**Algebra of complex numbers**

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

**Addition of two complex numbers**

Let z_{1} = a + ib and z_{2} = c + id be any two complex numbers. Then, the sum z_{1} + z_{2} is defined as follows:

z_{1} + z_{2} = (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 law**The sum of two complex numbers is a complex number, i.e., z_{1}+ z_{2}is a complex number for all complex numbers z_{1}and z_{2}.**The commutative law**For any two complex numbers z_{1}and z_{2}, z_{1}+ z_{2}= z_{2}+ z_{1}**The associative law**For any three complex numbers z_{1}, z_{2}, z_{3}, (z_{1}+ z_{2}) + z_{3}= z_{1}+ (z_{2}+ z_{3}).**The existence of additive identity**There 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 inverse**To 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 z_{1} and z_{2}, the difference z_{1} – z_{2} is defined as follows:

z_{1} – z_{2} = z_{1} + (– z_{2}).

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 z_{1} = a + ib and z_{2} = c + id be any two complex numbers. Then, the product z_{1} z_{2} is defined as follows:

z_{1} z_{2} = (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 law**The product of two complex numbers is a complex number, i.e., z_{1}z_{2}is a complex number for all complex numbers z_{1}and z_{2}.**The commutative law**For any two complex numbers z_{1}and z_{2}, z_{1}z_{2}= z_{2}z_{1}**The associative law**For any three complex numbers z_{1}, z_{2}, z_{3}, (z_{1}z_{2})z_{3}= z_{1}(z_{2}z_{3}).**The existence of multiplicative identity**There 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 inverse**For every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number (a/a^{2}+ b^{2}) + i (-b/a^{2}+ b^{2}) (denoted by 1/z or z^{-1}), called the multiplicative inverse of z such that z • 1/z = 1 (the multiplicative identity).**The distributive law**For any three complex numbers z_{1}, z_{2}, z_{3},

(a) z_{1}(z_{2}+ z_{3}) = z_{1}z_{2}+ z_{1}z_{3}

(b) (z_{1}+ z_{2}) z_{3}= z_{1}z_{3}+ z_{2}z_{3}

**Division of two complex numbers**

Given any two complex numbers z_{1} and z_{2}, where z_{2} ≠ 0, the quotient z_{1} / z_{1} is defined by

z_{1} / z_{2} = z_{1} • 1/z_{2}

For example, let z_{1} = 6 + 3i and z_{2} = 2 – i , Then

**Power of i** we know that

**The square roots of a negative real number**

Note that i^{2} = –1 and ( – i)^{2} = i^{2} = – 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 x^{2} + 1 = 0 or x^{2} = –1.

Similarly (√3 i)^{2} = (√3)^{2} i^{2} = 3(-1) = -3

(√3 i)^{2} = (√3)^{2} i^{2} = -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

i^{2} = √−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

(z_{1} + z_{2})^{2} = z_{1}^{2} + z_{2}^{2} + 2z_{1}z_{2} , for all complex numbers z_{1} and z_{2}.

**Proof** We have,

(z_{1} + z_{2})^{2} = (z_{1} + z_{2}) (z_{1} + z_{2}),

= (z_{1} + z_{2}) z_{1} + (z_{1} + z_{2}) z_{2} (Distributive law)

= z_{1}^{2} + z_{2}z_{1} + z_{1}z_{2} + z_{2}^{2} (Distributive law)

= z_{1}^{2} + z_{1}z_{2} + z_{1}z_{2} + z_{1}^{2} (Commutative law of multiplicatoin)

= z_{1}^{2} + z_{2}^{2} + 2z_{1}z_{2}

Similarly, we can prove the following identities:

(i) (z_{1} – z_{2})^{2} = z_{1}^{2} – 2z_{1}z_{2} + z_{2}^{2}

(ii) (z_{1} + z_{2})^{3} = z_{1}^{3} + 3z_{1}^{2}z_{2} + 3z_{1}z_{2}^{2} + z_{2}^{3}

(iii) (z_{1} – z_{2})^{3} = z_{1}^{3} – 3z_{1}^{2}z_{2} + 3z_{1}z_{2}^{2} – z_{2}^{3}

(iv) z_{1}^{2} – z_{2}^{2} = (z_{1} + z_{2}) (z_{1} – z_{2})

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: