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Lecture 1
Complex NumbersDefinitions.
Let i2 = −1.
∴ i =√−1.
Complex numbers are often denoted by z.
Just as R is the set of real numbers, C is the set of complex numbers. If z is a complexnumber, z is of the form
z = x + iy ∈ C, for some x, y ∈ R.
e.g. 3 + 4i is a complex number.
z = x + iy↑ ↖
real part imaginary part.If z = x + iy, x, y ∈ R,
the real part of z = �(z) = Re(z) = x
the imaginary part of z = �(z) = Im(z) = y.
eg. z = 3 + 4i
�(z) = 3
�(z) = 4.
If z = x + iy, then z (“z bar”) is given by
z = x − iy
and is called the complex conjugate of z.
eg. If z = 3 + 4i, then z = 3 − 4i.
Example. Solve x2 − 2x + 3 = 0.
x = −(−2)±√
(−2)2−4(1)(3)
2(1) = 2±√−8
2 = 2±2√−2
2 = 1 ±√
2 i. �
Lecture 2Complex Arithmetic.
Addition/Subraction.
Example 1. (2 + 3i) + (4 + i) = 6 + 4i.Example 2. (8 − 3i) − (−2 + 4i) = 10 − 7i.
Multiplication/Division.
Example 1. (2 + 3i)(1 + 2i) = 2 + 4i+ 3i− 6 = −4 + 7iExample 2. (3 − 2i)(3 + 2i) = 9 − (2i)2 = 9 + 4 = 13
∴ when we multiply two complex conjugates, we get a real number.
Example 3. 2+3i1+4i = 2+3i
1+4i × 1−4i1−4i = (2+3i)(1−4i)
(1+4i)(1−4i) = 2−8i+3i−12i2
1−(4i)2 = 14−5i17
(realising the denominator)
Lecture 3Theorem. If two complex numbers are equal then their real parts are equal and theirimaginary parts are equal, i.e., if a+ ib = c+ id where a, b, c, d ∈ R, then a = c and b = d.
Example 1. Find x, y if (3 + 4i)2 − 2(x− iy) = x+ iy.
Left hand side (LHS) = 9 − 16 + 24i− 2x+ i2y
= −7 − 2x+ i(24 + 2y)∴ −7 − 2x = x
3x = −7
x = − 73
& 24 + 2y = yy = −24 �
Example 2. Find x, y ifx
1 + i+
y
2 − i = 2 + 4i.
LHS =x
1 + i+
y
2 − i=
x
1 + i× 1 − i
1 − i +y
2 − i ×2 + i2 + i
=x(1 − i)1 + 1
+y(2 + i)4 + 1
=x(1 − i)
2+y(2 + i)
5
Nowx(1 − i)
2+y(2 + i)
5= 2 + 4i.
∴ 5x(1 − i) + 2y(2 + i) = 20 + 40i5x− i5x+ 4y + i2y = 20 + 40i
5x+ 4y + i(−5x+ 2y) = 20 + 40i
Equating real and imaginary part,
5x+ 4y = 20−5x+ 2y = 40
Solving simultaneously,6y = 60y = 10
& ∴ x = −4. �
Lecture 4Square Roots of Complex Numbers.
Example 1. Find the square root of 35 − 12i.
Let√
35 − 12i = a+ ib : − square both sides.
35 − 12i = (a+ ib)2
= a2 − b2 + i(2ab)
∴ a2 − b2 = 35and 2ab = −12
ab = −6.By inspection, solutions are a = 6& b = −1 or a = −6 or b = 1.or a2 − b2 = 35
ab = −6
b = −6a.
∴ a2 −(− 6a
)2
= 35
a2 − 36a2
= 35.
a4 − 36 = 35a2
a4 − 35a2 − 36 = 0.
(a2 − 36)(a2 + 1) = 0
a2 = 36 & a2 + 1 = 0 ⇒ a /∈ R
∴ a = ±6 & ∴ b = ±1.
