What is Complex Numbers, Imaginary unit, Conjugate Complex Number, Modulus and Argument, Square root of Complex Numbers, Cube root of Complex Numbers:
Imaginary number or imaginary unit =
√-1 = i
∴ i² = -1, i3= i² × i = -1 × i = -i, i4 = 1, i5
= i etc.
i4n + 1
= i, i4n + 2 = -1, i4n + 3 = -i, i4n + 1 = i
etc. where, n ϵ Z [in general i4n
= 1]
Note:-
√-a × √-b = √a × √-1 × √b × √-1 = √ab ×
√(-1²) = -√ab
But, √-a ×
√-b = √(-a × -b) = √ab is not valid.
Complex Number: A complex number is a
number that can be expressed in the form a + bi, where ‘a’ and ‘b’ are real
numbers and ‘i’ is the imaginary unit, that satisfies the equation i² = -1. In this
expression, ‘a’ is the real part and ‘b’ is the imaginary part of the complex
number and ‘i’ is the positive square root of -1.
Complex
Number a + ib is denoted by Z
∴ a + ib = Z
Where, a =
Real part, b = imaginary part
If b = 0,
complex number is real.
And if a =
0, b ≠ 0, complex number is purely imaginary.
Note:-
Two complex
numbers a + ib and c + id are equal if only a = c and b = d.
Conjugate Complex Number:
Complex
numbers a + ib and a - ib are said to be conjugate to each other,
Denoted by Z
= a + ib and Z̅ = a - ib
Properties of conjugate complex number: If
Z₁ and Z₂ be any two complex numbers, then
Example: 01
Reduce (1 + i)3/(2 + 3i) to the
form A = iB
Solution:
(1 + i)3/(2
+ 3i) = (1 + 3i +3i² + i3)/2 + 3i) = (-2 + 2i)/(2 + 3i)
= (-2 + 2i)
(2 - 3i) / (2 + 3i) (2 - 3i) = (2 + 10i) / (4 + 9) = (2 + 10i)/13
=> 2/13 +
i(10/13) = A + iB form
Where, A =
2/13 and B = 10/13.
Some Important Properties:
If Z₁ = a +
ib and Z₂ = c + id
Then, (i) Z₁
+ Z₂ = (a + c) + i(b + d)
(ii) Z₁ + Z₂ = (a - c) + i(b - d)
(iii) Z₁. Z₂ = (ac - bd) + (ad + bc)
(iv) Z₁/Z₂ = a + ib)/(c + id) = (a +
ib) (c - id) / (c + id) (c - id)
[Since, (a + b) (a - b) = a² - b²]
Modulus and Argument (Amplitude):
If Z = x +
iy, then modulus of Z, |Z| = √(x² + y²)
And = x + iy
= r (cosθ + i sinθ) [here, θ is called
argument]
∴ |Z| = √(x² + y²) = r
And, arg. Z
or amp. Z = θ
And, tan θ =
y/x
Note:-
1. |1| = 1,
arg.(1) = 0
2. |-1| = 1,
arg. (-1) = π
3. |i| = 1,
arg.(i) = π/2
4. |-i| = 1,
arg.(-i) = -π/2
Example: 02
Find mod. and, arg. of
(A) -1 + i
(B) -1 - √3 i
Solution: A
Let, (-1 + i)
= r (cosθ + i sinθ)
Then, r cosθ
= -1…………..eq.(i)
r sinθ = 1………..…..eq.(ii)
∴ Squaring both the eq. and adding,
r² = 2
[since cos²θ + sin²θ = 1]
=> r = √2
On dividing
eq.(ii) by eq.(i),
tanθ = -1
=> θ =
tan− -1 = 3π/4
∴ |-1 + i| = r = √2 and, arg.(-1 + i) = θ = 3π/4
Solution: B
Let, (-1 -√3
i) = r (cosθ + i sinθ)
Then, r cosθ
= -1……………eq.(i)
r sinθ = -√3………….eq.(ii)
∴ Squaring both the eq. and adding,
r² = 4
[since, cos²θ + sin²θ = 1]
=> r = 2
On dividing
eq.(ii) by eq.(i),
tanθ = √
=> θ = -2π/3
Hence, |-1 -
√3 i| = r = 2 and, arg. (-1 - √3 i) = θ = -2π/3
Note: If Z₁ and Z₂ are two complex
numbers, then |Z₁ + Z₂| ≤ |Z₁| + |Z₂|
Example: 03
Find the modulus of (3 - 4i) (1 + 7i) / (1
+ i) (12 + 5i)
Solution:
|(3 - 4i) (1 + 7i) / (1 + i) (12 + 5i)|
= |(3 - 4i)
(1 + 7i)| / |(1 + i) (12 + 5i)|
= |(3 - 4i)|.|
(1 + 7i)| / |(1 + i)|.|(12 + 5i)|
= √(9 + 16).
√(1 + 49) / √(1 + 1). √(144 + 25)
= 5. 5√2 /
√2. 13
= 25/13
Example: 04
Find the square root of 7 - 30√-2
Solution:
Let, √(7 -
30√-2) = x - iy
Then, (x -
iy)² = 7 - 30√-2
=> (x² -
y²) - 2ixy = 7 - 30√2i [since, √-2 = √2i]
=> (x² -
y²) = 7…….…eq.(i)
2xy = 30√2………..…..eq.(ii)
∴ eq.(i)² + eq.(ii)² = (x² - y²)² + (2xy)² = [(x² - y²)² + 4x²y²] =
49 + 1800
=> (x²)² + (y²)² - 2x²y² + 4x²y² = (x² + y²)²
= 49 + 1800
=> (x² + y²) = √(49 + 1800)
=> (x² + y²) = 43……………………eq.(iii)
∴ eq.(i) + eq.(iii) => 2x² = 50
=> x = ± 5
And,
eq.(iii) - eq.(i) = 2y² = 36 => y =
± 3√2
∴√(7 - 30√-2) = x
- iy = ± (5 - 3√2)
Hence, the
two roots are, (5 - 3√2) and, -(5 - 3√2)
Alternate Method:
(7 - 30√-2)
= 25 - 18 - 2. 5. 3√2i [since, 25 - 18
= 7, and, 2 × 5 × 3√2 = 30√2]
= (5 - 3√2i)²
∴ The square root of (7 - 30√-2) = ± (5 - 3√2)
Cube root of Complex Numbers:
Cube roots
of unity are, 1, -½ + i √3/2
and -½ - i √3/2 of which, 1 is real and the other are complex (conjugate to
each other)
If one of them is denoted by ω
and the other is denoted by ω²
∴ -½ + i √3/2 = ω
and, -½ - i √3/2 = ω²
Note:-
1. As the
sum of cube roots of unity is zero, so, we have, 1 + ω + ω² = 0
2. The
product of complex cube roots being unity, we have ω. ω² = 1 i.e. ω3
= 1
3. As a
consequence, of which one complex root is the reciprocal of the other,
I.e. ω = 1/ ω², ω² = 1/ ω
As, ω3
= 1, ω3n = 1, ω3n + 1 = ω, ω3n + 2 = ω² etc….
Example: 04
Find, [(√3 + i)/2]6 + [(i -
√3)/2]6
Solution:
(√3 + i)/2 =
(i √3 + i²)/2i = -i [(-1 + √3)/2] = -iω
(i - √3)/2 =
(i² - √3 i)/2i= -i [(-1 - √3 i)/2] = -iω²
∴ [(√3 + i)/2]6
+ [(i - √3)/2]6 =
(-iω)6 + (-iω²)6
= i6ω6 + i6ω12
= (-1) × (1) + (-1) × (1) = -2