For an odd integer $k\ge1$ , let $\mathbb{F}$ be the set of all entire functions f such that $f(x)=\vert{x^k}\vert$ for all $x\in(-1,1)$. Then the cardinality of $\mathbb{F}$ is
1.0
2.1
3.strictly greater than 1 but finite
4.infinite
f(x) = \begin{cases} x^k, & \text{if $x\in(0,1)$ } \\[2ex] -x^k, & \text{if $x\in(-1,0)$ } \end{cases}
consider convergent sequence {$\frac{1}{n}$} in (0,1)
Let g(z)=$z^k$ ,since $f(\frac{1}{n})=g(\frac{1}{n})$ on (0,1)
hence $f{(z)}=z^k$ on (-1,1) by using identity theorem.
similarly on (-1,1) i can show that $f{(z)}=-z^k$
but then how to conclude for final answer?
If $f(z)=z^k$ on $(0,1)$, then $f(z)=z^k$ on $ \mathbb C$.
A similar argument gives $f(z)=-z^k$ on $ \mathbb C$.
Conclusion: no such function exists.