Theta θ(g,x) and pi π(g,x) polynomials of hexagonal trapezoid system tb,aijcsa
A counting polynomial, called Omega Ω(G,x), was proposed by Diudea. It is defined on the ground of
“opposite edge strips” ops. Theta Θ(G,x) and Pi Π(G,x) polynomials can also be calculated by ops
counting. In this paper we compute these counting polynomials for a family of Benzenoid graphs that called
Hexagonal trapezoid system Tb,a.
Theta θ(g,x) and pi π(g,x) polynomials of hexagonal trapezoid system tb,aijcsa
A counting polynomial, called Omega Ω(G,x), was proposed by Diudea. It is defined on the ground of
“opposite edge strips” ops. Theta Θ(G,x) and Pi Π(G,x) polynomials can also be calculated by ops
counting. In this paper we compute these counting polynomials for a family of Benzenoid graphs that called
Hexagonal trapezoid system Tb,a.
HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2
/2
Ex.5 Evaluate : sin2
x dx .
d
If dx
[f(x)] =
(x) and a and b, are two values
0
/2
Sol. sin2 x dx
independent of variable x, then 0
b /2 FG1 cos 2xIJ
(x) dx = f(x) a = f(b) – f(a)
a
= zH 2 dx
1 LM sin 2x O /2
is called Definite Integral of (x) within limits
= x
2
2 PQ
a and b. Here a is called the lower limit and b is called the upper limit of the integral. The interval [a,b] is known as range of integration. It should be noted that every definite integral has
a unique value.
= 1 LM 0
0
= / 4 Ans.
1
x2
Ex.6 Evaluate : xe dx.
0
1
2 x2
Ex.1 Evaluate : x4 dx.
1
Sol. xe dx
0
2 Lx5 O2
1 ex2 1
Sol.
zx4 dx = MP= 32 – 1 = 31
=
Ans. 2 0
MN5 PQ1 5 5 5
/4
= 1 (e –1) Ans.
2
Ex.2 Evaluate : sec2 x dx.
0
1 x3
/4
Sol. sec2 x .dx = tan x /4 = tan / 4 – tan 0 = 1
0
Ex.7 Find the value of
0
1 x8
dx.
0
Ans.
2 1
Sol. Let x4 = t, then 4x3 dx = dt
Ex.3 Evaluate :
1
4 x2
dx.
I =
1 1 dt
4 =
1 [ sin–1 t] 1 =
4 8
2 1 L 2 0
Sol. z dx = Msin GJP
Ans.
1 4 x2
N H2KQ1
= sin–1 (1) – sin–1 (1/2)
z/3 cos x
= – =
Ans.
Ex.8 Evaluate :
0
3 4 sin x dx.
2 6 3
Sol. I =
/3 cos x 3 4 sin x dx.
Ex.4 Evaluate :
z2 1
2 dx
0 4 x2
0
Let 3 + 4 sin x = t 4 cos x. dx = dt
cos x dx = dt/4
Now in the given integral x lies between the
Sol.
4 x2 dx
limit x = 0 to x = / 3 . Now we will decide the limit of t.
1
= tan
2
1 x OP2
0
In 3 + 4 sin x = t, by putting lower limit of x as x = 0; and upper limit as x = / 3 . We
= 1 tan11 0 = / 8 Ans.
2
get lower and upper limit of t respectively.
Putting x = 0 3 + 4 sin 0 = t t = 3
z3 z2
z3 (x) dx
/3 cos x
dx =
zt3 2 3 1 dt
= 2 x2dx +
0
z3b3x 4gdx
0 3 4 sin x
t3 t 4
Fx3 I2 F3x2 I3
= 1 zt3 2 3 1 dt
= GJ+ G
4xJ
4 t3 t
H3 K H2
= 1 log t 32
4
= 8 +
3
27 – 12 – 6 + 8
2
= 1 [ log (3 + 2
4
) – log 3] Ans.
= 37/6 Ans.
Ex.9
sin(tan1 x)
2
dx equals-
Ex.11 Evaluate : |1 x|dx.