& ∴√
35 − 12i = 6 − i. � (By convention, sign(�(√z)) = sign(�(z)))
Example 2. Find the roots of z2 − (1 − i)z + 7i− 4 = 0 in the form a+ ib.
z =(1 − i) ±
√(1 − i)2 − 4(1)(7i− 4)
2
=(1 − i) ±
√1 − 1 − 2i− 28i+ 16
2
=(1 − i) ±
√16 − 30i
2From beside,
=(1 − i) ± (5 − 3i)
2
=1 − i+ 5 − 3i
2or
1 − i− (5 − 3i)2
= 3 − 2i or − 2 + i. �
√16 − 30i = (a+ ib)
16 − 30i = a2 − b2 + i(2ab)
a2 − b2 = 162ab = −30ab = −15a = 5 & b = −3
or a = −5 & b = 3
& ∴√
16 − 30i = 5 − 3i
∵ sign(16) = sign(5) = +
Lecture 5The Argand Diagram. (Note: Ordered pairs:- eg. 2 + i = (2, 1)
for 2 + i = x+ iy on (x, y)-plane)
Two methods: i. P (x, y) the point P on the (x, y)-planeii. Vector
−→OP
x-axis is called the real axis.y-axis is called the imaginary axis.
Eg. Plot the following on the Argand diagram:P = 2 + 3i;B = 3 − i;A = −2 − i;M = 4;E = 2i
z = x+ iy= r cos θ + ir sin θ
= r(cos θ + i sin θ)
Modulus (Distance OP )denoted by r,mod z, |z|, |x+ iy|by Pythagoras, r2 = x2 + y2
r =√x2 + y2
r = |z| = |x+ iy| =√x2 + y2.
Argument (angle θ)denoted by θ, arg z, arg(x+ iy) [or amp z, amp (x+ iy) {amplitude}]by definition, −180◦ < θ ≤ 180◦
For x �= 0, tan θ = yx .
The mod-arg form of a complex numberz = x+ iy
= r(cos θ + i sin θ)
( = r cis θ).
Complex ConjugateIf z = x+ iy, then the complex conjugate is z = x− iy
Radian measure (or circular measure)
eg. 360◦ = 2π radians = 2π rad = 2πc = 2π
180◦ = π
90◦ = π2
60◦ = π3
45◦ = π4
30◦ = π6
More on mod-arg forms.
Examples. Express the following in mod-arg form:-(a) 2 + 2i;(b) 2 + 5i;(c) −1 +
√3i;(d) 3i;(e) 1 − 3i
(a) 2 + 2i
r =√
22 + 22 =√
8 = 2√
2& tan θ = 2
2 = 1 & ∴ θ = π4
& ∴ 2 + 2i = 2√
2(cos π
4 + i sin π4
). �
(b) 2 + 5i
r =√
22 + 52 =√
29& tan θ = 5
2 & ∴ θ = tan−1 52 ≈ 68◦12′
& ∴ 2 + 5i =√
29(cos
(tan−1 5
2
)+ i sin
(tan−1 5
2
))≈
√29
(cos 68◦12′ + i sin 68◦12′
)�
(c) −1 +√
3i
r =√
12 + 3 =√
4 = 2tanα =
√3
1 & ∴ α = π3 & ∴ θ = π − π
3 = 2π3
& ∴ −1 +√
3i = 2(cos 2π
3 + i sin 2π3
). �
(d) 3i
By inspection, 3i = 3(cos π
2 + i sin π2
). �
(e) 1 − 3i
r =√
12 + 32 =√
10tan(−θ) = 3 & ∴ −θ = tan−1 3 & ∴ θ = − tan−1 3 ≈ 71◦34′.& ∴ 1 − 3i =
√10
(cos
(− tan−1 3
)+ i sin
(− tan−1 3
))=
√10
(cos
(tan−1 3
)− i sin
(tan−1 3
))≈
√10
(cos 71◦34′ − i sin 71◦34′
)�.
Lecture 6Axioms
An integral domain is a set of elements with two binary operations defined for them, whichobey the laws obeyed by the integers.
A set S is an integral domain if its elements a, b, c, . . . obey the following laws.
1. Closure Law for Addition, i.e., a + b ∈ S
2. Closure Law for Multiplication, i.e., a × b ∈ S
3. Commutative Law for Addition, i.e., a + b = b + a4. Commutative Law for Multiplication, i.e., a × b = b × a5. Associative Law for Addition, i.e., a + (b + c) = (a + b) + c6. Associative Law for Multiplication, i.e., a × (b × c) = (a × b) × c7. Distributive Law of Multiplication over Addition, i.e., a × (b + c) = a × b + a × c8. There exists an additive identity (or zero element) 0, such that for every a,
a + 0 = 0 + a = a (Note 0 ∈ S)9. There exists a multiplicative identity (or unity element) 1, such that for every a,
a × 1 = 1 × a = a (Note 1 ∈ S)10. There exists an additive inverse (or opposite), −a, for each member a of the set such
that a + (−a) = (−a) + a = 0.11. Cancellation Law. If ab = ac and a �= 0, then b = c.
