 5 

–

 2

i

Complex Conjugate Numbers

The complex conjugate

of a complex number 

is defined by

It is obtained geometrically by reflecting the point z in the real axis. Figure 322 shows
this for 

and its conjugate z

⫽ 5 ⫺ 2i.

z

⫽ 5 ⫹ 2i

z

⫽ x ⫺ iy.

z

⫽ x ⫹ iy

z

612

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

The complex conjugate is important because it permits us to switch from complex

to real. Indeed, by multiplication, 

(verify!). By addition and subtraction,

We thus obtain for the real part x and the imaginary part y

(not iy!) of 

the important formulas

(8)

If  z is real, 

then 

by the definition of 

and conversely. Working with

conjugates is easy, since we have

(9)

E X A M P L E   3

Illustration of (8) and (9)

Let 

and 

Then by (8),

Also, the multiplication formula in (9) is verified by

䊏

 z

1

z

2

⫽ (4 ⫺ 3i)(2 ⫺ 5i) ⫽ ⫺7 ⫺ 26i.

 (z

1

z

2

)

⫽ (4 ⫹ 3i)(2 ⫹ 5i) ⫽ (⫺7 ⫹ 26i) ⫽ ⫺7 ⫺ 26i,

Im z

1

⫽

1

2i

 

[(4

⫹ 3i) ⫺ (4 ⫺ 3i)] ⫽

3i

⫹ 3i
2i

⫽ 3.

z

2

⫽ 2 ⫹ 5i.

z

1

⫽ 4 ⫹ 3i

 (z

1

z

2

)

⫽ z

1

z

2

,

       

a

z

1

z

2

b ⫽

z

1

z

2

  

.

 (z

1

⫹ z

2

)

⫽ z

1

⫹ z

2

,

   

(z

1

⫺ z

2

)

⫽ z

1

⫺ z

2

,

z,

z

⫽ z

z

⫽ x,

Re z

⫽ x ⫽

1
2 

(z

⫹ z),

   

Im z

⫽ y ⫽

1

2i

 

(z

⫺ z).

z

⫽ x ⫹ iy

z

⫺ z ⫽ 2iy.

z

⫹ z ⫽ 2x,

zz

⫽ x

2

⫹ y

2

1. Powers of i. Show that 

and 

2. Rotation.  Multiplication by i is geometrically a

counterclockwise rotation through 

. Verify

p

>2 (90°)

1

>i ⫽ ⫺i, 1>i

2

⫽ ⫺1, 1

>i

3

⫽ i, Á .

i

5

⫽ i, Á

i

2

⫽ ⫺1, i

3

⫽ ⫺i, i

4

⫽ 1,

this by graphing z and iz and the angle of rotation for

3. Division. Verify the calculation in (7). Apply (7) to

(26

⫺ 18i)

>(6 ⫺ 2i).

z

⫽ 1 ⫹ i, z ⫽ ⫺1 ⫹ 2i, z ⫽ 4 ⫺ 3i.

P R O B L E M   S E T   1 3 . 1

c13.qxd  10/30/10  2:14 PM  Page 612

13.2

Polar Form of Complex Numbers. 
Powers and Roots

We gain further insight into the arithmetic operations of complex numbers if, in addition
to the xy-coordinates in the complex plane, we also employ the usual polar coordinates
r,

defined by

(1)

We see that then 

takes the so-called polar form

(2)

r is called the absolute value or modulus of z and is denoted by 

Hence

(3)

Geometrically, 

is the distance of the point z from the origin (Fig. 323). Similarly,

is the distance between 

and 

(Fig. 324).

is called the argument of z and is denoted by arg z. Thus 

and (Fig. 323)

(4)

Geometrically,  is the directed angle from the positive x-axis to OP in Fig. 323. Here, as
in calculus, all angles are measured in radians and positive in the counterclockwise sense.

u

(z

⫽ 0).

tan u

⫽

y
x

 

u

⫽ arg z

u

z

2

z

1

ƒ

z

1

⫺ z

2

ƒ

ƒ

z ƒ

ƒ

z ƒ

⫽ r ⫽ 2x

2

⫹ y

2

⫽ 1zz.

ƒ

z ƒ .

z

⫽ r(cos u ⫹ i sin u).

z

⫽ x ⫹ iy

x

⫽ r cos u,

     

y

⫽ r sin u.

u

SEC. 13.2

Polar Form of Complex Numbers. Powers and Roots

613

4. Law for conjugates. Verify (9) for 

5. Pure imaginary number. Show that 

is

pure imaginary if and only if 

6. Multiplication. If the product of two complex numbers

is zero, show that at least one factor must be zero.

7. Laws of addition and multiplication. Derive the

following laws for complex numbers from the cor-
responding laws for real numbers.

(Commutative laws)

(Associative laws)

(Distributive law)

z

⫹ (⫺z) ⫽ (⫺z) ⫹ z ⫽ 0,

       

z

#

1

⫽ z.

0

⫹ z ⫽ z ⫹ 0 ⫽ z,

z

1

(z

2

⫹ z

3

)

⫽ z

1

z

2

⫹ z

1

z

3

(z

1

z

2

)z

3

⫽ z

1

(z

2

z

3

)

(z

1

⫹ z

2

)

⫹ z

3

⫽ z

1

⫹ (z

2

⫹ z

3

),

z

1

⫹ z

2

⫽ z

2

⫹ z

1

, z

1

z

2

⫽ z

2

z

1

z

⫽ ⫺z.

z

⫽ x ⫹ iy

z

2

⫽ ⫺1 ⫹ 4i.

z

1

⫽ ⫺11 ⫹ 10i,

8–15

COMPLEX ARITHMETIC

Let 

Showing the details of

your work, find, in the form 

8.

9.

10.

11.

12.

13.

14.

15.

16–20

Let 

Showing details, find, in terms

of x and y:

16.

17.

18.

19.

20. Im (1

>z

2

)

Re (z

>z),

 

Im (z

>z)

Re [(1

⫹ i)

16

z

2

]

Re z

4

⫺ (Re z

2

)

2

Im (1

>z),

 

Im (1

>z

2

)

z

⫽ x ⫹ iy.

4 (z

1

⫹ z

2

)

>(z

1

⫺ z

2

)

z

1

>z

2

,

 

(z

1

>z

2

)

(z

1

⫹ z

2

)(z

1

⫺ z

2

),

 

z

1

2

⫺ z

2

2

z

1

>z

2

,

 

z

2

>z

1

(z

1

⫺ z

2

)

2

>16,

 

(z

1

>4 ⫺ z

2

>4)

2

Re (1

>z

2

2

),

 

1

>Re (z

2

2

)

Re (z

1

2

),

 

(Re z

1

)

2

z

1

z

2

,

 

(z

1

z

2

)

x

⫹ iy:

z

1

⫽ ⫺2 ⫹ 11i, z

2

⫽ 2 ⫺ i.

c13.qxd  10/30/10  2:14 PM  Page 613

For 

this angle  is undefined. (Why?) For a given 

it is determined only up

to integer multiples of 

since cosine and sine are periodic with period 

. But one

often wants to specify a unique value of arg z of a given 

. For this reason one defines

the principal value Arg z (with capital A!) of arg z by the double inequality

(5)

Then we have Arg 

for positive real 

which is practical, and Arg 

(not

) for negative real z, e.g., for 

The principal value (5) will be important in

connection with roots, the complex logarithm (Sec. 13.7), and certain integrals. Obviously,
for a given 

the other values of 

arg z

⫽ Arg z ⫾ 2n

p

 (n

⫽ ⫾1, ⫾2, Á ).

arg z are

z

⫽ 0,

z

⫽ ⫺4.

⫺

p

!

z

⫽

p

z

⫽ x,

z

⫽ 0

⫺

p

⬍ Arg z ⬉

p

.

z

⫽ 0

2

p

2

p

z

⫽ 0

u

z

⫽ 0

614

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

E X A M P L E   1

Polar Form of Complex Numbers. Principal Value Arg z

(Fig. 325) has the polar form 

. Hence we obtain

and

(the principal value).

Similarly, and 

CAUTION!

