If f: R rightarrow C is defined by f(x)=e^{2 | vedclass

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(D) Given the function $f: R \rightarrow C$ defined by $f(x) = e^{2 i x} = \cos(2x) + i \sin(2x)$.For $f$ to be one-one, $f(x_1) = f(x_2)$ must imply

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neither one-one nor onto

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(D) Given the function $f: R \rightarrow C$ defined by $f(x) = e^{2 i x} = \cos(2x) + i \sin(2x)$.For $f$ to be one-one, $f(x_1) = f(x_2)$ must imply $x_1 = x_2$. However, $f(x + \pi) = e^{2 i (x + \pi)} = e^{2 i x + 2 i \pi} = e^{2 i x} \cdot e^{2 i \pi} = e^{2 i x} \cdot 1 = f(x)$. Since $f(x) = f(x + \pi)$ for all $x \in R$, the function is many-one.For $f$ to be onto, the range of $f$ must be equal to the codomain $C$. The range of $f(x) = \cos(2x) + i \sin(2x)$ is the set of all complex numbers $z$ such that $|z| = 1$ (the unit circle in the complex plane). Since the codomain is the set of all complex numbers $C$, and the range is only a subset of $C$ (the unit circle), the function is not onto.Therefore, $f$ is neither one-one nor onto.

If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$, then $f$ is (where $C$ denotes the set of all complex numbers)

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