JEE MAIN - Mathematics (2003 - No. 35)
If $${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$ then :
x = 2n + 1, where n is any positive integer
x = 4n , where n is any positive integer
x = 2n, where n is any positive integer
x = 4n + 1, where n is any positive integer.
Explanation
$${\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$
$$ \Rightarrow {\left( i \right)^x} = 1$$
We know $$i = \sqrt { - 1} $$
$$\therefore$$ $${i^2} = - 1$$
$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$
$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$
So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1
$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.
$$ \Rightarrow $$ $${\left[ {{{\left( {1 + i} \right)\left( {1 + i} \right)} \over {\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{1 + 2i + {i^2}} \over {1 + 1}}} \right]^x} = 1$$
$$ \Rightarrow $$ $${\left[ {{{1 + 2i - 1} \over 2}} \right]^x} = 1$$
$$ \Rightarrow {\left( i \right)^x} = 1$$
We know $$i = \sqrt { - 1} $$
$$\therefore$$ $${i^2} = - 1$$
$$ \Rightarrow $$ $${i^3} = - 1 \times i = - i$$
$$ \Rightarrow $$ $${i^4} = - i \times i = - {i^2} = - \left( { - 1} \right) = 1$$
So when power of $$i$$ is 4 or multiple of 4 then it's value is = 1
$$\therefore$$ $${\left( i \right)^x} = 1$$ $$ = {\left( i \right)^{4n}}$$ where n is a positive integer.
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