Example 1. Z, the set of the integers, is an integral domain.
The elements of a field F obey the above axioms 1-10 for integral domains, (where a, b, care elements of F) and instead of the cancellation law, there is a law about the existenceof a multiplicative inverse (or reciprocal):
11′. If a−1 and 1 are elements of F, and a × a−1 = a−1 × a = 1, where a �= 0, then a−1 isthe multiplicative inverse of a.
Example 2. C, the set of complex numbers is a field.
Example 3. The additive inverse of z = 2 + 3i is −z = −2 − 3i
Example 4. The multiplicative inverse of z = 2 + 3i is z−1 = 12+3i = 1
2+3i
(2−3i2−3i
)= 2−3i
13 .
Lecture 7
(∗){
cos(A+B) = cosA cosB − sinA sinBsin(A+B) = sinA cosB + sinB cosA
Mod-arg theorems
i. If z1 = r1(cos θ1 + i sin θ1) & z2 = r2(cos θ2 + i sin θ2)then if z1 = z2 then r1 = r2 & θ1 = θ2.
ii. |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2 ± 2π.i.e., for example:
arg(z1z2) = 100◦ + 140◦ − 360◦
= −120◦
arg(
z1z2
)= arg z1 − arg z2 ± 2π.
Proof . If z1 = r1(cos θ1 + i sin θ1)
and z2 = r2(cos θ2 + i sin θ2)
then z1z2 = r1(cos θ1 + i sin θ1) · r2(cos θ2 + i sin θ2)
= r1r2(cos θ1 cos θ2 − sin θ1 sin θ2 + i sin θ2 cos θ1 + i sin θ1 cos θ2)
= r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2)) − (see (∗) above)
& ∴ |z1z2| = r1r2 = |z1||z2| and arg(z1z2) = θ1 + θ2 = arg(z1) + arg(z2).
Extended:
arg(z1z2 · · · zn) = arg z1 + arg z2 + · · · + arg zn ± 2πn.
|zn| = |z|n (eg., |z3| = |zzz| = |z||z||z| = |z|3).
and arg(zn) = n arg z ± 2πk.
∣∣ 1zn
∣∣ = 1|z|n and arg
(1
zn
)= arg 1 − arg(zn) = 0 − n arg z ± 2πk = −n arg z ± 2πk.
Example 1. Find the modulus and argument of z = (2 − i)(1 − 3i).
|z| = |2 − i||1 − 3i|=
√22 + 12 ·
√12 + 32
=√
5 ·√
10
=√
50
= 5√
2.
arg(z) = arg(2 − i) + arg(1 − 3i) = − tan−1 12 − tan−1 3 ≈ −98◦8′.
Example 2. z =(−1 + 2i)(1 + i)
−2 − 3i
|z| =| − 1 + 2i||1 + i|
| − 2 − 3i|
=√
5 ·√
2√13
=√
10√13
=
√1013.
arg(z) = arg(−1 + 2i) + arg(1 + i) − arg(−2 − 3i)
≈ 285◦15′ − 360◦
= −74◦45′
Lecture 8Triangle Inequalities.
Example 1. If z1 = 2 + i and z2 = −1 + 2i, z1 + z2 = 1 + 3i. �
Polygon Rule.
Subtraction of Complex Numbers.
z2 − z1 = z2 + (−z1):
Triangle Inequalities.|z1 + z2| ≤ |z1| + |z2|:
|z1 − z2| ≥ |z1| − |z2|:
Example 2. Verify the triangle inequalities if
z1 = 2 − 3i,z2 = −1 + 4i,
z1 + z2 = 1 + i,z1 − z2 = 3 − 7i.
|z1| =√
13
|z2| =√
17
|z1 + z2| =√
2
|z1 − z2| =√
58.
|z1 + z2| ≤ |z1| + |z2|√2 ≤
√13 +
√17 �
|z1 − z2| ≥ |z1| − |z2|√58 ≥
√13 −
√17 �.
