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1. Complex Numbers

1.1 The Algebra of Complex NumbersThe complex number z is defined to be

z = (x,y), (1.1)

where x and y are both real numbers. The numbers x and y are, moreover, known as thereal and imaginary parts of z, respectively; and we write

Re z = x, Im z = y

Two complex numbers are equal if and only if their x coordinates are equal and their ycoordinates are equal. In other words,

(x,y) = (u,v) iff x = u and y = v.

Definition 1.1.1 — Addition and Multiplication. If z1 = (x1,y1) and z2 = (x2,y2) arearbitrary complex numbers, then the sum and the product of z1 and z2 are defined asfollows:

(x1,y1)+(x2,y2) = (x1 + x2,y1 + y2). (1.2)

and

(x1,y1)(x2,y2) = (x1x2− y1y2,y1x2 + x1y2). (1.3)

2 Chapter 1. Complex Numbers

Note that the operations defined by (1.2) and (1.3) become the usual operations ofaddition and multiplication when restricted to the real numbers:

(x1,0)+(x2,0) = (x1 + x2,0). (1.4)

and

(x1,0)(x2,0) = (x1x2,0). (1.5)

Thus the complex number system is, therefore, a natural extension of the real numbersystem.

Any complex number z = (x,y) can be written z = (x,0)+(0,y), and it is easy to seethat (0,1)(y,0) = (0,y). Hence

z = (x,0)+(0,1)(y,0);

and if we think of a real number as either x or (x,0) and let i denote the imaginary number(0,1), it is clear that

z = x+ iy. (1.6)

We note that i2 = (0,1)(0,1) = (−1,0) or i2 =−1.In view of expression (1.6), (1.2) and (1.3) become

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

and

(x1 + iy1)(x2 + iy2) = (x1x2− y1y2)+ i(y1x2 + x1y2).

� Example 1.1 If z1 = (3,7) and z2 = (5,−6), thenz1 + z2 = (3,7)+(5,−6) = (8,1) andz1− z2 = (3,7)− (5,−6) = (−2,13).We can also use the notation z1 = 3+7i and z2 = 5−6i to obtainz1 + z2 = (3+7i)+(5−6i) = (8+ i) andz1− z2 = (3+7i)− (5−6i) = (−2+13i). �

� Example 1.2 If z1 = (3,7) and z2 = (5,−6), then

z1z2 = (3,7)(5,−6)= ((3 ·5)− (7 · (−6)),(3 · (−6))+(5 ·7))= (15+42,−18+35)= (57,17).

We can also use the notation z1 = 3+7i and z2 = 5−6i to obtain

z1z2 = (3+7i)(5−6i)= 15−18i+35i−42i2

= 15−42(−1)+(−18+35)i= 57+17i.

�

1.1 The Algebra of Complex Numbers 3

To motivate our definition for division, we proceed along the same lines as we did formultiplication, assuming that z2 6= 0.

z1

z2=

(x1,y1)

(x2,y2)

=(x1 + iy1)

(x2 + iy2)

=(x1 + iy1)(x2− iy2)

(x2 + iy2)(x2− iy2)

=x1x2 + y1y2 + i(−x1y2 + x2y1)

x22 + y2

2

=x1x2 + y1y2

x22 + y2

2+ i

(−x1y2 + x2y1)

x22 + y2

2

=(x1x2 + y1y2

x22 + y2

2,−x1y2 + x2y1

x22 + y2

2

).

Definition 1.1.2 — Division.

z1

z2=

(x1,y1)

(x2,y2)

=(x1x2 + y1y2

x22 + y2

2,−x1y2 + x2y1

x22 + y2

2

), for z2 6= 0.

� Example 1.3 If z1 = (3,7) and z2 = (5,−6), then

z1

z2=

(3,7)(5,−6)

=(15−42

25+36,18+3525+36

)=(−27

61,5361

).

