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Complex Numbers Lesson 1
www.MathAcademy.sg
c⃝ 2015 Math Academy www.MathAcademy.sg 1
Basic operations
Definition
Let i =
√
−1. A complex number is a number of the form
z = x + iy, where x, y ∈ R
x is the real part of z, denoted by Re(z).
y is the imaginary part of z, denoted by Im(z).
Powers of i
i0
1
i i
i2
−1
i3
−i
i4
1
i5
i
i6
−1
i7
−i
The pattern 1, i, −1, −i, . . . will repeat itself.
c⃝ 2015 Math Academy www.MathAcademy.sg 2
Basic operations
Definition
Let i =
√
−1. A complex number is a number of the form
z = x + iy, where x, y ∈ R
x is the real part of z, denoted by Re(z).
y is the imaginary part of z, denoted by Im(z).
Powers of i
i0
1
i i
i2
−1
i3
−i
i4
1
i5
i
i6
−1
i7
−i
The pattern 1, i, −1, −i, . . . will repeat itself.
c⃝ 2015 Math Academy www.MathAcademy.sg 3
Basic operations
Definition
Let i =
√
−1. A complex number is a number of the form
z = x + iy, where x, y ∈ R
x is the real part of z, denoted by Re(z).
y is the imaginary part of z, denoted by Im(z).
Powers of i
i0
1
i i
i2
−1
i3
−i
i4
1
i5
i
i6
−1
i7
−i
The pattern 1, i, −1, −i, . . . will repeat itself.
c⃝ 2015 Math Academy www.MathAcademy.sg 4
Operations of Complex Numbers
(i) Equality
a + bi = c + di ⇐⇒ a = c AND b = d
(ii) Addition/Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
(iii) Multiplication
(a + bi)(c + di) = ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
(iv) Division (Multiply denominator by its complex conjugate)
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
(a + bi)(c − di)
c2 + d2
c⃝ 2015 Math Academy www.MathAcademy.sg 5
Operations of Complex Numbers
(i) Equality
a + bi = c + di ⇐⇒ a = c AND b = d
(ii) Addition/Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
(iii) Multiplication
(a + bi)(c + di) = ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
(iv) Division (Multiply denominator by its complex conjugate)
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
(a + bi)(c − di)
c2 + d2
c⃝ 2015 Math Academy www.MathAcademy.sg 6
Operations of Complex Numbers
(i) Equality
a + bi = c + di ⇐⇒ a = c AND b = d
(ii) Addition/Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
(iii) Multiplication
(a + bi)(c + di) = ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
(iv) Division (Multiply denominator by its complex conjugate)
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
(a + bi)(c − di)
c2 + d2
c⃝ 2015 Math Academy www.MathAcademy.sg 7
Operations of Complex Numbers
(i) Equality
a + bi = c + di ⇐⇒ a = c AND b = d
(ii) Addition/Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
(iii) Multiplication
(a + bi)(c + di) = ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
(iv) Division (Multiply denominator by its complex conjugate)
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
(a + bi)(c − di)
c2 + d2
c⃝ 2015 Math Academy www.MathAcademy.sg 8
Operations of Complex Numbers
(i) Equality
a + bi = c + di ⇐⇒ a = c AND b = d
(ii) Addition/Subtraction
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
(iii) Multiplication
(a + bi)(c + di) = ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i
(iv) Division (Multiply denominator by its complex conjugate)
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
(a + bi)(c − di)
c2 + d2
c⃝ 2015 Math Academy www.MathAcademy.sg 9
Complex conjugate z∗
If z = x + iy, then its complex conjugate z∗
is defined as
z∗
= x − iy
Observe the following:
zz∗
= x2
+ y2
Thus, the product of a complex number and its complex conjugate is a real
number.
Proof:
zz∗
= (x + yi)(x − yi)
= x2
− (yi)2
= x2
− (−y2
)
= x2
+ y2
c⃝ 2015 Math Academy www.MathAcademy.sg 10
Complex conjugate z∗
If z = x + iy, then its complex conjugate z∗
is defined as
z∗
= x − iy
Observe the following:
zz∗
= x2
+ y2
Thus, the product of a complex number and its complex conjugate is a real
number.
