Class 9 – Complex Numbers

Take practice tests in Complex Numbers

laptop_img Online Tests

Topic Sub Topic Online Practice Test
Complex Numbers
  • Introduction to complex numbers
  • Operations on complex numbers
Take Test See More Questions
Complex Numbers
  • Conjugate of a complex number
Take Test See More Questions

file_img Study Material

Introduction: The equations of the form etc. are not solvable in R i.e. there is no real number whose square is a negative real number. Euler was the first mathematician to introduce the symbol (iota) for the square root of -1 with the property. He also called this symbol as the imaginary unit.

Integral Powers of Iota (i):

a) Positive integral powers of i:

We have,

and so on

in order to compute for, we divide n by 4 and obtain the remainder r. Let m be the quotient when n is divided by 4. Then,

, where

Thus, the value of for n > 4 is , where r is the remainder when n is divided by 4.

b) Negative integral powers of i: By the law of indices, we have

If n > 4, then , where r is the remainder when n is divided by 4. is defined as 1

Example: Evaluate the following.

(i) (ii)

(iii) (iv)


(i) , 135 leaves remainder as 3 when it is divided by 4.

(ii) , The remainder is 3 when 19 is divided by 4.

(iii) Wehave, = .

On dividing 999 by 4, we obtain 3 as the remainder.

(iv) Wehave,

Example: Show that

Solution: (i)We have,

(ii) Wehave,

(iii) Wehave,

Imaginary quantities:

Definition: The square root of a negative real number is called an imaginary quantity or an imaginary number.
For etc. are imaginary quantities.
A Useful Result: If a, b are positive real numbers,then
Proof: Wehave

Therefore, For any two real numbers is true only when at least one of a and b is either positive or zero.

In other words, is not valid if a and b both are negative.

For any positive real number a, we have .

Example: Computethe following.





Example: A student writes the formula. Then he substitutes and and finds 1 = –1.Explain where he is wrong?

Solution: Since a and b both are negative. Therefore, cannot be written as .In fact for a and b both negative, we have .
Is the following computation correct? If not give the correct computation:

Solution: The said computation is not correct, because -2 and -3 both are negative and is true when atleast one of a and b positive or zero. The correct computation is as givenbelow: