JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 19)
If $${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$$, (m, n
$$ \in $$ N) then the
greatest common divisor of the least values of
m and n is _______ .
Answer
4
Explanation
$${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$$
$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$
$$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$$ = i4 [ As i4 = 1]
$$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$$
$$ \Rightarrow $$ m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
$$ \therefore $$ GCD (8, 12) = 4
$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$
$$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$$ = i4 [ As i4 = 1]
$$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$$
$$ \Rightarrow $$ m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
$$ \therefore $$ GCD (8, 12) = 4
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