Proof:
zz∗
= (x + yi)(x − yi)
= x2
− (yi)2
= x2
− (−y2
)
= x2
+ y2
c⃝ 2015 Math Academy www.MathAcademy.sg 11
Complex conjugate z∗
If z = x + iy, then its complex conjugate z∗
is defined as
z∗
= x − iy
Observe the following:
zz∗
= x2
+ y2
Thus, the product of a complex number and its complex conjugate is a real
number.
Proof:
zz∗
= (x + yi)(x − yi)
= x2
− (yi)2
= x2
− (−y2
)
= x2
+ y2
c⃝ 2015 Math Academy www.MathAcademy.sg 12
Example (1)
Express the following in the form x + iy, where x, y ∈ R.
a) z = (3 − 3i) + (−3 + i)
c) z = (2 + 2i)(1 − 3i)
e) z = 2+10i
5−3i
c⃝ 2015 Math Academy www.MathAcademy.sg 13
Example (1)
Express the following in the form x + iy, where x, y ∈ R.
b) z = (2 −
√
2i) − (1 − 3i)
d) z = (
√
3 + 2i)(
√
3 − 2i)
f) z = 3+2i
4−i
c⃝ 2015 Math Academy www.MathAcademy.sg 14
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 15
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 16
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 17
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 18
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 19
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 20
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 21
Example (2)
Given that (2 − i)2
+ (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ.
(2 − i)2
+ (3λ + i)(µ − i) + 5i = 10
(4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10
4 + 3λµ + i(1 − 3λ + µ) = 10
Comparing real-coefficients,
4 + 3λµ = 10
λµ = 2 . . . (1)
Comparing imaginary-coefficients,
1 − 3λ + µ = 0
µ = 3λ − 1 . . . (2)
Substituting (2) into (1),
λ(3λ − 1) = 2
3λ2
− λ − 2 = 0
λ = 1 or −
2
3
=⇒ µ = 2 or − 3
c⃝ 2015 Math Academy www.MathAcademy.sg 22
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 23
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 24
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 25
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 26
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 27
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 28
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 29
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 30
Example (3)
Find the square root of 15 + 8i.
We want to solve for z such that z =
√
15 + 8i. Let z = x + yi.
z =
√
15 + 8i
z2
= 15 + 8i
(x + yi)2
= 15 + 8i
x2
− y2
+ 2xyi = 15 + 8i
Comparing real coefficient,
x2
− y2
= 15 . . . (1)
Comparing imaginary coefficient,
2xy = 8
y =
4
x
. . . (2)
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
c⃝ 2015 Math Academy www.MathAcademy.sg 31
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
From GC,
x = 4 or −4.
=⇒ y = 1 or −1.
∴ z = 4 + i or −4 − i.
c⃝ 2015 Math Academy www.MathAcademy.sg 32
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
From GC,
x = 4 or −4.
=⇒ y = 1 or −1.
∴ z = 4 + i or −4 − i.
c⃝ 2015 Math Academy www.MathAcademy.sg 33
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
From GC,
x = 4 or −4.
=⇒ y = 1 or −1.
∴ z = 4 + i or −4 − i.
c⃝ 2015 Math Academy www.MathAcademy.sg 34
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
From GC,
x = 4 or −4.
=⇒ y = 1 or −1.
∴ z = 4 + i or −4 − i.
c⃝ 2015 Math Academy www.MathAcademy.sg 35
Substitute (2) into (1),
x2
−
(
4
x
)2
= 15
x4
− 15x2
− 16 = 0
From GC,
x = 4 or −4.
=⇒ y = 1 or −1.
∴ z = 4 + i or −4 − i.
c⃝ 2015 Math Academy www.MathAcademy.sg 36
Example (4)
The complex numbers p and q are such that
p = 2 + ia, q = b − i,
where a and b are real numbers.