We can also use the notation z1 = 3+7i and z2 = 5−6i to obtain

z1z2 =3+7i5−6i

=3+7i5−6i

· 5+6i5+6i

=15+18i+35i+42i2

25+30i−30i−36i2

=15−42+(18+35)i

25+36

=−2761

+5361

i.

�


Properties of complex numbers(P1) Commutative law for addition : z1 + z2 = z2 + z1.(P2) Associative law for addition : z1 +(z2 + z3) = (z1 + z2)+ z3.(P3) Additive identity : ∀z ∈C ,∃w ∈C,z+w = z. The number w is obviously the order

pair (0,0).(P4) Additive inverse : ∀z ∈C, ∃!η ∈C,z+η = (0,0). Obviously, if z = (x,y) = x+ iy,

then η = (−x,−y) =−x− iy =−z.(P5) Commutative law for multiplication : z1z2 = z2z1.(P6) Associative law for multiplication : z1(z2z3) = (z1z2)z3.(P7) Multiplicative identity : ∀z ∈ C, ∃ζ ∈ C,zζ = z. (1,0) is the unique complex

number ζ having this property.(P8) Multiplicative inverse : ∀z ∈ C\{(0,0)}, ∃z−1 ∈ C,zz−1 = (1,0) = 1. The number

z−1 would be

z−1 =(1,0)

z=

1z=

1x+ iy

=x− iy

x2 + y2 =x

x2 + y2 + i−y

x2 + y2

=( x

x2 + y2 ,−y

x2 + y2

).

(P9) The distributive law : z1(z2 + z3) = z1z2 + z1z3.

1.1 The Algebra of Complex Numbers 5

Definition 1.1.3 The conjugate of z = x + iy, denoted z, is the complex number(x,−y) = x− iy.

� Example 1.4a) Re(−3+7i) =−3 and Re[(9,4)] = 9.b) Im(−3+7i) = 7 and Im[(9,4)] = 4.c) −3+7i =−3−7i and (9,4) = (9,−4).

�

Theorem 1.1.1 Suppose that z,z1 and z3 are arbitrary complex numbers. Then(1) z = z.(2) z1 + z2 = z1 + z2.(3) z1z2 = z1 · z2.

(4)(

z1z2

)= z1

z2if z2 6= 0.

(5) Re(z) = z+z2 .

(6) Im(z) = z−z2i .

(7) Re(iz) =−Im(z).(8) Im(iz) =Re(z).

Proof.


Exercises 1.11. Preform the required calculations and express your answers in the form a+bi

1.1 i2016

1.2i

25581.3 (i−3)3

1.4 Re(7+6i) + Im(5−4i)

1.5(4− i)(1−3i)−1+2i

1.6 (1+ i)−2

1.7 (1+ i)(2+ i)(3+ i)1.8 (1+

√3i)(i+

√3)

2. Show that zz̄ is always a real number.3. Let z1,z2 be arbitrary complex numbers. Prove or disprove the following.

3.1 Re(z1 + z2) = Re(z1)+ Re(z2)3.2 Re(z1z2) = Re(z1) Re(z2)3.3 Im(z1 + z2) = Im(z1)+ Im(z2)3.4 Im(z1z2) = Im(z1) Im(z2)

4. Verify that the complex number (0,0) has a multiplicative inverse or not?

5. Prove that if z = (x,y) where x,y ∈ R and not both 0, then z−1 =1z

.

1.2 The Geometry of Complex Numbers 7

1.2 The Geometry of Complex NumbersComplex numbers are ordered pairs of real numbers, so they can be represented by pointsin the plane.

We can represent the number z = x+ iy = (x,y) by a position vector in the xy planewhose tail is at the origin and whose head is at the point (x,y). When the xy plane isused for displaying complex numbers, it is called the complex plane, or more simply, thez plane.

