Algebra of complex numbers(Complex Number 1 NCERT)

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Algebra of complex numbers

In this Topic, we will see the algebra of complex numbers.

Let z1 = a + ib and z2 = c + id be any two complex numbers. Then, the sum z1 + z2 is defined as follows:
z1 + z2 = (a + c) + i (b + d), which is again a complex number.
For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8

The addition of complex numbers satisfy the following properties:

• The closure lawThe sum of two complex numbers is a complex number, i.e., z1 + z2 is a complex number for all complex numbers z1 and z2.
• The commutative lawFor any two complex numbers z1 and z2, z1 + z2 = z2 + z1
• The associative lawFor any three complex numbers z1, z2, z3, (z1 + z2) + z3 = z1 + (z2 + z3).
• The existence of additive identityThere exists the complex number 0 + i 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number z, z + 0 = z.
• The existence of additive inverseTo every complex number z = a + ib, we have the complex number – a + i(– b) (denoted as – z), called the additive inverse or negative of z. We observe that z + (–z) = 0 (the additive identity).

Difference of two complex numbers
Given any two complex numbers z1 and z2, the difference z1 – z2 is defined as follows:
z1 – z2 = z1 + (– z2).
For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i
and (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i

Multiplication of two complex numbers
Let z1 = a + ib and z2 = c + id be any two complex numbers. Then, the product z1 z2 is defined as follows:
z1 z2 = (ac – bd) + i(ad + bc)
For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28

The multiplication of complex numbers possesses the following properties, which we state without proofs.

• The closure lawThe product of two complex numbers is a complex number, i.e., z1z2 is a complex number for all complex numbers z1 and z2.
• The commutative lawFor any two complex numbers z1 and z2, z1z2 = z2z1
• The associative lawFor any three complex numbers z1, z2, z3, (z1z2)z3 = z1(z2z3).
• The existence of multiplicative identityThere exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, for every complex number z..
• The existence of multiplicative inverseFor every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number (a/a2 + b2) + i (-b/a2 + b2) (denoted by 1/z or z-1 ), called the multiplicative inverse of z such that z • 1/z = 1 (the multiplicative identity).
• The distributive lawFor any three complex numbers z1, z2, z3,
(a) z1 (z2 + z3) = z1 z2 + z1 z3
(b) (z1 + z2) z3 = z1 z3 + z2 z3

Division of two complex numbers
Given any two complex numbers z1 and z2, where z2 ≠ 0, the quotient z1 / z1 is defined by
z1 / z2 = z1 • 1/z2
For example, let z1 = 6 + 3i and z2 = 2 – i , Then

Power of i  we know that

The square roots of a negative real number
Note that i2 = –1 and ( – i)2 = i2 = – 1
Therefore, the square roots of – 1 are i, – i. However, by the symbol √− , we would mean i only. Now, we can see that i and –i both are the solutions of the equation x2 + 1 = 0 or x2 = –1.

Similarly (√3 i)2 = (√3)2 i2 = 3(-1) = -3
(√3 i)2 = (√3)2 i2 = -3
Therefore, the square roots of –3 are √3 i and −√3i .
Again, the symbol √−3 is meant to represent √3 i only, i.e., √-3 = √3 i .
Generally, if a is a positive real number, √−a = √ a √−1 = √a i ,
We already know that √a × √b = √ab for all positive real number a and b. This result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0? Let us examine.

Note that
i2 = √−1 √−1 √(−1)(−1) (by assuming √a × √b = √ab for all real numbers)
= √1 = 1, which is a contradiction to the fact that i = − .

Therefore,√a × √b ≠ √ab if both a and b are negative real numbers.
Further, if any of a and b is zero, then, clearly, √a × √b = √ab =0.

Identities
We prove the following identity
(z1 + z2)2 = z12 + z22 + 2z1z2 , for all complex numbers z1 and z2
Proof We have,
(z1 + z2)2 = (z1 + z2) (z1 + z2),
= (z1 + z2) z1 + (z1 + z2) z2 (Distributive law)
= z12 + z2z1 + z1z2 + z22 (Distributive law)
= z12 + z1z2 + z1z2 + z12 (Commutative law of multiplicatoin)
= z12 + z22 + 2z1z2

Similarly, we can prove the following identities:
(i) (z1 – z2)2 = z12 – 2z1z2 + z22
(ii) (z1 + z2)3 = z13 + 3z12z2 + 3z1z22 + z23
(iii) (z1 – z2)3 = z13 – 3z12z2 + 3z1z22 – z23
(iv) z12 – z22 = (z1 + z2) (z1 – z2
In fact, many other identities which are true for all real numbers, can be proved to be true for all complex numbers.

Example Express the following in the form of a + bi:

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