In using (4), we must pay attention to the quadrant in which z lies, since

has period  , so that the arguments of z and 

have the same tangent. Example:

for and  we 

have 

tan 

u

1

⫽ tan u

2

⫽ 1.

u

2

⫽ arg (⫺1 ⫺ i)

u

1

⫽ arg (1 ⫹ i)

⫺z

p

tan u

䊏

Arg z

⫽

1
3

p

.

z

⫽ 3 ⫹ 323i ⫽ 6 (cos 

1
3

p

⫹ i sin 

1
3

p

),  ƒ z ƒ

⫽ 6,

Arg z

⫽

1
4

p

ƒ

z ƒ

⫽ 22,

 

arg z

⫽

1
4

p

 

⫾

 

2n

p

 (n

⫽ 0, 1, Á ),

z

⫽ 22 (cos 

1
4

p

⫹ i sin 

1
4

p

)

z

⫽ 1 ⫹ i

Fig. 323.

Complex plane, polar form 

Fig. 324.

Distance between two 

of a complex number

points in the complex plane 

y

x

z

2

z

1

|

z

1 

– z

2

|

|

z

1

|

|

z

2

|

y

x

O

P

θ

|z

| = 

r

Imaginary

axis

Real
axis

z = x + iy

Triangle Inequality

Inequalities such as 

make sense for real numbers, but not in complex because there

is no natural way of ordering complex numbers. However, inequalities between absolute values
(which are real!), such as 

(meaning that 

is closer to the origin than  ) are of

great importance. The daily bread of the complex analyst is the triangle inequality

(6)

(Fig. 326)

which we shall use quite frequently. This inequality follows by noting that the three
points 0, 

and 

are the vertices of a triangle (Fig. 326) with sides 

and

and one side cannot exceed the sum of the other two sides. A formal proof is

left to the reader (Prob. 33). (The triangle degenerates if 

and 

lie on the same straight

line through the origin.)

z

2

z

1

ƒ

z

1

⫹ z

2

ƒ

,

ƒ

z

1

ƒ

,  ƒ z

2

ƒ

,

z

1

⫹ z

2

z

1

,

ƒ

z

1

⫹ z

2

ƒ

⬉ ƒ z

1

ƒ

⫹ ƒ z

2

ƒ

z

2

z

1

ƒ

z

1

ƒ

⬍ ƒ z

2

ƒ

x

1

⬍ x

2

y

x

1

1

1 

+ i

/4

π

2

Fig. 325.

Example 1

c13.qxd  10/30/10  2:14 PM  Page 614

By induction we obtain from (6) the generalized triangle inequality

(6*)

that is, the absolute value of a sum cannot exceed the sum of the absolute values of the terms.

E X A M P L E   2

Triangle Inequality

If 

and 

then (sketch a figure!)

Multiplication and Division in Polar Form

This will give us a “geometrical” understanding of multiplication and division. Let

Multiplication.

By (3) in Sec. 13.1 the product is at first

The addition rules for the sine and cosine [(6) in App. A3.1] now yield

(7)

Taking absolute values on both sides of (7), we see that the absolute value of a product
equals the product of the absolute values of the factors,

(8)

Taking arguments in (7) shows that the argument of a product equals the sum of the
arguments of the factors,

(9)

(up to multiples of 

).

Division.

We have 

Hence 

and by

division by 

(10)

(z

2

⫽ 0).

`

z

1

z

2

` ⫽

ƒ

z

1

ƒ

ƒ

z

2

ƒ

ƒ

z

2

ƒ

ƒ

z

1

ƒ

⫽ ƒ (z

1

>z

2

)

 

z

2

ƒ

⫽ ƒ z

1

>z

2

ƒ ƒ

z

2

ƒ

z

1

⫽ (z

1

>z

2

)z

2

.

2

p

arg (z

1

z

2

)

⫽ arg z

1

⫹ arg z

2

ƒ

z

1

z

2

ƒ

⫽ ƒ z

1

ƒ ƒ

z

2

ƒ

.

z

1

z

2

⫽ r

1

r

2

 

[cos

 

(u

1

⫹ u

2

)

⫹ i sin

 

(u

1

⫹ u

2

)].

z

1

z

2

⫽ r

1

r

2

[(cos u

1 

cos u

2

⫺ sin u

1 

sin u

2

)

⫹ i(sin u

1 

cos u

2

⫹ cos u

1 

sin u

2

)].

z

1

⫽ r

1

(cos u

1

⫹ i sin u

1

)

     

and

     

z

2

⫽ r

2

(cos u

2

⫹ i sin u

2

).

䊏

ƒ

z

1

⫹ z

2

ƒ

⫽ ƒ ⫺1 ⫹ 4i ƒ ⫽ 117 ⫽ 4.123 ⬍ 12 ⫹ 113 ⫽ 5.020.

z

2

⫽ ⫺2 ⫹ 3i,

z

1

⫽ 1 ⫹ i

ƒ

z

1

⫹ z

2 

⫹ Á ⫹ z

n

ƒ

⬉ ƒ z

1

ƒ

⫹ ƒ z

2

ƒ

 

⫹ Á ⫹

 

ƒ

z

n

ƒ

;

SEC. 13.2

Polar Form of Complex Numbers. Powers and Roots

615

y

x

z

2

z

1

z

1 

+ z

2

Fig. 326.

Triangle inequality

c13.qxd  10/30/10  2:14 PM  Page 615

Similarly, 

and by subtraction of arg 

(11)

(up to multiples of 

).

Combining (10) and (11) we also have the analog of (7),

(12)

To comprehend this formula, note that it is the polar form of a complex number of absolute
value 

and argument 

But these are the absolute value and argument of 

as we can see from (10), (11), and the polar forms of 

and 

E X A M P L E   3

Illustration of Formulas (8)–(11)

Let 

and 

Then 

. Hence (make a sketch)

and for the arguments we obtain 

.

E X A M P L E   4

Integer Powers of z. De Moivre’s Formula

From (8) and (9) with 

we obtain by induction for 

(13)

Similarly, (12) with 

and 

gives (13) for 

For 

formula (13) becomes

De Moivre’s formula

3

(13*)

We can use this to express 

and 

in terms of powers of 

and 

. For instance, for 

we

have on the left 

Taking the real and imaginary parts on both sides of 

with 

gives the familiar formulas

This shows that complex methods often simplify the derivation of real formulas. Try 

.

Roots

If 

then to each value of w there corresponds one value of z. We

shall immediately see that, conversely, to a given 

there correspond precisely n

distinct values of w. Each of these values is called an nth root of z, and we write

z

⫽ 0

z

⫽ w

n

 (n

⫽ 1, 2, Á ),

䊏

n

⫽ 3

cos 2u

⫽ cos

2

 u

⫺ sin

2

 u,

   

sin 2u

⫽ 2 cos u sin u.

n

⫽ 2

(13*)

cos

2

 u

⫹ 2i cos u sin u ⫺ sin

2

 u.

n

⫽ 2

sin u

cos u

sin nu

cos nu

(cos u

⫹ i sin u)

n

⫽ cos nu ⫹ i sin nu.

ƒ

z ƒ

⫽ r ⫽ 1,

n

⫽ ⫺1, ⫺2, Á .

z

2

⫽ z

n

z

1

⫽ 1

z

n

⫽ r

n

 (cos nu

⫹ i sin nu).

n

⫽ 0, 1, 2, Á

z

1

⫽ z

2

⫽ z

䊏

Arg (z

1

z

2

)

⫽ ⫺

 

3

p

4

⫽ Arg z

1

⫹ Arg z

2

⫺ 2

p

,

   

Arg 

a

z

1

z

2

b ⫽

p

4

⫽ Arg z

1

⫺ Arg z

2

Arg z

1

⫽ 3

p

>4, Arg z

2

⫽

p

>2,

ƒ

z

1

z

2

ƒ

⫽ 612 ⫽ 318 ⫽ ƒ z

1

ƒ ƒ

z

2

ƒ

,

   

ƒ

z

1

>z

2

ƒ

⫽ 212

>3 ⫽ ƒ z

1

ƒ

> ƒ z

2

ƒ

,

z

1

z

2

⫽ ⫺6 ⫺ 6i, z

1

>z

2

⫽

2
3

⫹ (

2
3

)i

z

2

⫽ 3i.

z

1

⫽ ⫺2 ⫹ 2i

z

2

.

z

1

z

1

>z

2

,

u

1

⫺ u

2

.

r

1

>r

2

z

1

z

2

⫽

r

1

r

2

 [cos (u

1

⫺ u

2

)

⫹ i sin (u

1

⫺ u

2

)].