∴ triangle inequalities hold. �
Product of Complex Numbers.
The triangle OQR is constructed similar to �AOP . A is the point (1, 0).
Multiplication by i, −1, −i.
Multiplication by i, rotation 90◦ (anticlockwise).
Multiplication by −1, rotation 180◦ anticlockwise.
Multiplication by −i, rotation 270◦ anticlockwise
Lecture 9Geometric Representation of Locus Problems.
General forms:- |z − z1| = a represents a circle, centre at z1 radius a units.
Example 1. |z| = 1.
Example 2. |z − 3| = 2.
Example 3. |z − i| = 1.
Example 4. |z − 1 − 2i| = 2
|z − (1 + 2i)| = 2 centre (1, 2), radius 2 units.
Example 5. |z| ≤ 3 (note:- if less than, it is inside, if it is greater than, it is outside.)
Example 6. 2 < |z| ≤ 3.
Example 7. |z| ≤ 4 and 0 ≤ arg z ≤ π3 .
Example 8. 1 ≤ �(z) ≤ 2 if z = x + iy,then �(z) = y (& ∴ 1 ≤ y ≤ 2)
Example 9. −π6 < arg z ≤ π
3 .
Example 10. 1 ≤ �(z) ≤ 2 and �(z) ≤ −1
Example 11. 1 ≤ �(z) ≤ 2 or �(z) ≤ −1
Example 12. |z| ≤ 4 or 0 ≤ arg z ≤ π3
Lecture 10Using Algebra to Represent Locus Problems
Example 1. Show algebraically that |z − 2− i| = 4 represents a circle with radius 4 unitsand centre (2, 1).
|z − 2 − i| = 4.
∴ |x + iy − 2 − i| = 4.
∴ |(x − 2) + i(y − 1)| = 4.
∴√
(x − 2)2 + (y − 1)2 = 4.
∴ (x − 2)2 + (y − 1)2 = 16.
which is a circle centre (2, 1), radius 4 units. �
Example 2. Sketch the curve: (i) �(z2) = 3 (ii) �(z2) = 4.
(i) �(z2) = 3
�((x + iy)2) = 3
�(x2 − y2 + 2ixy) = 3
x2 − y2 = 3.
(ii) �(z2) = 4.
∴ 2xy = 4.
∴ xy = 2.
Example 3. Describe in geometric terms, the curve described by 2|z| = z + z + 4.
2|z| = z + z + 4.
∴ 2|x + iy| = x + iy + x − iy + 4.
∴ 2√
x2 + y2 = 2x + 4 = 2(x + 2).
∴√
x2 + y2 = x + 2.
∴ x2 + y2 = (x + 2)2.
∴ x2 + y2 = x2 + 4x + 4.
∴ y2 = 4x + 4.
⇒ sideways parabola at vertex (−1, 0).
Example 4. Sketch the locus of �(z + iz) < 2.
�(x + iy + i(x + iy)) < 2.
∴ �(x + iy + ix − y) < 2.
∴ x − y < 2.
Example 5. If z1 = 1 + i & z2 = 2 + 3i find the locus of z if |z − z1| = |z − z2|.
|x + iy − (1 + i)| = |x + iy − (2 + 3i)|.∴ |(x − 1) + i(y − 1)| = |(x − 2) + i(y − 3)|.√
(x − 1)2 + (y − 1)2 =√
(x − 2)2 + (y − 3)2.
(x − 1)2 + (y − 1)2 = (x − 2)2 + (y − 3)2.
x2 − 2x + 1 + y2 − 2y + 1 = x2 − 4x + 4 + y2 − 6y + 9.
∴ 2x + 4y = 11.
N.B. |z − z1| = |z − z2| will always be a straight line. It will always be the perpendicularbisector of the interval joining z1 to z2.
Lecture 11(∗) Note. sin(A+B) = sinA cos B +sinB cos A & cos(A+B) = cos A cos B− sinA sinB.
De Moivres Theorem. (cos θ + i sin θ)n = cos nθ + i sinnθ.
Proof. (By mathematical induction for n = 0, 1, 2, . . . .)
Step 1. Test n = 0.
L.H.S. = (cos θ + i sin θ)0
= 1R.H.S. = cos 0 + i sin 0
= 1= L.H.S.
∴ it is true for n = 0.
