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# If $f:R \to C$ is defined by $f(x) = {e^{2ix}}$for $x \in R$, then $f$ is ?(where $C$ denotes the set of all complex numbers)A.One-oneB.OntoC.One-one and ontoD.Neither one-one nor onto.

Last updated date: 17th Jul 2024
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Hint: The given function asks us to find the nature of the given function whether it is injective(one-one) or surjective(onto ) or both. The function which is one one will always gives a unique value for every number in its domain which means that the functions that are cyclic or give same value for many values input are not one-one function, on the other hand onto function also known as surjective function occupy all the values in the codomain, hence for a function to be onto every element in its codomain has to be exhausted. Also important to remember the De Moivre's theorem for complex numbers.here ,
${e^{i\theta }} = \cos \theta + i\sin \theta$.

$f(x) = {e^{2ix}}$, the function can be written as the D’Moviers theorem as,
$f(x) = \cos 2x + i\sin 2x$.
$f(x) = \cos 2x + i\sin 2x$
Will give maximum value of $1$. But the codomain is in the whole complex aka argand plane, we can therefore safely say it will not cover every value in it because it is periodic and maximum value is $1$. Thus the function is not onto.