2

p

arg 

z

1

z

2

⫽ arg z

1

⫺ arg z

2

z

2

arg z

1

⫽ arg [(z

1

>z

2

)z

2

]

⫽ arg (z

1

>z

2

)

⫹ arg z

2

616

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

3

ABRAHAM DE MOIVRE (1667–1754), French mathematician, who pioneered the use of complex numbers

in trigonometry and also contributed to probability theory (see Sec. 24.8).

c13.qxd  10/30/10  2:14 PM  Page 616

(14)

Hence this symbol is multivalued, namely, n-valued. The n values of 

can be obtained

as follows. We write z and w in polar form

Then the equation 

becomes, by De Moivre’s formula (with 

instead of  ),

The absolute values on both sides must be equal; thus, 

so that 

where

is positive real (an absolute value must be nonnegative!) and thus uniquely determined.

Equating the arguments 

and  and recalling that  is determined only up to integer

multiples of 

, we obtain

where k is an integer. For 

we get n distinct values of w. Further integers

of  k would give values already obtained. For instance, 

gives 

, hence

the w corresponding to 

, etc. Consequently, 

for 

, has the n distinct values

(15)

where These 

n values lie on a circle of radius 

with center at the

origin and constitute the vertices of a regular polygon of n sides. The value of 

obtained

by taking the principal value of arg z and 

in (15) is called the principal value of

.

Taking 

in (15), we have 

and Arg 

Then (15) gives

(16)

These n values are called the nth roots of unity. They lie on the circle of radius 1 and
center 0, briefly called the unit circle (and used quite frequently!). Figures 327–329 show

2

3

1

⫽ 1, ⫺

1
2

⫾

1
2

23i, 2

4

1

⫽ ⫾1, ⫾i, and2

5

1.

k

⫽ 0, 1, Á , n ⫺ 1.

2

n

1

⫽ cos 

2k

p

n

⫹ i sin 

2k

p

n

,

z

⫽ 0.

ƒ

z ƒ

⫽ r ⫽ 1

z

⫽ 1

w

⫽ 1

n

z

k

⫽ 0

1

n

z

1

n

r

k

⫽ 0, 1, Á , n ⫺ 1.

1

n

z

⫽ 1

n

r 

acos 

u

⫹ 2k

p

n

⫹ i sin 

u

⫹ 2k

p

n

b

z

⫽ 0

1

n

z ,

k

⫽ 0

2k

p

>n ⫽ 2

p

k

⫽ n

k

⫽ 0, 1, Á , n ⫺ 1

n

␾ ⫽ u ⫹ 2k

p

,

     

thus

     

␾ ⫽

u
n ⫹

2k

p

n

2

p

u

u

n

␾

1

n

r

R

⫽ 1

n

r ,

R

n

⫽ r,

w

n

⫽ R

n

(cos n

␾ ⫹ i sin n␾) ⫽ z ⫽ r(cos u ⫹ i sin u).

u

␾

w

n

⫽ z

z

⫽ r(cos u ⫹ i sin u)

     

and

     

w

⫽ R(cos ␾ ⫹ i sin ␾).

1

n

z

w

⫽ 1

n

z

 

.  

SEC. 13.2

Polar Form of Complex Numbers. Powers and Roots

617

y

x

1

ω

2

ω

y

x

1

ω

2

ω

3

ω

1

y

x

ω

2

ω

4

ω

3

ω

Fig. 327.

Fig. 328.

Fig. 329.

2

5

1

2

4

1

2

3

1

c13.qxd  10/30/10  2:14 PM  Page 617

If 

denotes the value corresponding to 

in (16), then the n values of 

can be

written as

More generally, if 

is any nth root of an arbitrary complex number 

then the n

values of 

in (15) are

(17)

because multiplying 

by 

corresponds to increasing the argument of 

by 

.

Formula (17) motivates the introduction of roots of unity and shows their usefulness.

2k

p

>n

w

1

v

k

w

1

w

1

,

   

w

1

v,

   

w

1

v

2

,

   

Á

,

   

w

1

v

n

ⴚ1

1

n

z

z (

⫽ 0),

w

1

1, v, v

2

, Á , v

n

ⴚ1

.

2

n

1

k

⫽ 1

v

618

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

1–8

POLAR FORM

Represent in polar form and graph in the complex plane as
in Fig. 325. Do these problems very carefully because polar
forms will be needed frequently. Show the details.

1.

2.

3.

4.

5.

6.

7.

8.

9–14

PRINCIPAL ARGUMENT

Determine the principal value of the argument and graph it
as in Fig. 325.

9.

10.

11.

12.

13.

14.

15–18

CONVERSION TO 

Graph in the complex plane and represent in the form 

15.

16.

17.

18.

ROOTS

19. CAS PROJECT. Roots of Unity and Their Graphs.

Write a program for calculating these roots and for
graphing them as points on the unit circle. Apply the
program to 

with 

Then extend

the program to one for arbitrary roots, using an idea
near the end of the text, and apply the program to
examples of your choice.

n

⫽ 2, 3, Á , 10.

z

n

⫽ 1

250 (cos 

3
4

p

⫹ i sin 

3
4

p

)

28 (cos 

1
4

p

⫹ i sin 

1
4

p

)

6 (cos 

1
3

p

⫹ i sin 

1
3

p

)

3 (cos 

1
2

p

⫺ i sin 

1
2

p

)

x

⫹ iy:

x

ⴙ iy

⫺1 ⫹ 0.1i,

 

⫺1 ⫺ 0.1i

(1

⫹ i)

20

⫺

p

⫺

p

i

3 

⫾ 4i

⫺5,

 

⫺5 ⫺ i,

 

⫺5 ⫹ i

⫺1 ⫹ i

⫺4 ⫹ 19i

2

⫹ 5i

1

⫹

1
2

p

i

23 ⫺ 10i

⫺

1
2

23 ⫹ 5i

22 ⫹ i

>3

⫺28 ⫺ 2i

>3

⫺5

2i,

 

⫺2i

⫺4 ⫹ 4i

1

⫹ i

20. TEAM PROJECT. Square Root. (a) Show that

has the values

(18)

(b) Obtain from (18) the often more practical formula

( 1 9 )

where 

sign if sign  if and

all square roots of positive numbers are taken with
positive sign. Hint: Use (10) in App. A3.1 with 

(c) Find the square roots of 

and

by both (18) and (19) and comment on the

work involved.

(d) Do some further examples of your own and apply
a method of checking your results.

21–27

ROOTS

Find and graph all roots in the complex plane.

21.

22.

23.

24.

25.

26.

兹

8

1

苶

27.

28–31

EQUATIONS

Solve and graph the solutions. Show details.

28.

29.

30.

Using the solutions, factor 

into quadratic factors with real coefficients.

31. z

4

⫺ 6iz

2

⫹ 16 ⫽ 0

z

4

⫹ 324

z

4

⫹ 324 ⫽ 0.

z

2

⫹ z ⫹ 1 ⫺ i ⫽ 0

z

2

⫺ (6 ⫺ 2i)

 

z

⫹ 17 ⫺ 6i ⫽ 0

2

5

⫺1

2

4

i

2

4

⫺4

2

3

216

2

3

3

⫹ 4i

2

3

1

⫹ i

1

⫹ 248i

⫺14i, ⫺9 ⫺ 40i,

x

⫽ u

>2.

y

⬍ 0,

y

⫽ ⫺1

y

⭌ 0,

y

⫽ 1

2z

⫽ ⫾[2

1
2

( ƒ z ƒ

⫹ x) ⫹ (sign y)i2

1
2

( ƒ z ƒ

⫹ x)]

 

⫽ ⫺w

1

.

 w

2

⫽ 1r  c cos a

u

2

⫹

p

b ⫹ i sin a

u

2

⫹

p

b d

 w

1

⫽ 1r  ccos  

u

2

⫹ i sin 

u

2 d

,

w

⫽ 1z

P R O B L E M   S E T   1 3 . 2

c13.qxd  10/30/10  2:14 PM  Page 618

13.3

Derivative. Analytic Function

Just as the study of calculus or real analysis required concepts such as domain,
neighborhood, function, limit, continuity, derivative, etc., so does the study of complex
analysis. Since the functions live in the complex plane, the concepts are slightly more
difficult or different from those in real analysis. This section can be seen as a reference
section where many of the concepts needed for the rest of Part D are introduced.