Given that pq = 13 + 13i, find the possible values of a and b. [4]
[a = 3 or 10, b = 5 or 3
2
]
pq = (2 + ia)(b − i)
= 2b − 2i + abi + a
= 2b + a + i(ab − 2)
13 + 13i = 2b + a + i(ab − 2)
Comparing real and imaginary parts,
2b + a = 13
a = 13 − 2b − − − − − (1)
ab − 2 = 13
ab = 15 − − − − − (2)
Sub (1) into (2)
(13 − 2b)b = 15
−2b2
+ 13b − 15 = 0
b = 5 or 3
2
∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 37
Example (4)
The complex numbers p and q are such that
p = 2 + ia, q = b − i,
where a and b are real numbers.
Given that pq = 13 + 13i, find the possible values of a and b. [4]
[a = 3 or 10, b = 5 or 3
2
]
pq = (2 + ia)(b − i)
= 2b − 2i + abi + a
= 2b + a + i(ab − 2)
13 + 13i = 2b + a + i(ab − 2)
Comparing real and imaginary parts,
2b + a = 13
a = 13 − 2b − − − − − (1)
ab − 2 = 13
ab = 15 − − − − − (2)
Sub (1) into (2)
(13 − 2b)b = 15
−2b2
+ 13b − 15 = 0
b = 5 or 3
2
∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 38
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 39
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 40
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 41
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 42
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 43
Example (5)
Two complex numbers w and z are such that
2w + z = 12i and w∗
+ 2z =
−13 + 4i
2 − i
.
Find w and z, giving each answer in the form x + iy.
2w + z = 12i
z = 12i − 2w —(1)
w∗
+ 2z =
−13 + 4i
2 − i
—(2)
Sub (1) into (2):
w∗
+ 2(12i − 2w) =
−13 + 4i
2 − i
Let w = x + iy. Then w∗
= x − iy.
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 44
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − i
... (We group the terms with i together)
(−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i
Comparing coefficients,
−6x − 5y + 24 = −13 . . . (3)
3x − 10y + 48 = 4 . . . (4)
Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5.
Therefore,
w = 2 + 5i.
From (1),
z = 12i − 2w
= 12i − 2(2 + 5i)
= −4 + 2i
c⃝ 2015 Math Academy www.MathAcademy.sg 45
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − i
... (We group the terms with i together)
(−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i
Comparing coefficients,
−6x − 5y + 24 = −13 . . . (3)
3x − 10y + 48 = 4 . . . (4)
Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5.
Therefore,
w = 2 + 5i.
From (1),
z = 12i − 2w
= 12i − 2(2 + 5i)
= −4 + 2i
c⃝ 2015 Math Academy www.MathAcademy.sg 46
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − i
... (We group the terms with i together)
(−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i
Comparing coefficients,
−6x − 5y + 24 = −13 . . . (3)
3x − 10y + 48 = 4 . . . (4)
Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5.
Therefore,
w = 2 + 5i.
From (1),
z = 12i − 2w
= 12i − 2(2 + 5i)
= −4 + 2i
c⃝ 2015 Math Academy www.MathAcademy.sg 47
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − i
... (We group the terms with i together)
(−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i
Comparing coefficients,
−6x − 5y + 24 = −13 . . . (3)
3x − 10y + 48 = 4 . . . (4)
Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5.
Therefore,
w = 2 + 5i.
From (1),
z = 12i − 2w
= 12i − 2(2 + 5i)
= −4 + 2i
c⃝ 2015 Math Academy www.MathAcademy.sg 48
(x − iy) + 2[12i − 2(x + iy)] =
−13 + 4i
2 − i
... (We group the terms with i together)
(−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i
Comparing coefficients,
−6x − 5y + 24 = −13 . . . (3)
3x − 10y + 48 = 4 . . . (4)
Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5.
Therefore,
w = 2 + 5i.
From (1),
z = 12i − 2w
= 12i − 2(2 + 5i)
= −4 + 2i
c⃝ 2015 Math Academy www.MathAcademy.sg 49
Example (6)
By writing z = x + iy, x, y ∈ R, solve the simultaneous equations
z2
+ zw − 2 = 0 and z∗
=
w
1 + i
.
