Class 9 – Complex Numbers
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Complex Numbers 

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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

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) Solution: (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.
Solution:
(i)
(ii)
(iii)
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 .
Example: 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:
.