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__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, i

^{3}= i² × i = -1 × i = -i, i^{4}= 1, i^{5}= i etc.
i

^{4n + 1}= i, i^{4n + 2}= -1, i^{4n + 3}= -i, i^{4n + 1}= i etc. where, n Ïµ Z [in general i^{4n}= 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 + i^{3})/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}_{ }
= i

^{6}Ï‰^{6}+ i^{6}Ï‰^{12}
= (-1) × (1) + (-1) × (1) = -2

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