[z = 1 − i, w = 2i, z = −1 + i, w = −2i]
c⃝ 2015 Math Academy www.MathAcademy.sg 50
From second equation, w = z∗
(1 + i)
Substitute into first equation,
z2
+ z[z∗
(1 + i)] − 2 = 0
Let z = x + iy.
(x + iy)2
+ (x2
+ y2
)(1 + i) − 2 = 0
x2
+ 2xyi − y2
+ x2
+ y2
+ (x2
+ y2
)i − 2 = 0
2x2
− 2 + i(2xy + x2
+ y2
) = 0
Comparing coefficient of real and imaginary parts,
2x2
− 2 = 0
x2
= 1
x = ±1
2xy + x2
+ y2
= 0
(x + y)2
= 0
if x = 1, y = −1
if x = −1, y = 1
When z = 1 − i,
w = z∗
(1 + i)
= (1 + i)(1 + i)
= 1 + 2i − 1
= 2i
When z = −1 + i,
w = z∗
(1 + i)
= (−1 − i)(1 + i)
= −2i
c⃝ 2015 Math Academy www.MathAcademy.sg 51
Properties of Complex Conjugates
(i) z + z∗
= 2Re(z)
(ii) z − z∗
= 2iIm(z)
(iii) (z∗
)∗
= z
Bring ∗ into the brackets
(a) (z1 ± z2)∗
= z∗
1 ± z∗
2
(b) (z1z2)∗
= z∗
1 .z∗
2
(c)
(
z1
z2
)∗
=
z∗
1
z∗
2
(d) (kz)∗
= kz∗
, where k ∈ R
Example
Prove (i), (ii) and (a) in the above properties.
c⃝ 2015 Math Academy www.MathAcademy.sg 52
z + z∗ = 2Re(z)
Let z = x + yi. Then z∗
= x − yi.
c⃝ 2015 Math Academy www.MathAcademy.sg 53
z − z∗ = 2iIm(z)
Let z = x + yi. Then z∗
= x − yi.
c⃝ 2015 Math Academy www.MathAcademy.sg 54
(z1 ± z2)∗ = z∗
1 ± z∗
2
c⃝ 2015 Math Academy www.MathAcademy.sg 55

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Complex Numbers 1 - Math Academy - JC H2 maths A levels

  • 1. Complex Numbers Lesson 1 www.MathAcademy.sg c⃝ 2015 Math Academy www.MathAcademy.sg 1
  • 2. Basic operations Definition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 2
  • 3. Basic operations Definition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 3
  • 4. Basic operations Definition Let i = √ −1. A complex number is a number of the form z = x + iy, where x, y ∈ R x is the real part of z, denoted by Re(z). y is the imaginary part of z, denoted by Im(z). Powers of i i0 1 i i i2 −1 i3 −i i4 1 i5 i i6 −1 i7 −i The pattern 1, i, −1, −i, . . . will repeat itself. c⃝ 2015 Math Academy www.MathAcademy.sg 4
  • 5. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 5
  • 6. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 6
  • 7. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 7
  • 8. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 8
  • 9. Operations of Complex Numbers (i) Equality a + bi = c + di ⇐⇒ a = c AND b = d (ii) Addition/Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i (iii) Multiplication (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i (iv) Division (Multiply denominator by its complex conjugate) a + bi c + di = (a + bi)(c − di) (c + di)(c − di) = (a + bi)(c − di) c2 + d2 c⃝ 2015 Math Academy www.MathAcademy.sg 9
  • 10. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is defined as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 10
  • 11. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is defined as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 11
  • 12. Complex conjugate z∗ If z = x + iy, then its complex conjugate z∗ is defined as z∗ = x − iy Observe the following: zz∗ = x2 + y2 Thus, the product of a complex number and its complex conjugate is a real number. Proof: zz∗ = (x + yi)(x − yi) = x2 − (yi)2 = x2 − (−y2 ) = x2 + y2 c⃝ 2015 Math Academy www.MathAcademy.sg 12
  • 13. Example (1) Express the following in the form x + iy, where x, y ∈ R. a) z = (3 − 3i) + (−3 + i) c) z = (2 + 2i)(1 − 3i) e) z = 2+10i 5−3i c⃝ 2015 Math Academy www.MathAcademy.sg 13
