When quadratic equations come in action, you’ll be challenged with either entity or non-entity; the one whose name is written in the form – √-1, and it’s pronounced as the “square root of -1.”

So, we’ll be discussing in the context of the different algebraic complex numbers’ properties.

**1. When a + ib = 0 & a, b, c are the real numbers, then value of both a, b = 0, that is, a = 0, b = 0**

**Proof: **

As per the property,

a + ib = 0 = 0 + i **∙** 0,

Hence, considering the equality of 2 complex numbers’s definition, the conclusions states that the value of x & y = 0 i.e. x = 0 and y = 0.

**2. When a, b, c and d are real numbers and a + ib = c + id then a = c and b = d.**

**Proof:**

When a, b, c, d or x, y, p, q exist as real numbers, & as per the property, a + ib = c + id, or x + iy = p+iq, then accordingly, a = c & b = d or x = p & y = q.

**3. For z1, z2 and z3 complex numbers, the set must be satisfying the associative, commutative, and distributive laws. **

- z1
**∙**z2 = z2**∙**z1 (Commutative law for multiplication) - (z1z2)z3 = z1(z2z3) (Associative law for multiplication)
- z1 + z2 = z2+ z1 (Commutative law for addition).
- (z1 + z2) + z3 = z1 + (z2+ z3) (Associative law for addition)
- z1(z1 + z3) = z1z2 + z1z3 (Distributive law).

**4. The product and the sum of 2 complex conjugate quantities both exist as “real”**

**Proof:**

Let us assume,

z = x + iy, as the complex number where the real values are x, y.

Accordingly, the conjugate of z is equal to – “ = x − iy”

Now,

(x + iy)(x − iy) = x^{2} − i^{2}y^{2} = x^{2} + y^{2 =}^{ Z. }(Multiplication)

Which is real,

And

x + iy + x − iy = z + **(Addition)**

& the result is “2x”, which is also real.

Hence, Multiplication & Addition of 2 complex conjugate quantities are real.

**5. If the product and sum of 2 complex numbers or the quantities exist as real then, these complex quantities will be conjugate to each other.
**

**Proof:
**

Let us assume,

z_{1} = a + ib & z_{2} = c + id – these are the two complex quantities where the real values are a, b, c, d & b ≠ 0 plus d ≠0.

So, by the theory of assumption, z_{1} + z_{2} = a + ib + c + id which is equal to – (a + c) + i(b + d) and which is real.

Therefore, b + d = 0 which gives “d = -b” – Eq. 1

& z_{1}. z_{2} = (a + ib)( c + id) = (ac − bd) + i(ad + bc) exist for real.

As a result, ad + bc = 0 or −ab + bc = 0 ( ∵ d = -b) from the above Eq. 1

Or,

b(c − a) = 0 or c = a (Since b ≠ 0)

Therefore,

ad + bc = 0 or −ab + bc = 0 (Since d = -b)

Or, b(c − a) = 0 or c = a (Since b ≠ 0)

Thus,

z_{2} = c + id = a + i(-b) = a − ib = , which verifies that the values of z_{1} and z_{2} are conjugates of each other.

**6**. **For the 2 complex quantities – z _{1} and z_{2}, they show that:** |z

_{1}+ z

_{2}| ≤ |z

_{1}| + |z

_{2}|

**Proof:**

Let us take, z_{1} = r_{1}(cosx_{1} + isinx_{1} )

& z_{2} = r_{2}(cosx_{2} + isinx_{2} ),

Then, |z_{1} | = r_{1} and |z_{2} | = r_{2}

So,

z_{1} + z_{2} = r_{1}(cosx_{1}isinx_{1}) + r_{2}(cosx_{2} + isinx_{2})

= (r_{1}cosx_{1}+ r_{2}cosx_{2} )+ i(r_{1}sinx_{1}+ r_{2}sinx_{2})

Hence |z_{1}+ z_{2} | = √(r_{1}cosx_{1}+ r_{2}cosx_{2})_{2} + (r_{1}sinx_{1}+ r_{2}sinx_{2})_{2}

= √r_{1}2(cos_{2}x_{1}+ sin2x_{1}) + r_{2}2(cos2x_{2}+ sin2x_{2}) + 2r_{1}r_{2} (cosx_{1} cosx_{2}+ sinx_{1} sinx_{2})

= √r1_{2} + r2_{2} + 2r_{1}r_{2}cos (x_{1}– x_{2})

Now, |cos(x_{1}– x_{2})| ≤ 1

Therefore, |z_{1}+ z_{2}| ≤ √r1_{2} + r2_{2} + 2r_{1}r_{2} or |z_{1}+ z_{2} | ≤ |z_{1}| + |z_{2} |