Complex Numbers 2.pdf

ALGEBRA OF COMPLEX NUMBERS II
Dr. Gabriel Obed Fosu
Department of Mathematics
Kwame Nkrumah University of Science and Technology
Google Scholar: https://scholar.google.com/citations?user=ZJfCMyQAAAAJ&hl=en&oi=ao
ResearchGate ID: https://www.researchgate.net/profile/Gabriel_Fosu2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 1 / 36

Lecture Outline
1 Polar Form of Complex Numbers
Trigonometry Form
Exponential Form
The power of a complex number
2 Roots of Complex Numbers
3 Complex Logarithm

Polar Form of Complex Numbers
Outline of Presentation
Trigonometry Form
Exponential Form
3 Complex Logarithm

Polar Form of Complex Numbers Trigonometry Form
Definition (Trigonometric Form)
The trigonometric form of z = x +i y is z = r (cosθ +i sinθ) where ar g(z) = θ and ∥z∥ = r.

Trigonometry and complex number
1 For a complex number z = x + yi we can write the trigonometric representation z =
r(cosθ∗
+ i sinθ∗
), where r ∈ [0,∞) and θ∗
∈ [0,2π) are the polar coordinates of the
geometric image of z.

r(cosθ∗
+ i sinθ∗
2 The polar argument θ∗
of the geometric image of z is called the argument of z, denoted
by argz.

r(cosθ∗
+ i sinθ∗
by argz.
3 The polar radius r of the geometric image of z is equal to the modulus of z.

r(cosθ∗
+ i sinθ∗
by argz.
3 The polar radius r of the geometric image of z is equal to the modulus of z.
4 The set Ar gz = {θ : θ∗
+2kπ, k ∈ Z} is called the extended argument of the complex
number z. Then
z = r(cosθ +i sinθ) = z = r[cos
¡
θ∗
+2kπ
¢
+i sin
¡
θ∗
+2kπ
¢
] (1)

Example
Find the trigonometric representation of the following numbers and determine their extended
argument:
1. z1 = 2+2i
2. z2 = −1−i
3. z3 = −1+i
p
3
4. z4 = 1+i

Example
argument:
1. z1 = 2+2i
2. z2 = −1−i
3. z3 = −1+i
p
3
4. z4 = 1+i
r1 =
p
22 +22 = 2
p
2

Example
argument:
1. z1 = 2+2i
2. z2 = −1−i
3. z3 = −1+i
p
3
4. z4 = 1+i
r1 =
p
22 +22 = 2
p
2
θ∗
1 = tan−1
(1) =
π
4

Example
argument:
1. z1 = 2+2i
2. z2 = −1−i
3. z3 = −1+i
p
3
4. z4 = 1+i
r1 =
p
22 +22 = 2
p
2
θ∗
1 = tan−1
(1) =
π
4
z1 = 2
p
2
³
cos
π
4
+i sin
π
4
´

Example
argument:
1. z1 = 2+2i
2. z2 = −1−i
3. z3 = −1+i
p
3
4. z4 = 1+i
r1 =
p
22 +22 = 2
p
2
θ∗
1 = tan−1
(1) =
π
4
z1 = 2
p
2
³
cos
π
4
+i sin
π
4
´
Ar gz1 =
nπ
4
+2kπ; k ∈ Z
o

z2 = −1−i
1 r2 =
p
(−1)2 +(−1)2 =
p
2

z2 = −1−i
1 r2 =
p
(−1)2 +(−1)2 =
p
2
2 Because is in the third quadrant
θ∗
2 = tan−1 y
x
+π
= tan−1
(1)+π
=
π
4
+π =
5π
4

z2 = −1−i
1 r2 =
p
(−1)2 +(−1)2 =
p
2
θ∗
2 = tan−1 y
x
+π
= tan−1
(1)+π
=
π
4
+π =
5π
4
3 z2 =
p
2
µ
cos
5π
4
+i sin
5π
4
¶

z2 = −1−i
1 r2 =
p
(−1)2 +(−1)2 =
p
2
θ∗
2 = tan−1 y
x
+π
= tan−1
(1)+π
=
π
4
+π =
5π
4
3 z2 =
p
2
µ
cos
5π
4
+i sin
5π
4
¶
4 Ar gz2 =
½
5π
4
+2kπ; k ∈ Z
¾

z3 = −1+i
p
3

z3 = −1+i
p
3
r3 =
q
(−1)2 +
p
3
2
= 2

z3 = −1+i
p
3
r3 =
q
(−1)2 +
p
3
2
= 2
θ∗
3 = tan−1
(−
p
3)+π =
π
3
+π =
2π
3

z3 = −1+i
p
3
r3 =
q
(−1)2 +
p
3
2
= 2
θ∗
3 = tan−1
(−
p
3)+π =
π
3
+π =
2π
3
z3 = 2
µ
cos
2π
3
+i sin
2π
3
¶

z3 = −1+i
p
3
r3 =
q
(−1)2 +
p
3
2
= 2
θ∗
3 = tan−1
(−
p
3)+π =
π
3
+π =
2π
3
z3 = 2
µ
cos
2π
3
+i sin
2π
3
¶
Ar gz3 =
½
2π
3
+2kπ; k ∈ Z
¾

1 + i
z = 1+i has modulus
p
2 and argument arg(z) =
π
4
+2kπ,k ∈ Z.

1 + i
z = 1+i has modulus
p
2 and argument arg(z) =
π
4
+2kπ,k ∈ Z.
In other words, the trigonometric form of z can be any one of the following
z =
p
2(cosπ/4+i sinπ/4); k = 0 (2)
=
p
2(cos(π/4±2π)+i sin(π/4±2π)); k = ±1 (3)
=
p
2(cos(π/4±4π)+i sin(π/4±4π)); k = ±2 (4)
=
p
2(cos(π/4±6π)+i sin(π/4±6π)); k = ±3 (5)
=
.
.
. (6)
If there is no ambiguity, we choose the argument between 0 and 2π. Thus
z =
p
2(cosπ/4+i sinπ/4) (7)

Remarks
1 = cos0+i sin0

Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2

Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−1 = cosπ+i sinπ

Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−i = cos
3π
2
+i sin
3π
2

Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−i = cos
3π
2
+i sin
3π
2
If z1 = r1(cosθ1 +i sinθ1) and z2 = r2(cosθ2 +i sinθ2) then
z1z2 = r1r2 [cos(θ1 +θ2)+i sin(θ1 +θ2)]

Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−i = cos
3π
2
+i sin
3π
2
If z1 = r1(cosθ1 +i sinθ1) and z2 = r2(cosθ2 +i sinθ2) then
z1z2 = r1r2 [cos(θ1 +θ2)+i sin(θ1 +θ2)]
If z1 = r1(cosθ1 +i sinθ1) and z2 = r2(cosθ2 +i sinθ2) ̸= 0 then
z1
z2
=
r1
r2
[cos(θ1 −θ2)+i sin(θ1 −θ2)]

Example
Let z1 = 1−i and z2 =
p
3+i, then
z1 =
p
2
µ
cos
7π
4
+i sin
7π
4
¶
,

Example
p
3+i, then
z1 =
p
2
µ
cos
7π
4
+i sin
7π
4
¶
, z2 = 2
³
cos
π
6
+i sin
π
6
´

Example
p
3+i, then
z1 =
p
2
µ
cos
7π
4
+i sin
7π
4
¶
, z2 = 2
³
cos
π
6
+i sin
π
6
´
and
z1z2 = 2
p
2
·
cos
µ
7π
4
+
π
6
¶
+i sin
µ
7π
4
+
π
6
¶¸
= 2
p
2
µ
cos
23π
12
+i sin
23π
12
¶

Properties
Let z1 and z2 be two non-zero complex numbers. We have
1 z2 = z1 ⇐⇒ ∥z2∥ = ∥z1∥ and arg(z2) = arg(z1)+2kπ,k ∈ Z.
2 arg(z1z2) = arg(z1)+arg(z2)+2kπ.
3 arg(zn
1 ) = n arg(z1)+2kπ.
4 arg(1/z1) = −arg(z1)+2kπ.
5 arg( ¯
z1) = −arg(z1)+2kπ.

Polar Form of Complex Numbers Exponential Form
Euler and De Moivre’s Form
Definition (Euler’s equation)
The relation
eiθ
= cosθ +i sinθ (8)
is called Euler’s equation.

The relation
eiθ
is called Euler’s equation. Thus, if z ∈ C−{0}
(Trigonometric form) z = r(cosθ +i sinθ) ⇐⇒ (Exponential form): z = reiθ
.