  • 14. Example (1) Express the following in the form x + iy, where x, y ∈ R. b) z = (2 − √ 2i) − (1 − 3i) d) z = ( √ 3 + 2i)( √ 3 − 2i) f) z = 3+2i 4−i c⃝ 2015 Math Academy www.MathAcademy.sg 14
  • 15. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 15
  • 16. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 16
  • 17. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 17
  • 18. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 18
  • 19. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 19
  • 20. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 20
  • 21. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 21
  • 22. Example (2) Given that (2 − i)2 + (3λ + i)(µ − i) + 5i = 10, find the exact values of λ and µ. (2 − i)2 + (3λ + i)(µ − i) + 5i = 10 (4 − 4i − 1) + (3λµ − 3λi + µi + 1) + 5i = 10 4 + 3λµ + i(1 − 3λ + µ) = 10 Comparing real-coefficients, 4 + 3λµ = 10 λµ = 2 . . . (1) Comparing imaginary-coefficients, 1 − 3λ + µ = 0 µ = 3λ − 1 . . . (2) Substituting (2) into (1), λ(3λ − 1) = 2 3λ2 − λ − 2 = 0 λ = 1 or − 2 3 =⇒ µ = 2 or − 3 c⃝ 2015 Math Academy www.MathAcademy.sg 22
  • 23. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 23
  • 24. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 24
  • 25. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 25
  • 26. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 26
  • 27. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 27
  • 28. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 28
  • 29. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 29
  • 30. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 30
  • 31. Example (3) Find the square root of 15 + 8i. We want to solve for z such that z = √ 15 + 8i. Let z = x + yi. z = √ 15 + 8i z2 = 15 + 8i (x + yi)2 = 15 + 8i x2 − y2 + 2xyi = 15 + 8i Comparing real coefficient, x2 − y2 = 15 . . . (1) Comparing imaginary coefficient, 2xy = 8 y = 4 x . . . (2) Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 c⃝ 2015 Math Academy www.MathAcademy.sg 31
  • 32. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 32
  • 33. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 33
  • 34. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 34
  • 35. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 35
  • 36. Substitute (2) into (1), x2 − ( 4 x )2 = 15 x4 − 15x2 − 16 = 0 From GC, x = 4 or −4. =⇒ y = 1 or −1. ∴ z = 4 + i or −4 − i. c⃝ 2015 Math Academy www.MathAcademy.sg 36
  • 37. Example (4) The complex numbers p and q are such that p = 2 + ia, q = b − i, where a and b are real numbers. Given that pq = 13 + 13i, find the possible values of a and b. [4] [a = 3 or 10, b = 5 or 3 2 ] pq = (2 + ia)(b − i) = 2b − 2i + abi + a = 2b + a + i(ab − 2) 13 + 13i = 2b + a + i(ab − 2) Comparing real and imaginary parts, 2b + a = 13 a = 13 − 2b − − − − − (1) ab − 2 = 13 ab = 15 − − − − − (2) Sub (1) into (2) (13 − 2b)b = 15 −2b2 + 13b − 15 = 0 b = 5 or 3 2 ∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 37
