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) = x2 − i2y2 = x2 + y2 = 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,
z1 = a + ib & z2 = 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, z1 + z2 = 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
& z1. z2 = (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,
z2 = c + id = a + i(-b) = a − ib = 
, which verifies that the values of z1 and z2 are conjugates of each other.
6. For the 2 complex quantities – z1 and z2, they show that: |z1+ z2 | ≤ |z1 | + |z2 |
Proof:
Let us take, z1 = r1(cosx1 + isinx1 )
& z2 = r2(cosx2 + isinx2 ),
Then, |z1 | = r1 and |z2 | = r2
So,
z1 + z2 = r1(cosx1isinx1) + r2(cosx2 + isinx2)
= (r1cosx1+ r2cosx2 )+ i(r1sinx1+ r2sinx2)
Hence |z1+ z2 | = √(r1cosx1+ r2cosx2)2 + (r1sinx1+ r2sinx2)2
= √r12(cos2x1+ sin2x1) + r22(cos2x2+ sin2x2) + 2r1r2 (cosx1 cosx2+ sinx1 sinx2)
= √r12 + r22 + 2r1r2cos (x1– x2)
Now, |cos(x1– x2)| ≤ 1
Therefore, |z1+ z2| ≤ √r12 + r22 + 2r1r2 or |z1+ z2 | ≤ |z1| + |z2 |