Step 2. Assume true for n = k i.e., (cos θ + i sin θ)k = cos kθ + i sin kθ.
Test for n = k + 1.
i.e., L.H.S. = (cos θ + i sin θ)k+1 & R.H.S. = cos(k + 1)θ + i sin(k + 1)θ
= (cos θ + i sin θ)k(cos θ + i sin θ)1
= (cos kθ + i sin kθ)(cos θ + i sin θ)(since we have assumed it true for n = k)= cos kθ cos θ + i sin θ cos kθ + i sin kθ cos θ − sin kθ sin θ
= cos kθ cos θ − sin kθ sin θ + i(sin θ cos kθ + sin kθ cos θ)
= cos(kθ + θ) + i sin(kθ + θ) (see (∗) above)
= cos(k + 1)θ + i sin(k + 1)θ= R.H.S.
Step 3. If the result is true for n = 0, then true for n = 0 + 1, i.e., n = 1. If the result istrue for n = 1, then true for n = 1 + 1, i.e., n = 2 ans so on for all nonnegative integersn �
Example 1. Simplify:
(a) (cos θ − i sin θ)−4 (b) (sin θ − i cos θ)7 (c) (cos 2θ+i sin 2θ)3
(cos θ−i sin θ)4 .
(a) (cos θ − i sin θ)−4 = cos(−4θ) − i sin(−4θ)= cos 4θ + i sin 4θ �
(b) (sin θ − i cos θ)7 = (−i cos θ + sin θ)7
= −i7(cos θ − i sin θ)7
= i(cos 7θ − i sin 7θ)= sin 7θ + i cos 7θ �
(c) (cos 2θ+i sin 2θ)3
(cos θ−i sin θ)4 = (cos θ+i sin θ)6
(cos θ−i sin θ)4
= (cos θ+i sin θ)6
(cos(−θ)+i sin(−θ))4
= (cos θ+i sin θ)6
(cos θ+i sin θ)−4
= (cos θ + i sin θ)10
= cos 10θ + i sin 10θ �Example 2. Express in the form x + iy:
(a)(cos π
2 + i sin π2
)6(b)
(1 +
√3
)10.
(a) (cos π2 + i sin π
2 )6 = cos 6π2 + i sin 6π
2
= cos 3π + i sin 3π
= −1 + 0i
= −1 �(b) (1 +
√3)10 = (2(cos π
3 + i sin π3 )10
= 210(cos 10π3 + i sin 10π
3 )
= 210(− 1
2 − i√
32
)
= −512 − 512i√
3 �
Lecture 12De Moivre’s Theorem and the Argand Diagram
Example. If z =√
3 + i represent the following on the Argand Diagram:
z, iz, 1z ,−z, 2z, z, z2 + z, z3 − z
z = 2(cos π6 + i sin π
6 )
z−1 = (2(cos π6 + i sin π
6 ))−1
= 12 (cos−π
6 + i sin−π6 )
= 12 (cos π
6 − i sin π6 )
2z = 4(cos π6 + i sin π
6 )
z2 = (2(cos π6 + i sin π
6 ))2
= 4(cos π3 + i sin π
3 )z3 = (2(cos π
6 + i sin π6 ))3
= 8(cos π2 + i sin π
2 )
Solution on next page.
Lecture 13Trigonometric Identities and DeMoivre’s Theorem
Example. Obtain cos 6θ in terms of cos θ. Hence show that x = cos(2k + 1) π12 where
k = 0, 1, 2, 3, 4, 5 is a solution to the equation 32x6−48x4 +18x2−1 = 0 and hence deducethat cos π
12 . cos 5π12 = 1
4 .
cos 6θ + i sin 6θ = (cos θ + i sin θ)6.
Consider using Pascal’s Triangle:1
1 11 2 1
1 3 3 11 4 6 4 1
1 5 10 10 5 11 6 15 20 15 6 1
. . . . . . . . . . . . . . . . . .