  • 38. Example (4) The complex numbers p and q are such that p = 2 + ia, q = b − i, where a and b are real numbers. Given that pq = 13 + 13i, find the possible values of a and b. [4] [a = 3 or 10, b = 5 or 3 2 ] pq = (2 + ia)(b − i) = 2b − 2i + abi + a = 2b + a + i(ab − 2) 13 + 13i = 2b + a + i(ab − 2) Comparing real and imaginary parts, 2b + a = 13 a = 13 − 2b − − − − − (1) ab − 2 = 13 ab = 15 − − − − − (2) Sub (1) into (2) (13 − 2b)b = 15 −2b2 + 13b − 15 = 0 b = 5 or 3 2 ∴ a = 3 or 10c⃝ 2015 Math Academy www.MathAcademy.sg 38
  • 39. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 39
  • 40. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 40
  • 41. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 41
  • 42. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 42
  • 43. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 43
  • 44. Example (5) Two complex numbers w and z are such that 2w + z = 12i and w∗ + 2z = −13 + 4i 2 − i . Find w and z, giving each answer in the form x + iy. 2w + z = 12i z = 12i − 2w —(1) w∗ + 2z = −13 + 4i 2 − i —(2) Sub (1) into (2): w∗ + 2(12i − 2w) = −13 + 4i 2 − i Let w = x + iy. Then w∗ = x − iy. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − ic⃝ 2015 Math Academy www.MathAcademy.sg 44
  • 45. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coefficients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 45
  • 46. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coefficients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 46
  • 47. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coefficients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 47
  • 48. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coefficients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 48
  • 49. (x − iy) + 2[12i − 2(x + iy)] = −13 + 4i 2 − i ... (We group the terms with i together) (−6x − 5y + 24) + i(3x − 10y + 48) = −13 + 4i Comparing coefficients, −6x − 5y + 24 = −13 . . . (3) 3x − 10y + 48 = 4 . . . (4) Solving the simultaneous equations (3) and (4), we have x = 2 and y = 5. Therefore, w = 2 + 5i. From (1), z = 12i − 2w = 12i − 2(2 + 5i) = −4 + 2i c⃝ 2015 Math Academy www.MathAcademy.sg 49
  • 50. Example (6) By writing z = x + iy, x, y ∈ R, solve the simultaneous equations z2 + zw − 2 = 0 and z∗ = w 1 + i . [z = 1 − i, w = 2i, z = −1 + i, w = −2i] c⃝ 2015 Math Academy www.MathAcademy.sg 50
  • 51. From second equation, w = z∗ (1 + i) Substitute into first equation, z2 + z[z∗ (1 + i)] − 2 = 0 Let z = x + iy. (x + iy)2 + (x2 + y2 )(1 + i) − 2 = 0 x2 + 2xyi − y2 + x2 + y2 + (x2 + y2 )i − 2 = 0 2x2 − 2 + i(2xy + x2 + y2 ) = 0 Comparing coefficient of real and imaginary parts, 2x2 − 2 = 0 x2 = 1 x = ±1 2xy + x2 + y2 = 0 (x + y)2 = 0 if x = 1, y = −1 if x = −1, y = 1 When z = 1 − i, w = z∗ (1 + i) = (1 + i)(1 + i) = 1 + 2i − 1 = 2i When z = −1 + i, w = z∗ (1 + i) = (−1 − i)(1 + i) = −2i c⃝ 2015 Math Academy www.MathAcademy.sg 51
  • 52. Properties of Complex Conjugates (i) z + z∗ = 2Re(z) (ii) z − z∗ = 2iIm(z) (iii) (z∗ )∗ = z Bring ∗ into the brackets (a) (z1 ± z2)∗ = z∗ 1 ± z∗ 2 (b) (z1z2)∗ = z∗ 1 .z∗ 2 (c) ( z1 z2 )∗ = z∗ 1 z∗ 2 (d) (kz)∗ = kz∗ , where k ∈ R Example Prove (i), (ii) and (a) in the above properties. c⃝ 2015 Math Academy www.MathAcademy.sg 52
  • 53. z + z∗ = 2Re(z) Let z = x + yi. Then z∗ = x − yi. c⃝ 2015 Math Academy www.MathAcademy.sg 53
  • 54. z − z∗ = 2iIm(z) Let z = x + yi. Then z∗ = x − yi. c⃝ 2015 Math Academy www.MathAcademy.sg 54
  • 55. (z1 ± z2)∗ = z∗ 1 ± z∗ 2 c⃝ 2015 Math Academy www.MathAcademy.sg 55