Circles and Disks. Half-Planes

The unit circle

(Fig. 330) has already occurred in Sec. 13.2. Figure 331 shows a

general circle of radius  and center a. Its equation is

ƒ

z

⫺ a ƒ ⫽ r

r

ƒ

z ƒ

⫽ 1

SEC. 13.3

Derivative. Analytic Function

619

32–35

INEQUALITIES AND EQUALITY

32. Triangle inequality. Verify (6) for 

33. Triangle inequality. Prove (6).

z

2

⫽ ⫺2 ⫹ 4i

z

1

⫽ 3 ⫹ i,

34. Re and Im. Prove 

35. Parallelogram equality. Prove and explain the name

ƒ

z

1

⫹ z

2

ƒ

2

⫹ ƒ z

1

⫺ z

2

ƒ

2

⫽ 2 ( ƒ z

1

ƒ

2

⫹ ƒ z

2

ƒ

2

).

ƒ

Re z ƒ

⬉ ƒ z ƒ ,

 

ƒ

Im z ƒ

⬉ ƒ z ƒ .

y

x

1

y

x

ρ

a

a

y

x

1

ρ

2

ρ

Fig. 330.

Unit circle

Fig. 331.

Circle in the 

Fig. 332.

Annulus in the 

complex plane

complex plane

because it is the set of all z whose distance 

from the center a equals Accordingly,

its interior (“open circular disk”) is given by 

its interior plus the circle

itself (“closed circular disk”) by 

and its exterior by 

As an

example, sketch this for 

and 

to make sure that you understand these

inequalities.

An open circular disk 

is also called a neighborhood of a or, more precisely,

a -neighborhood of a. And a has infinitely many of them, one for each value of 
and a is a point of each of them, by definition!

In modern literature any set containing a  -neighborhood of a is also called a neigh-

borhood of a.

Figure 332 shows an open annulus (circular ring) 

which we shall

need later. This is the set of all z whose distance 

from a is greater than 

but

less than 

. Similarly, the closed annulus

includes the two circles.

Half-Planes.

By the (open) upper half-plane we mean the set of all points 

such that 

. Similarly, the condition 

defines the lower half-plane, the

right half-plane, and 

the left half-plane.

x

⬍ 0

x

⬎ 0

y

⬍ 0

y

⬎ 0

z

⫽ x ⫹ iy

r

1

⬉ ƒ z ⫺ a ƒ ⬉ r

2

r

2

r

1

ƒ

z

⫺ a ƒ

r

1

⬍ ƒ z ⫺ a ƒ ⬍ r

2

,

r

r (

⬎ 0),

r

ƒ

z

⫺ a ƒ ⬍ r

r

⫽ 2,

a

⫽ 1 ⫹ i

ƒ

z

⫺ a ƒ ⬎ r.

ƒ

z

⫺ a ƒ ⬉ r,

ƒ

z

⫺ a ƒ ⬍ r,

r.

ƒ

z

⫺ a ƒ

c13.qxd  10/30/10  2:14 PM  Page 619

For Reference: Concepts on Sets 
in the Complex Plane

To our discussion of special sets let us add some general concepts related to sets that we
shall need throughout Chaps. 13–18; keep in mind that you can find them here.

By a point set in the complex plane we mean any sort of collection of finitely many

or infinitely many points. Examples are the solutions of a quadratic equation, the
points of a line, the points in the interior of a circle as well as the sets discussed just
before.

A set S is called open if every point of S has a neighborhood consisting entirely of

points that belong to S. For example, the points in the interior of a circle or a square form
an open set, and so do the points of the right half-plane Re 

A set S is called connected if any two of its points can be joined by a chain of finitely

many straight-line segments all of whose points belong to S. An open and connected set
is called a domain. Thus an open disk and an open annulus are domains. An open square
with a diagonal removed is not a domain since this set is not connected. (Why?)

The complement of a set S in the complex plane is the set of all points of the complex

plane that do not belong to S. A set S is called closed if its complement is open. For example,
the points on and inside the unit circle form a closed set (“closed unit disk”) since its
complement is 

open.

A boundary point of a set S is a point every neighborhood of which contains both points

that belong to S and points that do not belong to S. For example, the boundary points of
an annulus are the points on the two bounding circles. Clearly, if a set S is open, then no
boundary point belongs to S; if S is closed, then every boundary point belongs to S. The
set of all boundary points of a set S is called the boundary of S.

A region is a set consisting of a domain plus, perhaps, some or all of its boundary points.

WARNING!

“Domain” is the modern term for an open connected set. Nevertheless, some

authors still call a domain a “region” and others make no distinction between the two terms.

Complex Function

Complex analysis is concerned with complex functions that are differentiable in some
domain. Hence we should first say what we mean by a complex function and then define
the concepts of limit and derivative in complex. This discussion will be similar to that in
calculus. Nevertheless it needs great attention because it will show interesting basic
differences between real and complex calculus.

Recall from calculus that a real function f defined on a set S of real numbers (usually an

interval) is a rule that assigns to every x in S a real number f(x), called the value of f at x.
Now in complex, S is a set of complex numbers. And a function f defined on S is a rule
that assigns to every z in S a complex number w, called the value of f at z. We write

Here  z varies in S and is called a complex variable. The set S is called the domain of
definition of f or, briefly, the domain of f. (In most cases S will be open and connected,
thus a domain as defined just before.)

Example: 

is a complex function defined for all z; that is, its domain

S is the whole complex plane.

The set of all values of a function f is called the range of f.

w

⫽ f(z) ⫽ z

2

⫹ 3z

w

⫽ f(z).

| z ƒ

⬎ 1

z

⫽ x ⬎ 0.

620

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

c13.qxd  10/30/10  2:14 PM  Page 620

w is complex, and we write 

where u and  v are the real and imaginary

parts, respectively. Now w depends on 

Hence u becomes a real function of x

and y, and so does v. We may thus write

This shows that a complex function  f (z) is equivalent to a pair of  real functions 
and 

, each depending on the two real variables x and y.

E X A M P L E   1

Function of a Complex Variable

Let Find 

u and v and calculate the value of f at .

Solution.

and Also,

This shows that 

and 

Check this by using the expressions for u and v.

E X A M P L E   2

Function of a Complex Variable

Let Find 

u and v and the value of f at 

Solution.

gives and  Also,

Check this as in Example 1.

Remarks on Notation and Terminology

1. Strictly speaking, f (z) denotes the value of f at  z, but it is a convenient abuse of

language to talk about the function f (z) (instead of the function f ), thereby exhibiting the
notation for the independent variable.

2. We assume all functions to be single-valued relations, as usual: to each z in S there

corresponds but one value 

(but, of course, several z may give the same value

just as in calculus). Accordingly, we shall not use the term “multivalued

function” (used in some books on complex analysis) for a multivalued relation, in which
to a z there corresponds more than one w.

Limit, Continuity

A function f (z) is said to have the limit l as z approaches a point z

0

, written

(1)

if  f is defined in a neighborhood of 

(except perhaps at z

0

itself) and if the values of 

f are “close” to l for all z “close” to 

in precise terms, if for every positive real  we can

find a positive real  such that for all 

in the disk 

(Fig. 333) we have

(2)

geometrically, if for every 

in that  -disk the value of f lies in the disk (2).

Formally, this definition is similar to that in calculus, but there is a big difference.

Whereas in the real case, x can approach an x

0

only along the real line, here, by definition,

d

z

⫽ z

0

ƒ

f

 

(z)

⫺ l ƒ ⬍ P;

ƒ

z

⫺ z

0

ƒ

⬍ d

z

⫽ z

0

d

P

z

0

;

z

0

lim

z:z

0

  

f

 

(z)

⫽ l,

w

⫽ f

 

(z),

w

⫽ f

 

(z)

䊏

f

 

(

1
2

⫹ 4i) ⫽ 2i(

1
2

⫹ 4i) ⫹ 6(

1
2

⫺ 4i) ⫽ i ⫺ 8 ⫹ 3 ⫺ 24i ⫽ ⫺5 ⫺ 23i.

v(x, y)

⫽ 2x ⫺ 6y.

u(x, y)

⫽ 6x ⫺ 2y

f

 

(z)

⫽ 2i(x ⫹ iy) ⫹ 6(x ⫺ iy)

z

⫽

1
2

⫹ 4i.

w

⫽ f

 

(z)

⫽ 2iz ⫹ 6z.