For example,(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
cos 6θ + i sin θ = (cos θ + i sin θ)6
= cos6 θ + 6 cos5 θi sin θ + 15 cos4 θ(i sin θ)2 + 20 cos3 θ(i sin θ)3 + 15 cos2 θ(i sin θ)4
+6 cos θ(i sin θ)5 + (i sin θ)6 - from Pascal’s Triangle= cos6 θ + 6i cos5 θ sin θ − 15 cos4 θ sin2 θ − 20i cos3 θ sin3 θ + 15 cos2 θ sin4 θ + 6i cos θ sin5 θ
− sin6 θ∴ cos 6θ = cos6 θ − 15 cos4 θ sin2 θ + 15 cos2 θ sin4 θ − sin6 θ − equating parts
= cos6 θ − 15 cos4 θ(1 − cos2 θ) + 15 cos2 θ(1 − cos2 θ)2 − (1 − cos2 θ)3
= cos6 θ − 15 cos4 θ + 15 cos6 θ + 15 cos2 θ(1 − 2 cos2 θ + cos4 θ)
− (1 − 3 cos2 θ + 3 cos4 θ − cos6 θ)
= cos6 θ − 15 cos4 θ + 15 cos6 θ + 15 cos2 θ − 30 cos4 θ + 15 cos6 θ − 1 + 3 cos2 θ
− 3 cos4 θ + cos6 θ
= 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1
If cos 6θ = 0, then 6θ = ±π2 ,± 3π
2 ,± 5π2 ,± 7π
2 ,± 9π2 , etc.
∴ θ = ± π12 ,± 3π
12 ,± 5π12 ,± 7π
12 ,± 9π12 , etc.
= 2k+112 π for k = 0,±1,±2,±3,±4,±5, . . .
∴ these are the roots of 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 = 0.
Now if x = cos θ, then 32x6 − 48x4 + 18x2 − 1 = 0 has roots
x = cos π12 , cos 3π
12 , cos 5π12 , cos 7π
12 , cos 9π12 , cos 11π
12
= cos 2k+112 π for k = 0, 1, 2, 3, 4, 5 (six roots because we have degree six).
Product of roots of 32x6 − 48x4 + 18x2 − 1 = 0 is − 132 .
∴ cos π12 · cos 3π
12 · cos 5π12 · cos 7π
12 · cos 9π12 · cos 11π
12 = − 132
∴ cos π12 · 1√
2· cos 5π
12 · cos 7π12 · (− 1√
2) · cos 11π
12 = − 132
∴ cos π12 · cos 5π
12 · cos 7π12 · cos 11π
12 = 116
But cos 11π12 = − cos π
12 and cos 7π12 = − cos 5π
12 .
∴ cos2 π12 · cos2 5π
12 = 116
∴ cos π12 · cos 5π
12 = 14 �
Lecture 14
Example. If z = cos θ + i sin θ, show that zn + 1zn = 2 cos nθ. Hence or otherwise obtain
an expression for cos5 θ in terms of cos nθ and then evaluate∫ π
20
cos5 θ dθ.
z = cos θ + i sin θ
zn = (cos θ + i sin θ)n
= cos nθ + i sinnθ
1zn
= z−n = (cos θ + i sin θ)−n
= cos−nθ + i sin−nθ
= cos nθ − i sinnθ.
∴ zn +1zn
= (cos nθ + i sinnθ) + (cos nθ − i sinnθ)
= 2 cos nθ.
Pascal’s �:1
1 11 2 1
1 3 3 11 4 6 4 1
1 5 10 10 5 1
(z +
1z
)5
= z5 + 5z4
(1z
)+ 10z3
(1z
)2
+ 10z2
(1z
)3
+ 5z
(1z
)4
+(
1z
)5
= z5 + 5z3 + 10z +10z
+5z3
+1z5
= z5 +1z5
+ 5(
z3 +1z3
)+ 10
(z +
1z
)
Now : − zn +1zn
= 2 cos nθ.
∴ z +1z
= 2 cos θ,
z3 +1z3
= 2 cos 3θ,
& z5 +1z5
= 2 cos 5θ.