The relation
eiθ
is called Euler’s equation. Thus, if z ∈ C−{0}
(Trigonometric form) z = r(cosθ +i sinθ) ⇐⇒ (Exponential form): z = reiθ
.
Also, because any two arguments for a give complex number differ by an integer multiple
of 2π we will sometimes write the exponential form as,
z = rei(θ+2πk)
, k = 0,±1,±2,··· (9)

Forms
1 Standard form : z = x +i y
2 Trigonometric form : z = r(cosθ +i sinθ) where r = ∥z∥ and arg(z) = θ +2kπ.
3 Exponential form: z = reiθ
where r = ∥z∥ and arg(z) = θ +2kπ.

Forms
1 Standard form : z = x +i y
2 Trigonometric form : z = r(cosθ +i sinθ) where r = ∥z∥ and arg(z) = θ +2kπ.
3 Exponential form: z = reiθ
where r = ∥z∥ and arg(z) = θ +2kπ.
Example
z = 2+2i
p
3 has modulus ∥z∥ = 4 and argument θ = π/3.
Therefore,
z = 4eiπ/3
(10)
is its exponential form.

De Moivre’s and Euler
De Moivre’s theorem
For all real number θ and all integer n,
(cosθ +i sinθ)n
= cos(nθ)+i sin(nθ). (11)

De Moivre’s and Euler
De Moivre’s theorem
For all real number θ and all integer n,
(cosθ +i sinθ)n
= cos(nθ)+i sin(nθ). (11)
Definition (Euler’s formula)
cosθ =
eiθ
+e−iθ
2
cos(nθ) =
einθ
+e−inθ
2
(12)
sinθ =
eiθ
−e−iθ
2i
sin(nθ) =
einθ
−e−inθ
2i
. (13)

Trigonometric identities
Let us use De Moivre’s formula to express cos2θ in terms of cosθ and sinθ

cos2θ +i sin2θ = (cosθ +i sinθ)2
(by the theorem) (14)
= cos2
θ −sin2
θ +2i cosθsinθ (15)

= cos2
θ −sin2
We observe that,
cos2θ = cos2
θ −sin2
θ (16)
and
sin2θ = 2cosθsinθ (17)

= cos2
θ −sin2
We observe that,
cos2θ = cos2
θ −sin2
θ (16)
and
sin2θ = 2cosθsinθ (17)
Exercise
Express cos3θ and sin4θ in terms of cosθ and sinθ.

Polar Form of Complex Numbers The power of a complex number
De Moivre’s Revisited
For z = r(cosθ +i sinθ) and n ∈ N, we have
zn
= rn
(cosθ +i sinθ)n
zn
= rn
(cosnθ +i sinnθ)

Example
If z = 2+2i
p
3 find z3
This has modulus ∥z∥ = 4 and argument θ = π/3. so

Example
If z = 2+2i
p
3 find z3
zn
= rn
(cosnθ +i sinnθ) (18)
z3
= 43
(cos3(π/3)+i sin3(π/3)) (19)
z3
= 64(cos(π)+i sin(π)) (20)
z3
= −64 (21)

Example
If z = 2+2i
p
3 find z3
zn
= rn
(cosnθ +i sinnθ) (18)
z3
= 43
(cos3(π/3)+i sin3(π/3)) (19)
z3
= 64(cos(π)+i sin(π)) (20)
z3
= −64 (21)
Thus
(2+2i
p
3)3
= −64

Example
Let us compute (1+i)1000

Example
The trigonometric representation of 1+i is
p
2(cosπ/4+i sinπ/4). Applying de Moivre’s
formula we obtain

Example
The trigonometric representation of 1+i is
p
2(cosπ/4+i sinπ/4). Applying de Moivre’s
formula we obtain
(1+i)1000
=
p
2
1000
(cos1000(π/4)+i sin1000(π/4))
= 2500
(cos250π+i sin250π)
= 2500

Roots of Complex Numbers
Trigonometry Form
Exponential Form
3 Complex Logarithm

Definition (Complex polynomials)
Polynomials with complex coefficients and unknown are called complex polynomials.
P(z) = iz4
−2z +i
p
2+4 is a complex polynomial of degree 4.

P(z) = iz4
−2z +i
p
Theorem (Fundamental Theorem of Algebra)
Every complex polynomial equation of degree n has exactly n complex roots.

P(z) = iz4
−2z +i
p
Theorem (Fundamental Theorem of Algebra)
Every complex polynomial equation of degree n has exactly n complex roots.
The roots of the complex polynomial z3
−3z2
+2z are 0,1,2. Note degree = 3 and number
of roots = 3.