䊏

v(1, 3)

⫽ 15.

u(1, 3)

⫽ ⫺5

f

 

(1

⫹ 3i) ⫽ (1 ⫹ 3i)

2

⫹ 3(1 ⫹ 3i) ⫽ 1 ⫺ 9 ⫹ 6i ⫹ 3 ⫹ 9i ⫽ ⫺5 ⫹ 15i.

v

⫽ 2xy ⫹ 3y.

u

⫽ Re f

 

(z)

⫽ x

2

⫺ y

2

⫹ 3x

z

⫽ 1 ⫹ 3i

w

⫽ f

 

(z)

⫽ z

2

⫹ 3z.

v(x, y)

u(x, y)

w

⫽ f

 

(z)

⫽ u(x, y) ⫹ iv(x, y).

z

⫽ x ⫹ iy.

w

⫽ u ⫹ iv,

SEC. 13.3

Derivative. Analytic Function

621

c13.qxd  10/30/10  2:14 PM  Page 621

Derivative

The derivative of a complex function f at a point 

is written 

and is defined by

(4)

provided this limit exists. Then f is said to be differentiable at  . If we write 

,

we have 

and (4) takes the form

Now comes an important point. Remember that, by the definition of limit, f (z) is defined

in a neighborhood of 

and z in ( ) may approach 

from any direction in the complex

plane. Hence differentiability at z

0

means that, along whatever path z approaches , 

the

quotient in ( ) always approaches a certain value and all these values are equal. This is
important and should be kept in mind.

E X A M P L E   3

Differentiability. Derivative

The function 

is differentiable for all z and has the derivative 

because

䊏

lim

¢

z:0

 

z

2

⫹ 2z ¢z ⫹ (¢z)

2

⫺ z

2

¢

z

⫽ lim

¢

z:0

 

(2z

⫹ ¢z) ⫽ 2z.

f

r

(z)

⫽ lim

¢

z:0

 

(z

⫹ ¢z)

2

⫺ z

2

¢

z

⫽

f

r

(z)

⫽ 2z

f

 

(z)

⫽ z

2

4

r

z

0

z

0

4

r

z

0

f

r

(z

0

)

⫽ lim

z:z

0

 

 

f (z)

⫺ f(z

0

)

z

⫺ z

0

.

(4

r

)

z

⫽ z

0

⫹ ¢z

¢

z

⫽ z ⫺ z

0

z

0

f

r

(z

0

)

⫽ lim

¢

z:0

 

f

 

(z

0

⫹ ¢z) ⫺ f(z

0

)

¢

z

f

r

(z

0

)

z

0

y

x

v

u

z

z

0

δ

f (z)

l

Œ

Fig. 333.

Limit

z may approach 

from any direction in the complex plane. This will be quite essential

in what follows.

If a limit exists, it is unique. (See Team Project 24.)

A function f (z) is said to be continuous at 

if 

is defined and

(3)

Note that by definition of a limit this implies that f (z) is defined in some neighborhood
of .

f (z) is said to be continuous in a domain if it is continuous at each point of this domain.

z

0

lim

z:z

0

 

 

f

 

(z)

⫽ f

 

(z

0

).

f

 

(z

0

)

z

⫽ z

0

z

0

622

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

c13.qxd  10/30/10  2:14 PM  Page 622

The differentiation rules are the same as in real calculus, since their proofs are literally
the same. Thus for any differentiable functions f and g and constant c we have

as well as the chain rule and the power rule 

(n integer).

Also, if f(z) is differentiable at z

0

, it is continuous at  . (See Team Project 24.)

E X A M P L E   4

not Differentiable

It may come as a surprise that there are many complex functions that do not have a derivative at any point. For
instance, 

is such a function. To see this, we write 

and obtain

(5)

If 

this is 

. If 

this is 

Thus (5) approaches 

along path I in Fig. 334 but 

along

path II. Hence, by definition, the limit of (5) as 

does not exist at any z. 

䊏

¢

z : 0

⫺1

⫹1

⫺1.

¢

x

⫽ 0,

⫹1

¢

y

⫽ 0,

f

 

(z

⫹ ¢z) ⫺ f

 

(z)

¢

z

⫽

(z

⫹ ¢z) ⫺ z

¢

z

⫽

¢

z

¢

z

⫽

¢

x

⫺ i

 

¢

y

¢

x

⫹ i

 

¢

y

 

.

¢

z

⫽ ¢x ⫹ i

 

¢

y

f

 

(z)

⫽ z ⫽ x ⫺ iy

z

z

0

(z

n

)

r

⫽ nz

n

ⴚ1

(cf )

r

⫽ cf

r

,

 

( f

⫹ g)

r

⫽ f

r

⫹ g

r

,

 

( fg)

r

⫽ f

r

g

⫹ fg

r

,

 

a

f

g b

r

⫽

f

r

g

⫺ fg

r

g

2

SEC. 13.3

Derivative. Analytic Function

623

Fig. 334.

Paths in (5)

y

x

ΙΙ

Ι

z +

 Δz

z

Surprising as Example 4 may be, it merely illustrates that differentiability of a complex
function is a rather severe requirement.

The idea of proof (approach of z from different directions) is basic and will be used

again as the crucial argument in the next section.

Analytic Functions

Complex analysis is concerned with the theory and application of “analytic functions,”
that is, functions that are differentiable in some domain, so that we can do “calculus in
complex.” The definition is as follows.

D E F I N I T I O N

Analyticity

A function 

is said to be analytic in a domain D if f (z) is defined and differentiable

at all points of D. The function f (z) is said to be analytic at a point

in D if

f (z) is analytic in a neighborhood of  .

Also, by an analytic function we mean a function that is analytic in some domain.

Hence analyticity of f (z) at 

means that f (z) has a derivative at every point in some

neighborhood of 

(including 

itself since, by definition, 

is a point of all its

neighborhoods). This concept is motivated by the fact that it is of no practical interest
if a function is differentiable merely at a single point 

but not throughout some

neighborhood of  . Team Project 24 gives an example.

A more modern term for analytic in D is holomorphic in D.

z

0

z

0

z

0

z

0

z

0

z

0

z

0

z

⫽ z

0

f (z)

c13.qxd  10/30/10  2:14 PM  Page 623

E X A M P L E   5

Polynomials, Rational Functions

The nonnegative integer powers 

are analytic in the entire complex plane, and so are polynomials,

that is, functions of the form

where 

are complex constants.

The quotient of two polynomials 

and 

is called a rational function. This f is analytic except at the points where 

here we assume that common

factors of g and h have been canceled.

Many further analytic functions will be considered in the next sections and chapters.

The concepts discussed in this section extend familiar concepts of calculus. Most
important is the concept of an analytic function, the exclusive concern of complex
analysis. Although many simple functions are not analytic, the large variety of remaining
functions will yield a most beautiful branch of mathematics that is very useful in
engineering and physics.

䊏

h(z)

⫽ 0;

f

 

(z)

⫽

g(z)

h(z)

,

h(z),

g(z)

c

0

, Á , c

n

f

 

(z)

⫽ c

0

⫹ c

1

z

⫹ c

2

z

2

⫹ Á ⫹ c

n

z

n

1, z, z

2

, Á

624

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

1–8

REGIONS OF PRACTICAL INTEREST

Determine and sketch or graph the sets in the complex plane
given by

1.

2.

3.

4.

5.

6.

7.

8.

9. WRITING PROJECT. Sets in the Complex Plane.

Write a report by formulating the corresponding
portions of the text in your own words and illustrating
them with examples of your own.

COMPLEX FUNCTIONS AND THEIR DERIVATIVES

10–12

Function Values. Find Re f, and Im f and their

values at the given point z.

10.

11.

12.

13. CAS PROJECT. Graphing Functions. Find and graph

Re  f, Im  f, and 

as surfaces over the z-plane. Also

graph the two families of curves 

and

Re f

 

(z)

⫽ const

ƒ

f ƒ

f

 

(z)

⫽ (z ⫺ 2)

>(z ⫹ 2) at 8i

f

 

(z)

⫽ 1

>(1 ⫺ z) at 1 ⫺ i

f

 

(z)

⫽ 5z

2

⫺ 12z ⫹ 3 ⫹ 2i at 4

 

⫺

 

3i

ƒ

z

⫹ i ƒ ⭌ ƒ z ⫺ i ƒ

Re 

 

z

⭌ ⫺1

Re (1

>z) ⬍ 1

ƒ

arg 

 

z ƒ

⬍

1
4

p

⫺

p

⬍ Im 

 

z

⬍

p

p

⬍ ƒ z ⫺ 4 ⫹ 2i ƒ ⬍ 3

p

0

⬍ ƒ z ƒ ⬍ 1

ƒ

z

⫹ 1 ⫺ 5i ƒ ⬉

3
2

in the same figure, and the curves

in another figure, where (a)

(b)

, (c) 

14–17

Continuity. Find out, and give reason, whether

f (z) is continuous at 

and for 

the

function f is equal to:

14.