0
0 1 x
Sol. Put tan
x = t, then 1 dx = dt
Sol. |1 x| =
RST1 x, when
0 x 1
–1
(1 x2 )
I =
x 1, when 1 x 2
1b1 xgdx + 2 bx 1gdx
/ 2
I = sin t dt [– cos t] /2 = 1 Ans.
0 1
L x2 O1 Lx2 O2
0 Mx
P M xP
0 = MN
2 PQ+
MN2
PQ1
= b1/ 2 0 + b0 1/ 2 = 1 Ans.
z z
i.e. the value of a definite integral remains unchanged if its variable is placed by any other symbol.
[P-4] f(x) dx = f(a x) dx .
0 0
Note :
[P-2]
b
f(x) dx
a
a
= – f(x) dx
b
This property can be used only when lower limit is zero. It is generally used for those complicated
i.e. the interchange of limits of a definite integral
changes only its sign.
zb zc zb
integrals whose denominators are unchanged when x is replaced by a– x. With the hel
On Triplet of Positive Integers Such That the Sum of Any Two of Them is a Per...inventionjournals
In this article we discussed determination of distinct positive integers a, b, c such that a + b, a + c, b + c are perfect squares. We can determine infinitely many such triplets. There are such four tuples and from them eliminating any one number we obtain triplets with the specific property. We can also obtain infinitely many such triplets from a single triplet.
Properties of Functions
Odd and Even Functions
Periodic Functions
Monotonic Functions
Bounded Functions
Maxima and Minima of Functions
Inverse Function
Sequence and Series
HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2
/2
Ex.5 Evaluate : sin2
x dx .
d
If dx
[f(x)] =
(x) and a and b, are two values
0
/2
Sol. sin2 x dx
independent of variable x, then 0
b /2 FG1 cos 2xIJ
(x) dx = f(x) a = f(b) – f(a)
a
= zH 2 dx
1 LM sin 2x O /2
is called Definite Integral of (x) within limits
= x
2
2 PQ
a and b. Here a is called the lower limit and b is called the upper limit of the integral. The interval [a,b] is known as range of integration. It should be noted that every definite integral has
a unique value.
= 1 LM 0
0
= / 4 Ans.
1
x2
Ex.6 Evaluate : xe dx.
0
1
2 x2
Ex.1 Evaluate : x4 dx.
1
Sol. xe dx
0
2 Lx5 O2
1 ex2 1
Sol.
zx4 dx = MP= 32 – 1 = 31
=
Ans. 2 0
MN5 PQ1 5 5 5
/4
= 1 (e –1) Ans.
2
Ex.2 Evaluate : sec2 x dx.
0
1 x3
/4
Sol. sec2 x .dx = tan x /4 = tan / 4 – tan 0 = 1
0
Ex.7 Find the value of
0
1 x8
dx.
0
Ans.
2 1
Sol. Let x4 = t, then 4x3 dx = dt
Ex.3 Evaluate :
1
4 x2
dx.
I =
1 1 dt
4 =
1 [ sin–1 t] 1 =
4 8
2 1 L 2 0
Sol. z dx = Msin GJP
Ans.
1 4 x2
N H2KQ1
= sin–1 (1) – sin–1 (1/2)
z/3 cos x
= – =
Ans.
Ex.8 Evaluate :
0
3 4 sin x dx.
2 6 3
Sol. I =
/3 cos x 3 4 sin x dx.
Ex.4 Evaluate :
z2 1
2 dx
0 4 x2
0
Let 3 + 4 sin x = t 4 cos x. dx = dt
cos x dx = dt/4
Now in the given integral x lies between the
Sol.
4 x2 dx
limit x = 0 to x = / 3 . Now we will decide the limit of t.
1
= tan
2
1 x OP2
0
In 3 + 4 sin x = t, by putting lower limit of x as x = 0; and upper limit as x = / 3 . We
= 1 tan11 0 = / 8 Ans.
2
get lower and upper limit of t respectively.