Recall that Re(z) = x and Im(z) = y. Geometrically, Re(z) is the projection of z = (x,y)on the x-axis, and Im(z) is the projection of z on to the y-axis. We call the x−axis thereal axis and the y−axis the imaginary axis

Y Imaginary axis

0

2 i− +

x Real axis

z x iy= +

Figure 1.1: The complex plane

Definition 1.2.1 — Modulus of Complex Numbers. The Modulus, or absolutevalue, of the complex number z = x+ iy is a nonnegative real number denoted by|z| and defined by the relation

|z|=√

x2 + y2. (1.7)

We have the following results.

Theorem 1.2.1 For any complex number z = x+ iy, the following are true.(a) The number |z| is the distance between the origin and the point z = (x,y).(b) |z|= 0 iff z = 0(c) |x|= | Re (z)| ≤ |z| and |y|= | Im (z)| ≤ |z|,(d) |z|2 = zz.

Proof.


R (a) The difference z1− z2 represents the displacement vector from z2 to z1, so thedistance between z1 and z2 is given by |z1− z2|. Therefore

|z1− z2|=√

(x1− x2)2 +(y1− y2)2.

(b) If z = (x,y) = x+ iy, then −z = (−x,−y) =−x− iy is the reflection of z throughthe origin and z = (x,−y) = x− iy is the reflection of z through the x-axis.

Theorem 1.2.2 — The triangle inequality. If z1 and z2 are arbitrary complex numbers,then

|z1 + z2| ≤ |z1|+ |z2|.

Proof.

1.2 The Geometry of Complex Numbers 9

Theorem 1.2.3 Let z1 and z2 be arbitrary complex numbers. Then(a) ||z1|− |z2|| ≤ |z1− z2|(b) |z1z2|= |z1||z2|(c) |z1/z2|= |z1|/|z2| whenever z2 6= 0.

Proof.


Exercises 1.21. Evaluate the following quantities.

1.1 |(1+ i)(2+ i)|

1.2∣∣∣∣4−3i

2− i

∣∣∣∣1.3 |(1+ i)50|1.4 |z−1|2 where z = x+ yi

2. Sketch the sets of points determined by the following relations2.1 |z+1−2i|= 22.2 Re(z+1) = 02.3 |z+2i| ≤ 12.4 Im(z−2i)> 6

3. Prove that√

2|z| ≥ |Re(z)|+ |Im(z)|.4. Prove that |z1− z2| ≤ |z1|+ |z2|.5. Prove that |z|= 0 iff z = 0.6. Show that the nonzero vectors z1 and z2 are parallel iff Im(z1z2) = 0.7. Show that the nonzero vectors z1 and z2 are perpendicular iff Re(z1z2) = 0.8. Show that |zn|= |z|n where n is a natural number.9. Suppose that either |z|= 1 or |w|= 1. Prove that |z−w|= |1− z̄w|.

10. Prove the Cauchy-Schwarz inequality:∣∣∣∣∣ n

∑k=1

zkwk

∣∣∣∣∣≤√

n

∑k=1|zk|2

√n

∑k=1|wk|2.

11. Show that ||z1|− |z2|| ≤ |z1− z2|.12. Prove that |z1− z2|2 = |z1|2−2Re(z1z2)+ |z2|2.

1.3 Complex Numbers in Polar Coordinates 11

1.3 Complex Numbers in Polar CoordinatesLet z = x+ iy and r = |z|. Let θ be the angle that the line from the origin to the complexnumber z make with the positive x-axis. Then the figure below shows

z = (r cosθ , r sinθ) = r(cosθ + isinθ). (1.8)

Note : The number θ is undefined if z = 0.

Definition 1.3.1 — Polar representation. Equation (1.8) is called as a polar repre-sentation of z, and the value r and θ are called polar coordinates of z.

� Example 1.5 If z = 1+ i, then r =√

2 and

cosθ =1√2

and sinθ =1√2.

Therefore θ = π

4 . This implies that

z =√

2(cosπ

4+ isin

π

4)

is a polar representation of z. �

The number θ can be any value for which the identities

cosθ = x/r and sinθ = y/r

hold. For z 6= 0, the collection of all values of θ for which z = r(cosθ + isinθ) is denotedby arg z and is called the set of arguments of z. Then, we have the following definitions.