15.

16.

17.

18–23

Differentiation. Find the value of the derivative

of

18.

19.

20.

at any z. Explain the result.

21.

at 0

22.

at 2i

23.

24. TEAM PROJECT. Limit, Continuity, Derivative

(a) Limit. Prove that (1) is equivalent to the pair of
relations

(b) Limit. If 

exists, show that this limit is

unique.

(c) Continuity. If 

are complex numbers for

which and 

if 

f(z) is continuous at 

show that  lim

n:

ⴥ 

 

f

 

(z

n

)

⫽ f

 

(a).

z

⫽ a,

lim

n:

ⴥ 

z

n

⫽ a,

z

1

, z

2

, Á

lim

z:z

0

 

 

f

 

(x)

lim

z:z

0

 

Re f

 

(z)

⫽ Re l,

   

lim

z:z

0

 

Im f

 

(z)

⫽

 

Im l.

z

3

>(z ⫹ i)

3 

at i

(iz

3

⫹ 3z

2

)

3

i(1

⫺ z)

n

(1.5z

⫹ 2i)

>(3iz ⫺ 4)

(z

⫺ 4i)

8 

at

⫽ 3 ⫹ 4i

(z

⫺ i)

>(z ⫹ i) at

 

i

(Re 

 

z)

>(1 ⫺ ƒ z ƒ )

(Im z

2

)

> ƒ z ƒ

2

ƒ

z ƒ

2 

Im (1

>z)

(Re z

2

)

> ƒ z ƒ

z

⫽ 0

z

⫽ 0 if f

 

(0)

⫽ 0

f

 

(z)

⫽ z

4

.

f

 

(z)

⫽ 1

>z

f (z)

⫽ z

2

,

ƒ

f

 

(z) ƒ

⫽ const

Im f

 

(z)

⫽ const

P R O B L E M   S E T   1 3 . 3

c13.qxd  10/30/10  2:14 PM  Page 624

13.4

Cauchy–Riemann Equations. 
Laplace’s Equation

As we saw in the last section, to do complex analysis (i.e., “calculus in the complex”) on
any complex function, we require that function to be analytic on some domain that is
differentiable in that domain.

The Cauchy–Riemann equations are the most important equations in this chapter

and one of the pillars on which complex analysis rests. They provide a criterion (a test)
for the analyticity of a complex function

Roughly, f is analytic in a domain D if and only if the first partial derivatives of u and 
satisfy the two Cauchy–Riemann equations

4

(1)

everywhere in D; here 

and 

(and similarly for v) are the usual

notations for partial derivatives. The precise formulation of this statement is given in
Theorems 1 and 2.

Example: 

is analytic for all z (see Example 3 in Sec. 13.3),

and 

and 

satisfy (1), namely, 

as well as 

. More examples will follow.

T H E O R E M   1

Cauchy–Riemann Equations

Let

be defined and continuous in some neighborhood of a

point

and differentiable at z itself. Then, at that point, the first-order

partial derivatives of u and v exist and satisfy the Cauchy–Riemann equations (1).

Hence, if

is analytic in a domain D, those partial derivatives exist and satisfy

(1) at all points of D.

f (z)

z

⫽ x ⫹ iy

f (z)

⫽ u(x, y) ⫹ iv(x, y)

⫺2y ⫽ ⫺v

x

u

y

⫽

u

x

⫽ 2x ⫽ v

y

v

⫽ 2xy

u

⫽ x

2

⫺ y

2

f (z)

⫽ z

2

⫽ x

2

⫺ y

2

⫹ 2ixy

u

y

⫽ 0u

>0y

u

x

⫽ 0u

>0x

u

x

⫽ v

y

,

         

u

y

⫽ ⫺v

x

v

w

⫽ f

 

(z)

⫽ u(x, y) ⫹ iv(x, y).

SEC. 13.4

Cauchy–Riemann Equations. Laplace’s Equation

625

(d) Continuity. If 

is differentiable at 

show that

f (z) is continuous at 

(e) Differentiability. Show that 

is not

differentiable at any z. Can you find other such functions?

(f) Differentiability. Show that 

is dif-

ferentiable only at 

hence it is nowhere analytic.

z

⫽ 0;

f

 

(z)

⫽ ƒ z ƒ

2

f

 

(z)

⫽ Re z ⫽ x

z

0

.

z

0

,

f

 

(z)

25. WRITING PROJECT. Comparison with Calculus.

Summarize the second part of this section beginning with
Complex Function, and indicate what is conceptually
analogous to calculus and what is not.

4

The French mathematician AUGUSTIN-LOUIS CAUCHY (see Sec. 2.5) and the German mathematicians

BERNHARD RIEMANN (1826–1866) and KARL WEIERSTRASS (1815–1897; see also Sec. 15.5) are the
founders of complex analysis. Riemann received his Ph.D. (in 1851) under Gauss (Sec. 5.4) at Göttingen, where
he also taught until he died, when he was only 39 years old. He introduced the concept of the integral as it is
used in basic calculus courses, and made important contributions to differential equations, number theory, and
mathematical physics. He also developed the so-called Riemannian geometry, which is the mathematical
foundation of Einstein’s theory of relativity; see Ref. [GenRef9] in App. 1.

c13.qxd  10/30/10  2:14 PM  Page 625

P R O O F

By assumption, the derivative 

at z exists. It is given by

(2)

The idea of the proof is very simple. By the definition of a limit in complex (Sec. 13.3),
we can let 

approach zero along any path in a neighborhood of z. Thus we may choose

the two paths I and II in Fig. 335 and equate the results. By comparing the real parts we
shall obtain the first Cauchy–Riemann equation and by comparing the imaginary parts the
second. The technical details are as follows.

We write 

. Then 

and in terms of u

and v the derivative in (2) becomes

(3)

.

We first choose path I in Fig. 335. Thus we let 

first and then 

. After 

is zero, 

. Then (3) becomes, if we first write the two u-terms and then the two

v-terms,

f

r

(z)

⫽ lim

¢

x:0

 

u(x

⫹ ¢x, y) ⫺ u(x, y)

¢

x

⫹ i

 

lim

¢

x:0

 

v(x

⫹ ¢x, y) ⫺ v(x, y)

¢

x

 

.

¢

z

⫽ ¢x

¢

y

¢

x : 0

¢

y : 0

f

r

(z)

⫽ lim

¢

z:0 

[u(x

⫹ ¢x, y ⫹ ¢y) ⫹ iv(x ⫹ ¢x, y ⫹ ¢y)] ⫺ [u(x, y) ⫹ iv(x, y)]

¢

x

⫹ i

 

¢

y

z

⫹ ¢z ⫽ x ⫹ ¢x ⫹ i(y ⫹ ¢y),

¢

z

⫽ ¢x ⫹ i ¢y

¢

z

f

r

(z)

⫽ lim

¢

z:0

 

f

 

(z

⫹ ¢z) ⫺ f

 

(z)

¢

z

 

.

f

r

(z)

626

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

y

x

ΙΙ

Ι

z +

 Δz

z

Fig. 335.

Paths in (2)

Since 

exists, the two real limits on the right exist. By definition, they are the partial

derivatives of u and v with respect to x. Hence the derivative 

of f (z) can be written

(4)

Similarly, if we choose path II in Fig. 335, we let 

first and then 

. After

is zero, 

, so that from (3) we now obtain

Since 

exists, the limits on the right exist and give the partial derivatives of u and v

with respect to y; noting that 

we thus obtain

(5)

The existence of the derivative 

thus implies the existence of the four partial derivatives

in (4) and (5). By equating the real parts 

and 

in (4) and (5) we obtain the first

v

y

u

x

f

r

(z)

f

r

(z)

⫽ ⫺iu

y

⫹ v

y

.

1

>i ⫽ ⫺i,

f

r

(z)

f

r

(z)

⫽ lim

¢

y:0

 

u(x, y

⫹ ¢y) ⫺ u(x, y)

i

 

¢

y

⫹ i lim

¢

y:0

 

v(x, y

⫹ ¢y) ⫺ v(x, y)

i

 

¢

y

 

.