Putting x = 0 3 + 4 sin 0 = t t = 3
z3 z2
z3 (x) dx
/3 cos x
dx =
zt3 2 3 1 dt
= 2 x2dx +
0
z3b3x 4gdx
0 3 4 sin x
t3 t 4
Fx3 I2 F3x2 I3
= 1 zt3 2 3 1 dt
= GJ+ G
4xJ
4 t3 t
H3 K H2
= 1 log t 32
4
= 8 +
3
27 – 12 – 6 + 8
2
= 1 [ log (3 + 2
4
) – log 3] Ans.
= 37/6 Ans.
Ex.9
sin(tan1 x)
2
dx equals-
Ex.11 Evaluate : |1 x|dx.
0
0 1 x
Sol. Put tan
x = t, then 1 dx = dt
Sol. |1 x| =
RST1 x, when
0 x 1
–1
(1 x2 )
I =
x 1, when 1 x 2
1b1 xgdx + 2 bx 1gdx
/ 2
I = sin t dt [– cos t] /2 = 1 Ans.
0 1
L x2 O1 Lx2 O2
0 Mx
P M xP
0 = MN
2 PQ+
MN2
PQ1
= b1/ 2 0 + b0 1/ 2 = 1 Ans.
z z
i.e. the value of a definite integral remains unchanged if its variable is placed by any other symbol.
[P-4] f(x) dx = f(a x) dx .
0 0
Note :
[P-2]
b
f(x) dx
a
a
= – f(x) dx
b
This property can be used only when lower limit is zero. It is generally used for those complicated
i.e. the interchange of limits of a definite integral
changes only its sign.
zb zc zb
integrals whose denominators are unchanged when x is replaced by a– x. With the hel
On Triplet of Positive Integers Such That the Sum of Any Two of Them is a Per...inventionjournals
In this article we discussed determination of distinct positive integers a, b, c such that a + b, a + c, b + c are perfect squares. We can determine infinitely many such triplets. There are such four tuples and from them eliminating any one number we obtain triplets with the specific property. We can also obtain infinitely many such triplets from a single triplet.
Properties of Functions
Odd and Even Functions
Periodic Functions
Monotonic Functions
Bounded Functions
Maxima and Minima of Functions
Inverse Function
Sequence and Series
Matrix Arithmetic and Properties
Some Special Matrices
Nonsingular Matrices
Symmetric Matrices
Orthogonal and Orthonormal Matrix
Complex Matrices
Hermitian Matrices
Lecture Notes on Number Theory : Introduction to Numbers, Natural Numbers, Integers, Rational and Irrational Numbers, Real Number
Mathematical Induction.
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Complex Numbers 2.pdf
1. 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
2. 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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 2 / 36
3. Polar Form of Complex Numbers
Outline of Presentation
1 Polar Form of Complex Numbers
Trigonometry Form
Exponential Form
The power of a complex number
2 Roots of Complex Numbers
3 Complex Logarithm
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 3 / 36
4. 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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 4 / 36
5. Polar Form of Complex Numbers Trigonometry Form
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 5 / 36
6. Polar Form of Complex Numbers Trigonometry Form
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.
2 The polar argument θ∗
of the geometric image of z is called the argument of z, denoted
by argz.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 5 / 36
7. Polar Form of Complex Numbers Trigonometry Form
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.
2 The polar argument θ∗
of the geometric image of z is called the argument of z, denoted
by argz.
3 The polar radius r of the geometric image of z is equal to the modulus of z.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 5 / 36
8. Polar Form of Complex Numbers Trigonometry Form
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.
