Complex Numbers - ObjectiveBooks

# Complex Numbers

### 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 Ïµ 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 (cosÎ¸ + i sinÎ¸)    [here, Î¸ is called argument]
|Z| = √(x² + y²) = r
And, arg. Z or amp. Z = Î¸
And, tan Î¸ = y/x

Note:-
1. |1| = 1, arg.(1) = 0
2. |-1| = 1, arg. (-1) = Ï€
3. |i| = 1, arg.(i) = Ï€/2
4. |-i| = 1, arg.(-i) = -Ï€/2

Example: 02
Find mod. and, arg. of
(A) -1 + i
(B) -1 - √3 i

Solution: A
Let, (-1 + i) = r (cosÎ¸ + i sinÎ¸)
Then, r cosÎ¸ = -1…………..eq.(i)
r sinÎ¸ = 1………..…..eq.(ii)

Squaring both the eq. and adding,
r² = 2   [since cos²Î¸ + sin²Î¸ = 1]
=> r = √2
On dividing eq.(ii) by eq.(i),
tanÎ¸ = -1
=> Î¸ = tan -1 = 3Ï€/4
|-1 + i| = r = √2 and, arg.(-1 + i) = Î¸ = 3Ï€/4

Solution: B
Let, (-1 -√3 i) = r (cosÎ¸ + i sinÎ¸)
Then, r cosÎ¸ = -1……………eq.(i)
r sinÎ¸ = -√3………….eq.(ii)

Squaring both the eq. and adding,
r² = 4   [since, cos²Î¸ + sin²Î¸ = 1]
=> r = 2
On dividing eq.(ii) by eq.(i),
tanÎ¸ = √
=>   Î¸ = -2Ï€/3
Hence, |-1 - √3 i| = r = 2 and, arg. (-1 - √3 i) = Î¸ = -2Ï€/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 Ï‰ and the other is denoted by Ï‰²
-½ + i √3/2 = Ï‰ and, -½ - i √3/2 = Ï‰²

Note:-
1. As the sum of cube roots of unity is zero, so, we have, 1 + Ï‰ + Ï‰² = 0
2. The product of complex cube roots being unity, we have Ï‰. Ï‰² = 1 i.e. Ï‰3 = 1
3. As a consequence, of which one complex root is the reciprocal of the other,
I.e. Ï‰ = 1/ Ï‰²,    Ï‰² = 1/ Ï‰
As, Ï‰3 = 1,   Ï‰3n = 1,   Ï‰3n + 1 = Ï‰,    Ï‰3n + 2 = Ï‰² etc….

Example: 04
Find, [(√3 + i)/2]6 + [(i - √3)/2]6

Solution:
(√3 + i)/2 = (i √3 + i²)/2i = -i [(-1 + √3)/2] = -iÏ‰
(i - √3)/2 = (i² - √3 i)/2i= -i [(-1 - √3 i)/2] = -iÏ‰²

[(√3 + i)/2]6 + [(i - √3)/2]6  = (-iÏ‰)6 + (-iÏ‰²)6
= i6Ï‰6 + i6Ï‰12

= (-1) × (1) + (-1) × (1) = -2
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