The nth
Roots of Unity
Consider a positive integer n ≥ 2 and a complex number z0 ̸= 0. As in the field of real
numbers, the equation
Zn
− z0 = 0 =⇒ Z = n
p
z0 (22)
is used for defining the nth
roots of number z0.
Hence we call any solution Z of the equation (22) an nth
root of the complex number z0.

Definition (Trigonometric)
Let z0 = r(cosθ +i sinθ) be a complex number with r > 0 and θ ∈ [0,2π), then the number z0
has n distinct nth
roots given by the formulas
Zk = n
p
r
µ
cos
θ +2kπ
n
+i sin
θ +2kπ
n
¶
(23)
where k = 0,1,··· ,n −1

Definition (Trigonometric)
Let z0 = r(cosθ +i sinθ) be a complex number with r > 0 and θ ∈ [0,2π), then the number z0
has n distinct nth
roots given by the formulas
Zk = n
p
r
µ
cos
θ +2kπ
n
+i sin
θ +2kπ
n
¶
(23)
where k = 0,1,··· ,n −1
Alternatively definition for Exponential
If we had considered z = reθi
, the roots should have been
Zk = n
p
re
θ+2kπ
n
i
= n
p
r exp
µ
θ +2kπ
n
i
¶
; k = 0,1,··· ,n −1 (24)

Example
Let us find the third roots of the number z = 1+i

Example
The trigonometric representation of z = 1+i is
z =
p
2(cosπ/4+i sinπ/4)

Example
z =
p
The cube roots of the number z are
Zk =
3
q
p
2
·
cos
µ
π
12
+k
2π
3
¶
+i sin
µ
π
12
+k
2π
3
¶¸
;k = 0,1,2

Example
z =
p
The cube roots of the number z are
Zk =
3
q
p
2
·
cos
µ
π
12
+k
2π
3
¶
+i sin
µ
π
12
+k
2π
3
¶¸
;k = 0,1,2
or in explicit form as
Z0 =
6
p
2
³
cos
π
12
+i sin
π
12
´
Z1 =
6
p
2
µ
cos
3π
4
+i sin
3π
4
¶
Z2 =
6
p
2
µ
cos
17π
12
+i sin
17π
12
¶

Exercise
Find the argument and the standard form of the fourth roots of a = cos
2π
3
+i sin
2π
3
.
Solution
arg(z) =
½
π
6
,
4π
6
,
7π
6
,
10π
6
¾

The nth
roots of unity
The roots of the equation Zn
−1 = 0 are called the nth
roots of unity. Since
1 = cos0+i sin0, from the formulas for the nth
roots of a complex number (23) we derive
that the nth
roots of unity are
Zk = cos
2kπ
n
+i sin
2kπ
n
, k = 0,1,2,··· ,n −1 (25)

The nth
roots of unity
that the nth
roots of unity are
Zk = cos
2kπ
n
+i sin
2kπ
n
, k = 0,1,2,··· ,n −1 (25)
Simplified as:
z0 = 1, z1 = e
2
n
πi
, z2 = e
4
n
πi
, ...,zn−1 = e
2(n−1)
n
πi
(26)

The nth
roots of unity
that the nth
roots of unity are
Zk = cos
2kπ
n
+i sin
2kπ
n
, k = 0,1,2,··· ,n −1 (25)
Simplified as:
z0 = 1, z1 = e
2
n
πi
, z2 = e
4
n
πi
, ...,zn−1 = e
2(n−1)
n
πi
(26)
The nth of unity in Exponential form
Zk = e
2kπ
n
i
= exp
µ
2kπ
n
i
¶
k = 0,1,2,··· ,n −1 (27)

Example
If Z is a complex number, find the roots of Z3
= 1
Solution
1 Zk = e
2kπ
n
i
= exp
³
2kπ
n i
´
k = 0,1,2,··· ,n −1

Example
= 1
Solution
1 Zk = e
2kπ
n
i
= exp
³
2kπ
n i
´
k = 0,1,2,··· ,n −1
2 Z0 = exp
³
2(0)π
3 i
´
= 1

Example
= 1
Solution
1 Zk = e
2kπ
n
i
= exp
³
2kπ
n i
´
k = 0,1,2,··· ,n −1
2 Z0 = exp
³
2(0)π
3 i
´
= 1
3 Z1 = exp
³
2(1)π
3 i
´
= cos(2/3π)+i sin(2/3π)

