# Thread: Find all entire functions f

1. ## Find all entire functions f

Find all the entire functions f such that
$f(0) = i$ and $\left |f(z)- \cos z \right |\geq{\sqrt{2}}$
for all ${z\in{\mathbb{C}}}$.

My attempt at this question...
Let $g(z) = \frac{1}{f(z)-\cos z}$.
Since $f(z)$ and $\cos z$ are entre and $f(z) - \cos z \neq 0, \forall {z\in{\mathbb{C}}}$ (since $\left |f(z)- cos z \right |\geq{\sqrt{2}}$).
$\Rightarrow g(z)$ is entire & $\left |g(z) \right |= \frac{1}{\left |f(z)-\cos z\right |} \leq \frac{1}{\sqrt{2}}, \forall {z\in{\mathbb{C}}}$.
By Liouville's Theorem, $g(z) \equiv k$ for some ${k\in{\mathbb{C}}$.
$\Rightarrow \frac{1}{f(z)-\cos z} \equiv k$ for some ${k\in{\mathbb{C}}, k \neq 0$, since $f(z) - \cos z \neq 0, \forall {z\in{\mathbb{C}}}$.
Hence $f(z) = \frac{1}{k} + \cos z$
Now it is given that $f(0) = i$, and $cos(0) = 1$, hence $\frac{1}{k} = i-1$.
$\Rightarrow f(z) = i - 1 + \cos z$

Is this correct? Are there any other solutions? Thanks in advance!.

2. ## Re: Find all entire functions f

Originally Posted by alphabeta89
Find all the entire functions f such that
$f(0) = i$ and $\left |f(z)- \cos z \right |\geq{2}$
for all ${z\in{\mathbb{C}}}$.
$\text{Umm... } f(0) = i \text{ and } \cos(0) = \frac{e^{i(0)} + e^{-i(0)}}{2} = \frac{1 + 1}{2} = 1,$

$\text{so } 2 \le \lVert f(0)- \cos(0) \rVert = \lVert -1 + i \rVert = \sqrt{2} < 2.$

The answer to this problem is that there are no such functions.

3. ## Re: Find all entire functions f

Originally Posted by johnsomeone
$\text{Umm... } f(0) = i \text{ and } \cos(0) = \frac{e^{i(0)} + e^{-i(0)}}{2} = \frac{1 + 1}{2} = 1,$

$\text{so } 2 \le \lVert f(0)- \cos(0) \rVert = \lVert -1 + i \rVert = \sqrt{2} < 2.$

The answer to this problem is that there are no such functions.
Hi, there is a typo in my question. I have made the necessary changes. Sorry for that. Is my solution still correct?

4. ## Re: Find all entire functions f

Yes - that looks absolutely correct to me. And because your reasoning works backwards from assuming such an entire f, there are no other solutions. Well done.