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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-one
B.Onto
C.One-one and onto
D.Neither one-one nor onto.

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Last updated date: 17th Jul 2024
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Answer
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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 $.

Complete step-by-step answer:
The function given is
 \[f(x) = {e^{2ix}}\], the function can be written as the D’Moviers theorem as,
$f(x) = \cos 2x + i\sin 2x$.
The trigonometric function is a periodic function, because the function is the trigonometric function, and will repeat after a specific value. But the values given by the function must be unique for it to be a one one function and thus this function is not one-one.
For check for onto we know that the function
$f(x) = \cos 2x + i\sin 2x$
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.
The correct option will therefore be D, which is that it is neither one one nor onto.
So, the correct answer is “Option D”.

Note: There is another way to check for the surjectivity, (property of a function being an onto function) which is the range of the function in question should be equal to its codomain, because if a function gives value in all of its codomain we can say it will occupy whole of the codomain and will be onto, our given function had codomain of complex numbers but the range only lied between a small set of points and hence the result of being not an onto function.