Suppose $f$ is holomorphic in an open neighbhourhood of $z \in \mathbb C$. Given that the series $f(z)=\sum_{n=0}^\infty{f^{(n)}(z)}$ converges absolutely, we can conclude that
$f$ is constant
$f$ is a polynomial
$f$ can be extended to an entire function
$f(x) \in \mathbb R$ for all $x \in \mathbb R$
My attempt : I take $f(z) = i$ for option D. as option D is clearly false. I don't know the other option pliz help me and tell me the solution…
On
By the absolute convergence of the series on the right, we can differentiate summand-wise. But then $$ \begin{align}f^{(k)}(z)=&\frac{\mathrm d^k}{\mathrm dz^k}\sum_{n=0}^\infty f^{(n)}(z)\\&=\sum_{n=0}^\infty \frac{\mathrm d^k}{\mathrm dz^k}f^{(n)}(z)\\&=\sum_{n=0}^\infty f^{(n+k)}(z)\\&=\sum_{n=k}^\infty f^{(n)}(z)\\&=\sum_{n=0}^\infty f^{(n)}(z)\quad-\sum_{n=0}^{k-1}f^{(n)}(z)\\ &= f(z)-\sum_{n=0}^{k-1}f^{(n)}(z)\\\end{align}.$$ For $k=1$, this gives us $f'(z)=f(z)-f(z)=0$, and for $k>1$, $$ \begin{align}f^{(k)}(z)&=f(z)-\sum_{n=0}^{k-1}f^{(n)}(z)\\&=f(z)-\sum_{n=0}^{k-2}f^{(n)}(z)-f^{(k-1)}(z)\\&=f^{(k-1)}(z)-f^{(k-1)}(z)\\&=0\end{align}$$ We conclude that the Taylor series of $f$, developed around $z$, has all coefficients $=0$ except for the constant term, i.e., $f$ is constant. Therefore claim (1) is true, as are claims (2) and (3), but not necessarily claim (4).
The function $f$ can be extended to the entire function$$F(w)=\sum_{n=0}^\infty\frac{f^{(n)}(z)}{n!}(w-z)^n.$$This series converges abolutely, for every $w\in\mathbb C$, because you can apply Abel's test to the series$$\sum_{n=0}^\infty\bigl|f^{(n)}(z)\bigr|\frac{|w-z|^n}{n!}.$$Note that the series that defines $F$ is the Taylor series of $f$; therefore, $F(w)=f(w)$ in the neighborhood of $z$.