Complex Numbers

The mathematical need to have solution for negative discriminant, these new kind of numbers we call it as complex numbers.
According to the quadratic formula ,
if b² - 4ac ≥ 0 then roots are real but if b² - 4ac < 0 then the roots are not real.
For example : x² + 4 = 0
x² = - 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
i² = -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
z₁ = z₂ , if a₁ = a₂ and b₁= b₂
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)

11th grade math

From complex numbers to Home page

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.