# Complex Numbers

The mathematical need to have solution for negative discriminant, these new kind of numbers we call it as complex numbers.
,
if b
2 - 4ac ≥ 0 then roots are real but if b 2 - 4ac < 0 then the roots are not real.
For example : x
2 + 4 = 0
x
2 = - 4
∴ x = $\pm$ √ (-4)
∴ x = $\pm$ 2 √ (-1)
We assume that the square root of (-1) is denoted by the symbol ‘i’, called imaginary unit. The symbol ‘i’ was first introduced by the famous Swiss mathematician, Leonhard Euler in 1748, possibly because ‘i’ is the first letter of the Latin word ‘imaginarius’. Thus , for any two real numbers ‘a’ and ‘b’, we can form a new number
a + ib . This number a + ib is called a complex number. The set of all complex numbers is denoted by ‘C’.
The complex-numbers is of the form ‘a + ib’, where a and b are real numbers and ‘i’ is the imaginary unit which has the property of
i
2 = -1 .So in a given complex-numbers ‘a’ is a real part and ‘b’ is its imaginary part.
Example : 1) 4 + 7i      2) -1/2 + i
In the 1st example 4 is a real part and 7 is an imaginary part.
In the 2nd example -1/2 is a real part and 1 is an imaginary part.
A complex-number is denoted by single letter such as z, w etc.
z = a + ib are denoted as a = Re z and b = Im z .
Two complex-numbers
z1 = a1 + ib1 and z2 = a2 + ib2
z
1 = z 2 , if a 1 = a 2 and b 1 = b 2
A complex-number ‘z’ is said to be zero if its both real and imaginary parts are zero. In other words
z= a + ib = 0 if and only if a = 0 and b = 0
If z = a + ib, the number a – ib is called the complex conjugate(or simply conjugate) of a + ib and is denoted by
z̄

## Examples on complex Numbers

1) Write the following as complex-numbers.
i) √(-27) ⇒ √(-1 x 27) = √(-1) √27 = i√27

ii) 4 - √(-5) ⇒ 4 - √(-1 x 5) = 4 - √5 √(-1) = 4 - i√5
2) Write the real and imaginary part of the following numbers.
i) 2 + i √2
Solution Let z = 2 + i√2
Re z = 2 and Im z = √2
ii) √3/ 5 i
Solution : Let z = √3/5 i = 0 + √3/5 i
Re z = 0 and Im z √3/5
3) Find a and b such that 2a + i 4b and 2i represents the same complex-numbers.
Solution : 2a + i 4b = 2i
2a + i 4b = 0 + 2i
By the definition of equality,
2a = 0 ⇒ a = 0
And 4b = 2 ⇒ b = 2/4 = ½
∴ a = 0 and b = ½ 4) Find the conjugates of 2 + 5i and √5
Solution : By definition, the complex conjugate is obtained by reversing the sign of the imaginary part of the complex-numbers. Hence the required conjugates are
2 – 5i and √5 (as there is no imaginary part)