∴ (2 cos θ)5 = 2 cos 5θ + 5 · 2 cos 3θ + 10 · 2 cos θ
32 cos5 θ = 2 cos 5θ + 10 cos 3θ + 20 cos θ
∴ cos5 θ = 132
(2 cos 5θ + 10 cos 3θ + 20 cos θ
)= 1
16 (cos 5θ + 5 cos 3θ + 10 cos θ)
& ∴∫ π
2
0
cos5 θ dθ = 116
∫ π2
0
(cos 5θ + 5 cos 3θ + 10 cos θ) dθ
= 116
[sin 5θ
5 + 53 sin 3θ + 10 sin θ
]π2
0
= 116
((15 sin 5π
2 + 52 sin 3π
2 + 10 sin π2
)−
(0))
= 116
(15 − 5
3 + 10)
= 815 �
N.B. Similar expressions can be found for sin5 θ:
zn − 1zn
= (cos nθ + i sinnθ) − (cos nθ − i sinnθ)
= 2i sinnθ.(z − 1
z
)5
= z5 − 5 · z4 · 1z
+ 10z3
(1z
)2
− 10z2
(1z
)3
+ 5z
(1z
)4
−(
1z
)5
=(
z5 − 1z5
)− 5
(z3 − 1
z3
)+ 10
(z − 1
z
)
∴ (2i sin θ)5 = 2i sin 5θ − 5 · 2i sin 3θ + 10 · 2i sin θ
32i5 sin5 θ = 2i sin 5θ − 10i sin 3θ + 20i sin θ (Note : i5 = i)
sin5 θ = 132 (2 sin 5θ − 10 sin 3θ + 20 sin θ) (dividing by 32i)
= 116 (sin 5θ − 5 sin 3θ + 10 sin θ)(
& ∴∫
sin5 θ dθ= 116
∫(sin 5θ−5 sin 3θ+10 sin θ) dθ= 1
16
(− 1
5 cos 5θ+ 53 cos 3θ−10 cos θ)+C)
Lecture 15
Complex Roots of Unity.
If zn = ±1 has n roots, all lying on the unit circle in the argand diagram evenly spaced,for example:
For zn = 1:• n is odd, 1 real root and n − 1 non-real complex roots.• n even, 2 real roots, n − 2 non-real complex roots.zn = −1 has n complex roots.
Example. Solve z7 = 1 and show the roots on the argand diagram. Hence show thatcos 2π
7 + cos 4π7 + cos 6π
7 = − 12 .
If z = cos θ + i sin θ (modulus 1 because |z7| = |z|7 = 1 & ∴ |z| = 1),z7 = (cos θ + i sin θ)7 = 1
= cos 7θ + i sin 7θ = 1.
Equating real parts,cos 7θ = 1
7θ = 0, 2π, 4π, 6π, 8π, 10π, 12π, . . .
θ = 0, 2π7 , 4π
7 , 6π7 , 8π
7 , 10π7 , 12π
7 , . . .
∴ roots are z1 = cos 0 + i sin 0 = 1
z2 = cos 2π7 + i sin 2π
7 = α
z3 = cos 4π7 + i sin 4π
7 = α2
z4 = cos 6π7 + i sin 6π
7 = α3
z5 = cos 8π7 + i sin 8π
7 = α4
z6 = cos 10π7 + i sin 10π
7 = α5
z7 = cos 12π7 + i sin 12π
7 = α6
(7 solutions because degree of polynomial equation z7 = 1 is 7.)
cos 8π7 = cos 6π
7 sin 8π7 = − sin 6π
7
cos 10π7 = cos 4π
7 sin 10π7 = − sin 4π
7
cos 12π7 = cos 2π
7 sin 12π7 = − sin 2π
7
.
& ∴ the roots for − π < arg z ≤ π, are :z1 = 1
z2 = cos 2π7 + i sin 2π
7 = z7 = α
z3 = cos 4π7 + i sin 4π
7 = z6 = α2
z4 = cos 6π7 + i sin 6π
7 = z5 = α3
z5 = cos 6π7 − i sin 6π
7 = z4 = α−3
z6 = cos 4π7 − i sin 4π
7 = z3 = α−2
z7 = cos 2π7 − i sin 2π
7 = z2 = α−1
i.e., the complex roots of unity always occur as pairs
(Note: anxn + an−1xn−1 + an−2x
n−2 + · · · + a0 = 0, then sum of roots = − ba .)
z7 = 1
z7 − 1 = 0& ∴ sum of roots = 0 (since coefficients of z6 is 0.)
∴ z1 + z2 + z3 + z4 + z5 + z6 + z7 = 0
& ∴ 1 + 2 cos 2π7 + 2 cos 4π
7 + 2 cos 6π7 = 0
2 cos 2π7 + 2 cos 4π
7 + 2 cos 6π7 = −1
cos 2π7 + cos 4π
7 + cos 6π7 = − 1
2
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