IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

alokmittal

suppose R is set of reals and C is the set of complex numbers and a function is defined as f:R

C

f(t) = 1+ti/1-ti where t

R,

prove that f is injective

gcch29

let a,b be arbitrary numbers
if f(a) = f(b)
f is one-one if a=b
i+ai / 1-ai = 1+bi / 1-bi
on solving we get a = b
hence the given function is one - one
here it does not matter that domain is real no and co-domain is complex no since every real no is also a complex no with imaginary part =0

joyfrancis

Assume the function to be bijective.

so,

f( a ) = f( b ) now check whether a = b or not , if it is then the function is one-one otherwise it is many-one.

f( a ) = f (b )

.: 1 + ia / 1 - ia = 1 + ib / 1 - ib

Rationalising..

1 + a² / (1 - ia)² = 1 + b² / (1 - ib)²

=> ( 1 + a² )( 1 - ib )² = ( 1 + b² )( 1 - ia )²

=> a² - b² - i( 2b + 2a²b ) = b² - a² - i( 2a + 2ab² )

.:a² - b² = b² - a² ....(i)

&

2b + 2a²b = 2a + 2ab² .....(ii)

now, since it is given that the domain consists of only real numbers..

from (i)

a = b or a = -b

Out of these two a = b ONLY would satisfy equation (ii)

Hence the function is one-one or injective.

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