Example
= 1
Solution
1 Zk = e
2kπ
n
i
= exp
³
2kπ
n i
´
k = 0,1,2,··· ,n −1
2 Z0 = exp
³
2(0)π
3 i
´
= 1
3 Z1 = exp
³
2(1)π
3 i
´
= cos(2/3π)+i sin(2/3π)
4 Z2 = exp
³
2(2)π
3 i
´
= cos(4/3π)+i sin(4/3π)

Complex Logarithm
Trigonometry Form
Exponential Form
3 Complex Logarithm

Complex Logarithm
Complex Logarithm
Consider z = rei(θ+2kπ)
. By assuming that lnz exists, we use the properties of the logarithm
function to simplify z.
We get lnz = ln
¡
rei(θ+2kπ)
¢
= lnr +i(θ +2kπ).

Complex Logarithm
Complex Logarithm
Consider z = rei(θ+2kπ)
. By assuming that lnz exists, we use the properties of the logarithm
function to simplify z.
We get lnz = ln
¡
rei(θ+2kπ)
¢
= lnr +i(θ +2kπ).
Definition
The complex logarithm of a complex number z = rei arg(z)
is defined as
lnz = lnr +iθ (28)
where θ is the argument of z that lies in the range [−π, π].
In general, we use
ln(z) = lnr +i(θ+2kπ) (29)

Complex Logarithm
Example
ln(−i) = lne−iπ/2
(30)
= ln1−i
³π
2
+2kπ
´
(31)
= −
π
2
i (32)

Complex Logarithm
Example
ln(−i) = lne−iπ/2
(30)
= ln1−i
³π
2
+2kπ
´
(31)
= −
π
2
i (32)
Example
ln
³
3e
5πi
3
´
= ln3+i(
5π
3
+2kπ) (33)
= ln3+i
µ
5π
3
−2π
¶
(34)
= ln3−
π
3
i (35)

Complex Logarithm
Example
Find the trigonometric form of z = 2i

Complex Logarithm
Example
ln(z) = ln
³
2i
´
(36)
= i ln(2) = i ln2 (37)

Complex Logarithm
Example
ln(z) = ln
³
2i
´
(36)
= i ln(2) = i ln2 (37)
z = eln(z)
(38)
= ei ln2
(39)
= cos(ln2)+i sin(ln2) (40)

Complex Logarithm
Example
Find the trigonometric form of z = i2i+3
ln(z) = (2i +3)ln(i) (41)
ln(z) = (2i +3)i(
π
2
), from(32) ln(i) = i(
π
2
) (42)
ln(z) = −π+i(
3π
2
) (43)
z = eln(z)
= exp
µ
−π+i(
3π
2
)
¶
(44)
z = e−π
ei( 3π
2
)
(45)
z = e−π
µ
cos
3π
2
+i sin
3π
2
¶
. (46)

Complex Logarithm
Alternative Solution
z = i2i+3
(47)
=
³
e
π
2
i
´2i+3
, NB i = e
π
2
i
= cos(90)+i sin(90) (48)
= e−2 π
2
+ 3π
2
i
(49)
= e−π
e
3π
2
i
(50)
= e−π
µ
cos
3π
2
+i sin
3π
2
¶
. (51)

Complex Logarithm
Exercise
1 Find the square and cubic roots of the following complex numbers:
1 z = 1+i
2 z = i
3 z = 1/
p
2+i/
p
2
2 Find the fourth roots of the following complex numbers:
1 z = −2i
2 z =
p
3+i
3 z = −7+24i
3 Solve the equations:
1 z3
−125 = 0
2 z4
+16 = 0
3 z7
−2iz4
−iz3
−2 = 0;

Complex Logarithm
Exercise
1 Find the polar coordinates for the following points, given their cartesian coordinates
1 M1 = (−3,3);
2 M2 = (−4
p
3,−4);
3 M3 = (0,−5);
2 Find the cartesian coordinates for the following points, given their polar coordinates:
1 P1 = (2,π/3)
2 P2 = (3,−π)
3 P3 = (1,π/2)
3 Find polar representations for the following complex numbers
1 z1 = 6+6i
p
3
2 z2 = −4i
3 z3 = cosa −i sina, a ∈ [0,2π)

END OF LECTURE
THANK YOU

Complex Numbers 2.pdf

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