¢

z

⫽ i ¢y

¢

x

¢

y : 0

¢

x : 0

f

r

(z)

⫽ u

x

⫹ iv

x

.

f

r

(z)

f

r

(z)

c13.qxd  10/30/10  2:14 PM  Page 626

Cauchy–Riemann equation (1). Equating the imaginary parts gives the other. This proves
the first statement of the theorem and implies the second because of the definition of
analyticity.

Formulas (4) and (5) are also quite practical for calculating derivatives 

as we shall see.

E X A M P L E   1

Cauchy–Riemann Equations

is analytic for all z. It follows that the Cauchy–Riemann equations must be satisfied (as we have

verified above).

For 

we have 

and see that the second Cauchy–Riemann equation is satisfied,

but the first is not: 

We conclude that 

is not analytic, confirming

Example 4 of Sec. 13.3. Note the savings in calculation!

The Cauchy–Riemann equations are fundamental because they are not only necessary but
also sufficient for a function to be analytic. More precisely, the following theorem holds.

T H E O R E M   2

Cauchy–Riemann Equations

If two real-valued continuous functions 

and 

of two real variables x

and y have continuous first partial derivatives that satisfy the Cauchy–Riemann
equations in some domain D, then the complex function

is

analytic in D.

The proof is more involved than that of Theorem 1 and we leave it optional (see App. 4).

Theorems 1 and 2 are of great practical importance, since, by using the Cauchy–Riemann

equations, we can now easily find out whether or not a given complex function is analytic.

E X A M P L E   2

Cauchy–Riemann Equations. Exponential Function

Is analytic?

Solution.

We have 

and by differentiation

We see that the Cauchy–Riemann equations are satisfied and conclude that f (z) is analytic for all z. ( f (z) will
be the complex analog of 

known from calculus.)

E X A M P L E   3

An Analytic Function of Constant Absolute Value Is Constant

The Cauchy–Riemann equations also help in deriving general properties of analytic functions.

For instance, show that if 

is analytic in a domain D and in 

D, then 

in

D. (We shall make crucial use of this in Sec. 18.6 in the proof of Theorem 3.)

Solution.

By assumption, 

By differentiation,

Now use 

in the first equation and 

in the second, to get

(6)

(a)

(b)

 uu

y

⫺ vu

x

⫽ 0.

 uu

x

⫺ vu

y

⫽ 0,

v

y

⫽ u

x

v

x

⫽ ⫺u

y

 uu

y

⫹ vv

y

⫽ 0.

 uu

x

⫹ vv

x

⫽ 0,

ƒ

f ƒ

2

⫽ ƒ u ⫹ iv ƒ

2

⫽ u

2

⫹ v

2

⫽ k

2

.

f

 

(z)

⫽ const

ƒ

f

 

(z) ƒ

⫽ k ⫽ const

f

 

(z)

䊏

e

x

 u

y

⫽ ⫺e

x 

sin y,

     

 v

x

⫽ e

x 

sin y.

 u

x

⫽ e

x 

cos y,

     

 v

y

⫽ e

x 

cos y

u

⫽ e

x

 

cos y, v

⫽ e

x 

sin y

f

 

(z)

⫽ u(x, y) ⫹ iv(x, y) ⫽ e

x

(cos y

⫹ i sin y)

f

 

(z)

⫽ u(x, y) ⫹ iv(x, y)

v(x, y)

u(x, y)

䊏

f

 

(z)

⫽ z

u

x

⫽ 1 ⫽ v

y

⫽ ⫺1.

u

y

⫽ ⫺v

x

⫽ 0,

u

⫽ x, v ⫽ ⫺y

f

 

(z)

⫽ z ⫽ x ⫺ iy

f

 

(z)

⫽ z

2

f

r

(z),

䊏

SEC. 13.4

Cauchy–Riemann Equations. Laplace’s Equation

627

c13.qxd  10/30/10  2:14 PM  Page 627

To get rid of 

, multiply (6a) by u and (6b) by v and add. Similarly, to eliminate 

, multiply (6a) by 

and

(6b) by u and add. This yields

If then 

hence 

If then 

Hence, 

by 

the

Cauchy–Riemann equations, also 

Together this implies 

and 

; hence

We mention that, if we use the polar form 

and set 

, then the Cauchy–Riemann equations are (Prob. 1)

(7)

Laplace’s Equation. Harmonic Functions

The great importance of complex analysis in engineering mathematics results mainly from
the fact that both the real part and the imaginary part of an analytic function satisfy Laplace’s
equation, the most important PDE of physics. It occurs in gravitation, electrostatics, fluid
flow, heat conduction, and other applications (see Chaps. 12 and 18).

T H E O R E M   3

Laplace’s Equation

If

is analytic in a domain D, then both u and v satisfy

Laplace’s equation

(8)

(

read “nabla squared”) and

(9)

,

in D and have continuous second partial derivatives in D.

P R O O F

Differentiating 

with respect to x and 

with respect to y, we have

(10)

Now the derivative of an analytic function is itself analytic, as we shall prove later (in
Sec. 14.4). This implies that u and v have continuous partial derivatives of all orders; in
particular, the mixed second derivatives are equal: 

By adding (10) we thus

obtain (8). Similarly, (9) is obtained by differentiating 

with respect to y and

with respect to x and subtracting, using 

Solutions of Laplace’s equation having continuous second-order partial derivatives are called
harmonic functions and their theory is called potential theory (see also Sec. 12.11). Hence
the real and imaginary parts of an analytic function are harmonic functions.

䊏

u

xy

⫽ u

yx

.

u

y

⫽ ⫺v

x

u

x

⫽ v

y

v

yx

⫽ v

xy

.

u

xx

⫽ v

yx

,

 

       

u

yy

⫽ ⫺v

xy

.

u

y

⫽ ⫺v

x

u

x

⫽ v

y

ⵜ

2

v

⫽ v

xx

⫹ v

yy

⫽ 0

ⵜ

2

ⵜ

2

u

⫽ u

xx

⫹ u

yy

⫽ 0

f

 

(z)

⫽ u(x, y) ⫹ iv(x, y)

 v

r

⫽ ⫺

 

1

r

 u

u

(r

⬎ 0).

 u

r

⫽

1

r

 v

u

,

iv(r, u)

f

 

(z)

⫽ u(r, u) ⫹

z

⫽ r(cos u ⫹ i sin u)

䊏

f

⫽ const.

v

⫽ const

u

⫽ const

u

x

⫽ v

y

⫽ 0.

u

x

⫽ u

y

⫽ 0.

k

2

⫽ u

2

⫹ v

2

⫽ 0,

f

⫽ 0.

u

⫽ v ⫽ 0;

k

2

⫽ u

2

⫹ v

2

⫽ 0,

 (u

2

⫹ v

2

)u

y

⫽ 0.

 (u

2

⫹ v

2

)u

x

⫽ 0,

⫺v

u

x

u

y

628

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

c13.qxd  10/30/10  2:14 PM  Page 628

If two harmonic functions u and v satisfy the Cauchy–Riemann equations in a domain

D, they are the real and imaginary parts of an analytic function f in D. Then v is said to
be a harmonic conjugate function of u in D. (Of course, this has absolutely nothing to
do with the use of “conjugate” for 

E X A M P L E   4

How to Find a Harmonic Conjugate Function by the Cauchy–Riemann Equations

Verify that 

is harmonic in the whole complex plane and find a harmonic conjugate function

v of u.

Solution.

by direct calculation. Now 

and 

Hence because of the Cauchy–

Riemann equations a conjugate v of u must satisfy

Integrating the first equation with respect to y and differentiating the result with respect to x, we obtain

.

A comparison with the second equation shows that 

This gives 

. Hence 

(c any real constant) is the most general harmonic conjugate of the given u. The corresponding analytic function is

Example 4 illustrates that a conjugate of a given harmonic function is uniquely determined
up to an arbitrary real additive constant.

The Cauchy–Riemann equations are the most important equations in this chapter. Their

relation to Laplace’s equation opens a wide range of engineering and physical applications,
as shown in Chap. 18.

䊏

f

 

(z)

⫽ u ⫹ iv ⫽ x

2

⫺ y

2

⫺ y ⫹ i(2xy ⫹ x ⫹ c) ⫽ z

2

⫹ iz ⫹ ic.

v

⫽ 2xy ⫹ x ⫹ c

h(x)

⫽ x ⫹ c

dh

>dx ⫽ 1.

v

⫽ 2xy ⫹ h(x),

     

v

x

⫽ 2y ⫹

dh

dx

v

y

⫽ u

x

⫽ 2x,

     

v

x

⫽ ⫺u

y

⫽ 2y ⫹ 1.

u

y

⫽ ⫺2y ⫺ 1.

u

x

⫽ 2x

ⵜ

2

u

⫽ 0

u

⫽ x

2

⫺ y

2

⫺ y

z.)