Definition 1.3.2 — Set of Arguments. If z 6= 0, then denote the set of arguments arg zby

arg z = {θ : z = r(cosθ + isinθ)}. (1.9)

If θ ∈ arg z, we say that θ is an argument of z.

R Note that, if θ1,θ2 ∈ arg z, then there exists some integer n such that

θ1 = θ2 +2nπ.


� Example 1.6 Since 1+ i =√

2(cos π

4 + isin π

4 ), we have

arg (1+ i) ={

π

4+2nπ : n is an integer

}=

{...,−7π

4,π

4,9π

4,17π

4, ...}.

�

Definition 1.3.3 — Principal Value of the Argument. Let z 6= 0 be a complex numbersuch that z = r(cosθ + isinθ) and −π < θ ≤ π. The number θ is called the principalvalue of the argument and is denoted by Arg z; that is θ = Arg z.

R θ = Arg z⇔ θ ∈ arg z and −π < θ ≤ π.

� Example 1.7 Arg(1+ i) = π

4 . �

� Example 1.8 Let z =−√

3− i. Thus r = |z|=√

3+1 = 2 and tanθ = −1−√

3= 1√

3. This

implies that

θ ∈ arctan1√3={

π

6+nπ : n ∈ Z

}.

Since −√

3− i is in the third quadrant, this implies that θ = −5π

6 . Therefore

−√

3− i = 2cos−5π

6+ i2sin

−5π

6

= 2cos(−5π

6+2nπ

)+ i2sin

(−5π

6+2nπ

)= 2

(cos(−5π

6+2nπ

)+ isin

(−5π

6+2nπ

)),

where n ∈ Z. Hence Arg(−√

3− i) = −5π

6 and arg(−√

3− i) ={−5π

6 +2nπ : n ∈ Z}. �

� Example 1.9 Let z = iy. Therefore Arg z = π

2 if Im z > 0, and Arg z =−π

2 if Im z <0. �

The following theorem will be proved in Chapter 5.

Theorem 1.3.1 — Euler’s formula. Let z = x+ iy. Then

ez = ex+iy = ex(cosy+ isiny) (1.10)

If we set x = 0 and let θ = y in the (1.10), we get a famous result known as Euler’sformula :

eiθ = cosθ + isinθ = (cosθ ,sinθ). (1.11)

R If θ is a real number, then eiθ will be located somewhere on the circle with radius 1centered at the origin. This assertion is easy to verify because∣∣∣eiθ

∣∣∣=√cos2 θ + sin2θ = 1.


Definition 1.3.4 — Exponential Representation. Let z = r(cosθ + isinθ) be anycomplex number. Using the Euler’s formula we can write z in its exponential form:

z = reiθ . (1.12)

Moreover we have arg z = {θ : z = reiθ}.

� Example 1.10 If z =−√

3− i, then z = 2ei(−5π/6).�

Proposition 1.3.2 Suppose that z1 = r1eiθ1 and z2 = r2eiθ2 . Therefore

z1z2 = r1eiθ1r2eiθ2

= r1r2ei(θ1+θ2)

= r1r2[cos(θ1 +θ2)+ isin(θ1 +θ2)].

Theorem 1.3.3 Suppose that z1 = r1eiθ1 6= 0 and z2 = r2eiθ2 6= 0. Thusarg z1z2 = arg z1+ arg z2.

Proof.


Using z = reiθ , we obtain the following results:(1)

z−1 =1z=

1reiθ =

1r

e−iθ =1r[cos(−θ)+ isin(−θ)],

(2)

z = r(cosθ − isinθ) = r[cos(−θ)+ isin(−θ)] = re−iθ ,

(3)

z1

z2=

r1

r2[cos(θ1−θ2)+ isin(θ1−θ2)] =

r1

r2ei(θ1−θ2).

