Question

If $f:R\to C$ is defined by $f\left(x\right)=e^{2ix}$ 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

Answer 1

Answer: (D)

Neither one-one nor Onto

$f\left(x\right)=e^{2ix}=\cos 2x+i\sin 2x$

We know that $\cos$ and $\sin$ functions are periodic.

Period of $\cos 2x$ $=$ Period of $\sin 2x=\frac{2\pi }{2}=\pi$

$\therefore f(\pi +x)=\cos 2(\pi +x)+i\sin 2(\pi +x)=\cos 2x+i\sin 2x=f\left(x\right)$

Thus, $f\left(x\right)$ is also periodic with a period of $\pi$.

Since periodic functions are not one-one, $f\left(x\right)$ is not one-one.

Magnitude of $f\left(x\right)=|f\left(x\right)|=|e^{2ix}|=\sqrt{{\cos }^{2}2x+{\sin }^{2}2x}=1$

But, in a complex plane, magnitude of a complex number can vary from 0 up to infinity.

Since for any value of $x$, the magnitude of $f\left(x\right)$ cannot be greater than or less than $1$ (i.e., $f\left(x\right)$ doesn't cover the entire complex plane), we can say that $f\left(x\right)$ is not onto.