SEC. 13.4

Cauchy–Riemann Equations. Laplace’s Equation

629

1. Cauchy–Riemann equations in polar form. Derive (7)

from (1).

2–11

CAUCHY–RIEMANN EQUATIONS

Are the following functions analytic? Use (1) or (7).

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12–19

HARMONIC FUNCTIONS

Are the following functions harmonic? If your answer
is yes, find a corresponding analytic function 

12.

13. u

⫽ xy

u

⫽ x

2

⫹ y

2

u(x, y)

⫹ iv(x, y).

f

 

(z) 

⫽

f

 

(z)

⫽ cos x cosh y ⫺ i sin x sinh y

f

 

(z)

⫽ ln  ƒ z ƒ ⫹ i Arg z

f

 

(z)

⫽ 3

p

2

>(z

3

⫹ 4

p

2

z)

f

 

(z)

⫽ Arg 2

p

z

f

 

(z)

⫽ i

>z

8

f

 

(z)

⫽ 1

>(z ⫺ z

5

)

f

 

(z)

⫽ Re (z

2

)

⫺ i Im (z

2

)

f

 

(z)

⫽ e

x

 (cos y

⫺ i sin y)

f

 

(z)

⫽ e

ⴚ2x 

(cos 2y

⫺ i

 

sin 2y)

f

 

(z)

⫽ izz

14.

15.

16.

17.

18.

19.

20. Laplace’s equation. Give the details of the derivative

of (9).

21–24

Determine a and b so that the given function is

harmonic and find a harmonic conjugate.

21.

22.

23.

24.

25. CAS PROJECT. Equipotential Lines. Write a

program for graphing equipotential lines 

of

a harmonic function u and of its conjugate v on the
same axes. Apply the program to (a)

(b)

26. Apply the program in Prob. 25 to 

and to an example of your own.

v

⫽ e

x 

sin y

u

⫽ e

x 

cos y,

u

⫽ x

3

⫺ 3xy

2

, v

⫽ 3x

2

y

⫺ y

3

.

v

⫽ 2xy,

u

⫽ x

2

⫺ y

2

,

u

⫽ const

u

⫽ cosh ax cos y

u

⫽ ax

3

⫹ bxy

u

⫽ cos ax cosh 2y

u

⫽ e

p

x 

cos av

v

⫽ e

x

 sin 2y

u

⫽ x

3

⫺ 3xy

2

v

⫽ (2x ⫹ 1)y

u

⫽ sin x cosh y

u

⫽ x

>(x

2

⫹ y

2

)

v

⫽ xy

P R O B L E M   S E T 1 3 . 4

c13.qxd  10/30/10  2:14 PM  Page 629

13.5

Exponential Function

In the remaining sections of this chapter we discuss the basic elementary complex
functions, the exponential function, trigonometric functions, logarithm, and so on. They
will be counterparts to the familiar functions of calculus, to which they reduce when 
is real. They are indispensable throughout applications, and some of them have interesting
properties not shared by their real counterparts.

We begin with one of the most important analytic functions, the complex exponential

function

also written

exp z.

The definition of 

in terms of the real functions 

, and 

is

(1)

This definition is motivated by the fact the 

extends the real exponential function 

of

calculus in a natural fashion. Namely:

(A)

for real 

because 

and 

when

(B)

is analytic for all z. (Proved in Example 2 of Sec. 13.4.)

(C) The derivative of 

is  , that is,

(2)

This follows from (4) in Sec. 13.4,

REMARK. 

This definition provides for a relatively simple discussion. We could define 

by the familiar series 

with x replaced by z, but we would

then have to discuss complex series at this very early stage. (We will show the connection
in Sec. 15.4.)

Further Properties.

A function 

that is analytic for all z is called an entire function.

Thus, e

z

is entire. Just as in calculus the functional relation

(3)

e

z

1

⫹z

2

⫽ e

z

1

e

z

2

f

 

(z)

1

⫹ x ⫹ x

2

>2! ⫹ x

3

>3! ⫹ Á

e

z

(e

z

)

r

⫽ (e

x 

cos y)

x

⫹ i(e

x 

sin y)

x

⫽ e

x 

cos y

⫹ ie

x 

sin y

⫽ e

z

.

(e

z

)

r

⫽ e

z

.

e

z

e

z

e

z

y

⫽ 0.

sin y

⫽ 0

cos y

⫽ 1

z

⫽ x

e

z

⫽ e

x

e

x

e

z

e

z

⫽ e

x

(cos y

⫹ i sin y).

sin y

e

x

, cos y

e

z

e

z

,

z

⫽ x

630

CHAP. 13

Complex Numbers and Functions. Complex Differentiation

27. Harmonic conjugate. Show that if u is harmonic and

v is a harmonic conjugate of u, then u is a harmonic
conjugate of  v.

28. Illustrate Prob. 27 by an example.

29. Two further formulas for the derivative. Formulas (4),

(5), and (11) (below) are needed from time to time. Derive

(11)

f

r

(z)

⫽ u

x

⫺ iu

y

,

     

f

r

(z)

⫽ v

y

⫹ iv

x

.

⫺

30. TEAM PROJECT. Conditions for 

. Let

be analytic. Prove that each of the following

conditions is sufficient for 

.

(a)

(b)

(c)

(d)

(see Example 3)

ƒ

f (z) ƒ

⫽ const

f

r

(z)

⫽ 0

Im f (z)

⫽ const

Re f

 

(z)

⫽ const

f (z)

⫽ const

f

 

(z)

f

 

(z)

⫽ const

c13.qxd  10/30/10  2:14 PM  Page 630

holds for any 

and 

. Indeed, by (1),

Since for 

these 

real functions, by an application of the addition formulas

for the cosine and sine functions (similar to that in Sec. 13.2) we see that

as asserted. An interesting special case of (3) is 

; then

(4)

Furthermore, for 

we have from (1) the so-called Euler formula

(5)

Hence the polar form of a complex number, 

, may now be written

(6)

From (5) we obtain

(7)

as well as the important formulas (verify!)

(8)

Another consequence of (5) is

(9)

That is, for pure imaginary exponents, the exponential function has absolute value 1, a
result you should remember. From (9) and (1),

(10)

Hence

,

since 

shows that (1) is actually 

in polar form.

From 

in (10) we see that

(11)

for all z.

So here we have an entire function that never vanishes, in contrast to (nonconstant)
polynomials, which are also entire (Example 5 in Sec. 13.3) but always have a zero, as
is proved in algebra.

e

x

⫽ 0

ƒ

e

z

ƒ

⫽ e

x

⫽ 0

e

z

ƒ

e

z

ƒ

⫽ e

x

(n

⫽ 0, 1, 2, Á )

arg e

z

⫽ y ⫾ 2n

p

ƒ

e

z

ƒ

⫽ e

x

.

ƒ

e

iy

ƒ

⫽ ƒ cos y ⫹ i sin y ƒ ⫽ 2cos

2

 y

⫹ sin

2

 y

⫽ 1.

e

p

i

>2

⫽ i,

     

e

p

i

⫽ ⫺1,

     

e

ⴚ

p

i

>2

⫽ ⫺i,

     

e

ⴚ

p

i

⫽ ⫺1.

e

2

p

i

⫽ 1

z

⫽ re

iu

.

z

⫽ r(cos u ⫹ i sin u)

e

iy

⫽ cos y  ⫹ i sin y.

z

⫽ iy

e

z

⫽ e

x

e

iy

.

z

1

⫽ x, z

2

⫽ iy

e

z

1

e

z

2

⫽ e

x

1

⫹x

2

[cos (

 

y

1

⫹ y

2

)

⫹ i sin (

 

y

1

⫹ y

2

)]

⫽ e

z

1

⫹z

2

e

x

1

e

x

2

⫽  e

x

1

⫹x

2

e

z

1

e

z

2

⫽ e

x

1

(cos y

1

⫹ i sin y

1

)

 

e

x

2

(cos y

2

⫹ i sin y

2

).

z

2

⫽ x

2

⫹ iy

2

z

1

⫽ x

1

⫹ iy

1

SEC. 13.5

Exponential Function

631

c13.qxd  10/30/10  2:14 PM  Page 631