2 The polar argument θ∗
of the geometric image of z is called the argument of z, denoted
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 5 / 36
9. Polar Form of Complex Numbers Trigonometry Form
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 6 / 36
10. Polar Form of Complex Numbers Trigonometry Form
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
r1 =
p
22 +22 = 2
p
2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 6 / 36
11. Polar Form of Complex Numbers Trigonometry Form
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
r1 =
p
22 +22 = 2
p
2
θ∗
1 = tan−1
(1) =
π
4
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 6 / 36
12. Polar Form of Complex Numbers Trigonometry Form
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
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
´
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 6 / 36
13. Polar Form of Complex Numbers Trigonometry Form
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
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 6 / 36
14. Polar Form of Complex Numbers Trigonometry Form
z2 = −1−i
1 r2 =
p
(−1)2 +(−1)2 =
p
2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 7 / 36
15. Polar Form of Complex Numbers Trigonometry Form
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 7 / 36
16. Polar Form of Complex Numbers Trigonometry Form
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
3 z2 =
p
2
µ
cos
5π
4
+i sin
5π
4
¶
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 7 / 36
17. Polar Form of Complex Numbers Trigonometry Form
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
3 z2 =
p
2
µ
cos
5π
4
+i sin
5π
4
¶
4 Ar gz2 =
½
5π
4
+2kπ; k ∈ Z
¾
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 7 / 36
18. Polar Form of Complex Numbers Trigonometry Form
z3 = −1+i
p
3
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 8 / 36
19. Polar Form of Complex Numbers Trigonometry Form
z3 = −1+i
p
3
r3 =
q
(−1)2 +
p
3
2
= 2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 8 / 36
20. Polar Form of Complex Numbers Trigonometry Form
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 8 / 36
21. Polar Form of Complex Numbers Trigonometry Form
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
¶
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 8 / 36
22. Polar Form of Complex Numbers Trigonometry Form
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 8 / 36
23. Polar Form of Complex Numbers Trigonometry Form
1 + i
z = 1+i has modulus
p
2 and argument arg(z) =
π
4
+2kπ,k ∈ Z.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 9 / 36
24. Polar Form of Complex Numbers Trigonometry Form
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 9 / 36
25. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
26. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
27. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−1 = cosπ+i sinπ
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
28. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−1 = cosπ+i sinπ
−i = cos
3π
2
+i sin
3π
2
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
29. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−1 = cosπ+i sinπ
−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)]
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
30. Polar Form of Complex Numbers Trigonometry Form
Remarks
1 = cos0+i sin0
i = cos
π
2
+i sin
π
2
−1 = cosπ+i sinπ
−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)]
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 10 / 36
31. Polar Form of Complex Numbers Trigonometry Form
Example
Let z1 = 1−i and z2 =
p
3+i, then
z1 =
p
2
µ
cos
7π
4
+i sin
7π
4
¶
,
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 11 / 36
32. Polar Form of Complex Numbers Trigonometry Form
Example
Let z1 = 1−i and z2 =
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
´
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 11 / 36
33. Polar Form of Complex Numbers Trigonometry Form
Example
Let z1 = 1−i and z2 =
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
¶
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 11 / 36
34. Polar Form of Complex Numbers Trigonometry Form
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 12 / 36
35. 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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 13 / 36
36. 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. Thus, if z ∈ C−{0}
(Trigonometric form) z = r(cosθ +i sinθ) ⇐⇒ (Exponential form): z = reiθ
.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 13 / 36
37. 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. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 13 / 36
38. Polar Form of Complex Numbers Exponential Form
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 14 / 36
39. Polar Form of Complex Numbers Exponential Form
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 14 / 36
40. Polar Form of Complex Numbers 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 15 / 36
41. Polar Form of Complex Numbers 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)
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 15 / 36
42. Polar Form of Complex Numbers Exponential Form
Trigonometric identities
Let us use De Moivre’s formula to express cos2θ in terms of cosθ and sinθ
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 16 / 36
43. Polar Form of Complex Numbers Exponential Form
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 16 / 36
44. Polar Form of Complex Numbers Exponential Form
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)
We observe that,
cos2θ = cos2
θ −sin2
θ (16)
and
sin2θ = 2cosθsinθ (17)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 16 / 36
45. Polar Form of Complex Numbers Exponential Form
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)
We observe that,
cos2θ = cos2
θ −sin2
θ (16)
and
sin2θ = 2cosθsinθ (17)
Exercise
Express cos3θ and sin4θ in terms of cosθ and sinθ.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 16 / 36
46. Polar Form of Complex Numbers The power of a complex number
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 17 / 36
47. Polar Form of Complex Numbers The power of a complex number
Example
If z = 2+2i
p
3 find z3
This has modulus ∥z∥ = 4 and argument θ = π/3. so
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 18 / 36
48. Polar Form of Complex Numbers The power of a complex number
Example
If z = 2+2i
p
3 find z3
This has modulus ∥z∥ = 4 and argument θ = π/3. so
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 18 / 36
49. Polar Form of Complex Numbers The power of a complex number
Example
If z = 2+2i
p
3 find z3
This has modulus ∥z∥ = 4 and argument θ = π/3. so
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 18 / 36
50. Polar Form of Complex Numbers The power of a complex number
Example
Let us compute (1+i)1000
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 19 / 36
51. Polar Form of Complex Numbers The power of a complex number
Example
Let us compute (1+i)1000
The trigonometric representation of 1+i is
p
2(cosπ/4+i sinπ/4). Applying de Moivre’s
formula we obtain
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 19 / 36
52. Polar Form of Complex Numbers The power of a complex number
Example
Let us compute (1+i)1000
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 19 / 36
53. Roots of Complex Numbers
Outline of Presentation
1 Polar Form of Complex Numbers
Trigonometry Form
Exponential Form
The power of a complex number
2 Roots of Complex Numbers
3 Complex Logarithm
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 20 / 36
54. Roots of Complex Numbers
Roots of Complex Numbers
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 21 / 36
55. Roots of Complex Numbers
Roots of Complex Numbers
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.
Theorem (Fundamental Theorem of Algebra)
Every complex polynomial equation of degree n has exactly n complex roots.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 21 / 36
56. Roots of Complex Numbers
Roots of Complex Numbers
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.
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 21 / 36
57. Roots of Complex Numbers
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.
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 22 / 36
58. Roots of Complex Numbers
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 23 / 36
59. Roots of Complex Numbers
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 23 / 36
60. Roots of Complex Numbers
Example
Let us find the third roots of the number z = 1+i
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 24 / 36
61. Roots of Complex Numbers
Example
Let us find the third roots of the number z = 1+i
The trigonometric representation of z = 1+i is
z =
p
2(cosπ/4+i sinπ/4)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 24 / 36
62. Roots of Complex Numbers
Example
Let us find the third roots of the number z = 1+i
The trigonometric representation of z = 1+i is
z =
p
2(cosπ/4+i sinπ/4)
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 24 / 36
63. Roots of Complex Numbers
Example
Let us find the third roots of the number z = 1+i
The trigonometric representation of z = 1+i is
z =
p
2(cosπ/4+i sinπ/4)
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
¶
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 24 / 36
64. Roots of Complex Numbers
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 25 / 36
65. Roots of Complex Numbers
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 26 / 36
66. Roots of Complex Numbers
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)
Simplified as:
z0 = 1, z1 = e
2
n
πi
, z2 = e
4
n
πi
, ...,zn−1 = e
2(n−1)
n
πi
(26)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 26 / 36
67. Roots of Complex Numbers
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)
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 26 / 36
68. Roots of Complex Numbers
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
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 27 / 36
69. Roots of Complex Numbers
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
2 Z0 = exp
³
2(0)π
3 i
´
= 1
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 27 / 36
70. Roots of Complex Numbers
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
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 27 / 36
71. Roots of Complex Numbers
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
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 27 / 36
72. Complex Logarithm
Outline of Presentation
1 Polar Form of Complex Numbers
Trigonometry Form
Exponential Form
The power of a complex number
2 Roots of Complex Numbers
3 Complex Logarithm
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 28 / 36
73. 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).
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 29 / 36
74. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 29 / 36
75. Complex Logarithm
Example
ln(−i) = lne−iπ/2
(30)
= ln1−i
³π
2
+2kπ
´
(31)
= −
π
2
i (32)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 30 / 36
76. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 30 / 36
78. Complex Logarithm
Example
Find the trigonometric form of z = 2i
ln(z) = ln
³
2i
´
(36)
= i ln(2) = i ln2 (37)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 31 / 36
79. Complex Logarithm
Example
Find the trigonometric form of z = 2i
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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 31 / 36
80. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 32 / 36
81. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 33 / 36
82. 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;
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 34 / 36
83. 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)
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 35 / 36
84. END OF LECTURE
THANK YOU
Dr. Gabby (KNUST-Maths) Algebra of Complex Numbers 36 / 36