Let $f$ be an entire function such that image of $f$ lies in $L=\{2+iy:y\in R\}$ if $f(2+i)=2+i$ then show that $f(z)=2+i$ for all $z \in C$

Question

Let $~f~$ be an entire function such that image of $~f~$ lies in $L=\{2+iy:y\in R\}$ if $f(2+i)=2+i$ then show that $f(z)=2+i$ for all $z \in C$

Answer 1

Hint: Let $y=\dfrac1n+1$ and use identity theorem.

Answer 2

Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$. $Re(f(x))=2$ for $x \in \mathbb{R}$. Hence \begin{equation} \sum_{n=0}^{\infty} Re(a_n) x^n = 2. \end{equation} Hence $Re(a_0) = 2$ and $Re(a_n) = 0$ for $n \geq 1$.

Hence $f(z) = 2+ \sum_{n=0}^{\infty} ib_n z^n$, $b_n \in \mathbb{R}$.

Now set $z=ix$. $f(z) = 2+ \sum_{n=0}^{\infty} ib_n i^n x^n$

Hence $Re(f(z)) = 2+ \sum_{n=0,n \text{odd}}^{\infty} b_n (-1)^{\frac{n+1}{2}} x^n = 2$.

Hence $b_n = 0$, $\forall n $ odd.

Now set $z = e^{i \frac{\pi}{4}} x$ to conclude $b_n = 0$ $\forall n \mod 4 \neq 0$. Now repeat the argument with $z = e^{i \frac{\pi}{8}} x$ and so on with $z=e^{i \frac{\pi}{2^k}} x$.

Hence $f(z) = 2+ib_0$ and by $f(2+i) = 2+i$, we have $f(z) = 2+i$.