� Example 1.11 Let z = 1+ i, then r = |z|=√

2 and θ = Argz = π

4 . Therefore

z−1 =1√2[cos(−π

4)+ isin(−π

4)] =

1√2

[√22− i

√2

2

].

�

� Example 1.12 Let z1 = 8i, and z2 = 1+ i√

3, then z1 = 8(cos π

2 + isin π

2 ) and z2 =2(cos π

3 + isin π

3 ). Therefore

z1

z2=

82

[cos(

π

2− π

3

)+ isin

(π

2− π

3

)]= 4

(cos

π

6+ isin

π

6

)= 2

√3+2i.

�


Exercises 1.31. Find Argz for the following values of z.

1.1 (1− i)−2

1.2 (−√

3+ i)2

1.32

1+ i√

3

1.41+ i√

3(1+ i)2

2. Use exponential notation to show that2.1 (

√3− i)(1+ i

√3) = 2

√3+2i

2.2 (1+ i)3 =−2+2i2.3 2i(

√3+ i)(1+ i

√3) =−8

2.48

1+ i= 4+4i

3. Represent the following complex numbers in polar form.3.1 −2

√3−2i

3.21

(1− i)2

3.36

1+√

33.4 (5+5i)3

4. Show that arg z1 = arg z2 iff z2 = cz1 for some positive real number c.5. Show that if Arg z1, Arg z2 ∈

[−π

2,π

2

], then Arg (z1z2) = Arg z1+ Arg z2.

6. Show that arg(z1z2) = arg z1 – arg z2.7. Show that Arg z−w =−Arg(z−w) iff z−w is not a negative real number.


1.4 De Moivre’s FormulaThe important players in this regard are the exponential and polar forms of a nonzerocomplex number z = reiθ = r(cosθ + isinθ). By the laws of exponentiation that we willproved in Chapter 5, we have the following results:

zn = (reiθ )n = rneinθ = rn[cos(nθ)+ isin(nθ)]

andz−n = (reiθ )−n = r−ne−inθ = r−n[cos(−nθ)+ isin(−nθ)].

� Example 1.13 Show that (−√

3− i)3 =−8i �

Proof.

� Example 1.14 Evaluate (−√

3− i)30. �

Solution.

An interesting application of the laws of exponents comes from putting the equation(eiθ )n = einθ in its polar form. Doing so gives

(cosθ + isinθ)n = cosnθ + isinnθ , (1.13)

which is known as De Moivre’s formula.

� Example 1.15 Use De Moivre’s formula to show that

cos5θ = cos5θ −10cos3

θ sin2θ +5cosθ sin4

θ .

�

1.4 De Moivre’s Formula 17

Theorem 1.4.1 — Fundamental Theorem of Algebra. If P(z) is a polynomial ofdegree n(n > 0) with complex coefficients, then the equation P(z) = 0 has precisely n(not necessarily distinct) solutions

Proof. Refer to Chapter 6.

� Example 1.16 Find roots of the polynomial P(z) = z3 +(2−2i)z2 +(−1−4i)z−2. �

Theorem 1.4.2 — nth roots of unity. The n solutions of zn = 1 can be expressed as

zk = ei 2kπ

n = cos2kπ

n+ isin

2kπ

n, for k = 0,1,2, ...,n−1. (1.14)

(They are called the nth roots of unity.)When k = 1 we call

ωn = z1 = ei 2π

n = cos2π

n+ isin

2π

n

the primitive nth root of unity.

R By De Moivre’s formula, the nth roots of unity can be expressed as

1,ωn,ω2n , ...,ω

n−1n .

� Example 1.17 The solutions of the equation z8 = 1 are

zk = ei 2kπ

8 = cos2kπ

8+ isin

2kπ

8, for k = 0,1,2, ...,7.

