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Relations And Functions

Question

Verify that

$space space space space space space space space space space space space space top enclose open parentheses fraction numerator 3 plus 4 straight i over denominator 1 minus 2 straight i end fraction close parentheses end enclose space equals space fraction numerator top enclose open parentheses 3 plus 4 straight i close parentheses end enclose over denominator top enclose open parentheses 1 minus 2 straight i close parentheses end enclose end fraction$

Solution

space space space space space space space fraction numerator 3 plus 4 straight i over denominator 1 minus 2 straight i end fraction equals left parenthesis 3 plus 4 straight i right parenthesis fraction numerator left parenthesis 1 plus 2 straight i right parenthesis over denominator left parenthesis 1 right parenthesis squared plus left parenthesis 2 right parenthesis squared end fraction

space space space space space space space space space space space open parentheses because fraction numerator 1 over denominator straight a plus ib end fraction equals fraction numerator straight a minus ib over denominator straight a squared plus straight b squared end fraction close parentheses

space space space space space space space space space space space equals space space space fraction numerator left parenthesis 3 plus 4 straight i right parenthesis left parenthesis 1 plus 2 straight i right parenthesis over denominator 5 end fraction equals fraction numerator 3 plus 10 straight i plus 8 straight i squared over denominator 5 end fraction equals fraction numerator negative 5 plus 10 straight i over denominator 5 end fraction equals negative 1 plus 2 straight i

rightwards double arrow

space space space space space space space top enclose open parentheses fraction numerator 3 plus 4 straight i over denominator 1 minus 2 straight i end fraction close parentheses end enclose space equals space top enclose open parentheses negative 1 plus 2 straight i close parentheses end enclose space equals space minus 1 minus 2 straight i

....(i)
Also,

WiredFaculty

space space space space space space equals space fraction numerator negative 5 minus 10 straight i over denominator 5 end fraction equals negative 1 minus 2 straight i

....(ii)
Hence, from (i) and (ii),

space space space space space space space space space space space space space space space space space space space space top enclose open parentheses fraction numerator 3 plus 4 straight i over denominator 1 minus 2 straight i end fraction close parentheses end enclose space equals space top enclose fraction numerator open parentheses 3 plus 4 straight i close parentheses over denominator top enclose open parentheses 1 minus 2 straight i close parentheses end enclose end fraction end enclose

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