Therefore z0 = cos0+ isin0 = 1,z1 = cos 2π

8 + isin 2π

8 =√

22 + i

√2

2 ,

z2 = cos 4π

8 + isin 4π

8 = i,

z3 = cos 6π

8 + isin 6π

8 =−√

22 + i

√2

2 ,

z4 = cos 8π

8 + isin 8π

8 =−1,

z5 = cos 10π

8 + isin 10π

8 =−√

22 − i

√2

2 ,

z6 = cos 12π

8 + isin 12π

8 =−i,

z7 = cos 14π

8 + isin 12π

8 =√

22 − i

√2

2 .


The primitive 8th root of unity is

ω8 = ei 2π

8 = ei π

4 = cosπ

4+ isin

π

4=

√2

2+ i

√2

2.

�

Theorem 1.4.3 — Roots of the Nonzero Complex Number. The n solutions of zn =c, where c is a nonzero complex number can be expressed as

zk = ρ1n ei φ+2kπ

n = ρ1n

[cos

φ +2kπ

n+ isin

φ +2kπ

n

], (1.15)

for k = 0,1,2, . . . ,n−1, where ρ = |c| and φ ∈ arg c.

� Example 1.18 Find all cube roots of 8i = 8(cos π

2 + isin π

2 ) �

Solution.

1.4 De Moivre’s Formula 19

Exercises 1.41. Calculate the following.

1.1 (1− i√

3)3(√

3+ i)2

1.2(1+ i)3

(1− i)5

1.3 (√

3+ i)6

2. Show that (√

3+ i)4 =−8+ i8√

3.3. Find all roots in both polar and rectangular form for each expression.

3.1 (−2+2i)13

3.2 (−1)15

3.3 (−64)14

3.4 (16i)14

4. Let z be any nonzero complex number and n be an integer. Show that zn +(z̄)n is areal number.

5. Let z = i be a root of z4−4z3 +6z2−4z+5 = 0. Find another roots of the equation.6. Solve the equation (z+1)3 = z3.7. Find the three solutions to z

32 = 4

√2+ i4

√2.


1.5 The Topology of Complex NumbersDefinition 1.5.1 — Curve in the Complex Plane. A curve is the range of a continu-ous complex-valued function z(t) defined on the interval [a,b]. That is a curve C is therange of a function given by

z(t) = (x(t),y(t)) = x(t)+ iy(t), for a≤ t ≤ b,

where both x(t) and y(t) are continuous real-valued functions.We specify a curve C as

C : z(t) = (x(t),y(t)) = x(t)+ iy(t), for a≤ t ≤ b,

and say that z(t) is a parametrization of a curve C. Note that, with this parametriza-tion, we are specifying a direction for the curve C, saying that C is a curve thatgoes from the initial point z(a) = (x(a),y(a)) = x(a)+ iy(a) to the terminal pointz(b) = (x(b),y(b)) = x(b)+ iy(b).

If we had another function whose range was the same set of point z(t) but whose initialand final points were reversed, we would indicate the curve that this function defines by−C.

� Example 1.19 Find parametrizations for C and−C, where C is the straight-line segmentbegining at z0 = (x0,y0) and ending at z1 = (x1,y1). �

Solution.

Definition 1.5.2 — Closed Curve and Simple Curve. (a) A curve C having theproperty that z(a) = z(b) is said to be a closed curve.(b) The curve C : z(t) for a ≤ t ≤ b, is said to be simple provided that z(t1) 6= z(t2)whenever t1 6= t2, except possibly when t1 = a and t2 = b.

� Example 1.20 Show that the circle C with center z0 = x0 + iy0 and radius R can beparametrized to form a simple closed curve. �

1.5 The Topology of Complex Numbers 21

Definition 1.5.3 — Open Disk and ε neighborhood. An ε neighborhood is theopen disk of radius ε > 0 centered at the point z0 and is denoted by Dε(z0). That is

Dε(z0) = {z : |z− z0|< ε}.

� Example 1.21 (a) The solution set of the inequality |z|< 1 is the open disk D1(0).(b) The solution set of the inequality |z− i|< 2 is the open disk D2(i).(c) The solution set of the inequality |z+1+2i|< 3 is the open disk D3(−1−2i). �

Definition 1.5.4 — Closed Disk and Punctured Disk. (a) The disk Dε(z0) := {z :|z− z0| ≤ ε} is called the closed disk of radius ε centered at z0.(b) The disk D∗ε(z0) := {z : 0 < |z− z0| ≤ ε} is called the punctured disk of radius ε

centered at z0.

Definition 1.5.5 — Interior-Exterior-Boundary Points. (a) The point z0 is said to bean interior point of the set S provided that there exists ε > 0 such that

Dε(z0)⊂ S.

The set of all interior points of S is denoted by intS ; that is

intS = {z0 : there exists ε > 0 such that Dε(z0)⊂ S}

(b) The point z0 is called an exterior point of the set S if there exists ε > 0 suchthat

Dε(z0)⊂ C\S.

The set of all exterior points of S is denoted by extS ; that is

extS = {z0 : there exists ε > 0 such that Dε(z0)⊂ C\S}.

(c) The point z0 is called a boundary point of the set S if for each ε > 0,

Dε(z0)∩S 6= /0 and Dε(z0)∩C\S 6= /0.


The set of all boundary points of S is denoted by bdS ; that is

bdS = {z0 : for each ε > 0,Dε(z0)∩S 6= /0 and Dε(z0)∩C\S 6= /0}.� Example 1.22 Let S = D1(0) = {z : |z| < 1}. Find the set of interior, exterior, andboundary points of S. �

Solution.

Definition 1.5.6 — Accumulation Point. The point z0 is called an accumulationpoint of a set S if, for each ε ,

D∗ε(z0)∩S 6= /0.

Definition 1.5.7 — Some Topological Structures of Sets. (a) A set S is called anopen set if every point of S is an interior point of S ; that is S = int S.(b) A set S is called a closed set if bd S⊂ S.(c) A set S is said to be a connected set if every pairs of points z1 and z2 contained in Scan be joined by a curve that lies entirely in S.(d) A set S is called the domain if S is a connected open set.

� Example 1.23 Show that the open disk D1(0) = {z : |z|< 1} is a connected open set. �

Solution.

1.5 The Topology of Complex Numbers 23

� Example 1.24 The annular A = {z : 1 < |z|< 2} is a connected open set. �

Solution.

� Example 1.25 The set B = {z : |z+2|< 1 or |z−2|< 1} is not a connected set. �

Solution.

� Example 1.26 Show that the right half-plane H = {z : Re(z)> 0} is a domain. �

Solution.


Exercises 1.51. Find a parametrization of the line that

1.1 joins the origin to the point 1+ i.1.2 joins the point 1 to the point 1+ i.1.3 joins the point i to the point 1+ i.1.4 joins the point 2 to the point 1+ i.

2. Sketch the curve z(t) = t2 +2t + i(t +1)2.1 for −1≤ t ≤ 0.2.2 for 1≤ t ≤ 2.

3. Show that D1(0) is a domain but {z : |z| ≤ 1} is not.4. Let S be a set of finite complex numbers. Show that S is bounded.5. Which of the following terms apply to the sets listed below: open; connected;

domain; region; closed region; bounded.5.1 {z : Re(z)> 1} .5.2 {z :−1 < Im(z)≤ 2}.5.3 {z : |z−2− i| ≤ 2}.5.4 {z : |z+3i|> 1}.5.5 {reiθ : 0 < r < 1 and − π

2< θ <

π

2}.

5.6 {reiθ : r > 1 andπ

4< θ <

π

3}.

6. Let S be the open set consisting of all points z such that |z+2| < 1 or |z−2| < 1.Show that S is not connected.

7. Prove that the only accumulation point of{

in

: n ∈ N}

is 0.

2. Complex Functions

2.1 Functions and Linear MappingsExercises 2.1

2.2 Limits and Continuity of Complex FunctionsExercises 2.2

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