MHB No Non-Constant Entire Function $f(z)$ with $Re(f(z))<0$

[Amer]

Show that there is no non constant entire function f(z) such that $Re(f(z)) < 0 $

My solution, suppose there exist a non constant function with $Re(f(z)) < 0 $
Take the circle $|w - w_0| = r $ we choose $w_0, r$ such that the circle is in the right half plane like this
View attachment 2297

$f(z) $ is entire, and $|f(z) - w_0| > r $ for all z
Let
$g(z) = \dfrac{1}{f(z) - w_0} $, g is entire function since the denominator dose not vanish ( =/= zero ) and g(z) is bounded
$ |g(z) | = \dfrac{1}{ |f(z) - w_0|} < \dfrac{1}{r}$
By Liouville's theorem, any bounded entire function must be constant so g(z) is constant, f(z) is constant Contradiction.

But my Pro said my solution is not Ok, where is the mistake ?

Thanks

 

[ThePerfectHacker]

For one thing your proof is a sloppy. What is $w_0$? You never define it. What is $r$? How are you defining them?

Here is a more precise way to say what you want to say. We will greatly generalize your theorem.

Definition: A subset $D$ of $\mathbb{C}$ is called dense if and only if the closure of $D$ is $\mathbb{C}$. In other words, every point of $p\in \mathbb{C}$ has the property that the disk $\Delta(p,r) $ intersects $D$ for every $r>0$.
In fancy language,
$$ \forall p\in \mathbb{C},~ \forall r > 0,~ \Delta(p,r) \cap D \not = \emptyset$$

Negating this statement tells us that if $D$ is not dense then,
$$ \exists p\in \mathbb{C}, \exists r > 0, ~ \Delta(p,r)\cap D = \emptyset $$

Theorem: If $f:\mathbb{C}\to \mathbb{C}$ is an entire function which is non-constant then the image of $f$ is dense set of $\mathbb{C}$.

Corollary: Given the same as above in theorem, if $\text{Re}(f) < 0$ then $f$ must be constant. (Your problem).

Now we prove the theorem.

Proof: Suppose it aint true that the image $I$ is dense. Then there is a $p\in \mathbb{C}$ and $r>0$ such that the disk $\Delta(p,r) \cap I = \emptyset$. This means for any $w\in I$ we have that $w\not \in \Delta(p,r) \implies |p-w|\geq r$. Since $I$ is the image set it follows that $|p-f(z)|\geq r$ for all $z\in \mathbb{C}$. Now define the function,
$$ g(z) = \frac{1}{f(z) - p} $$
This function $g$ is entire and as you said,
$$ |g(z)| = \frac{1}{|f(z)-p|} \leq \frac{1}{r} $$
Thus, $g$ is a bounded as you said forces $f(z)$ to be constant by Liouville which is a contradiction.

 

[Amer]

Thanks, I picked them in the picture!.
I picked $w_o $ and $r$ such the points $w$ that satisfies $|w - w_o| = r $ are in the right half plane.
for exmaple $w_o = (2,2) $ and $r = 1$

Thanks again.
In fact I build my work after I read that the Image of the Entire function is dense.
Your proof is very clean and clear.:D

 

[alyafey22]

Consider the following function

$$g(z)=e^{f(z)}=e^{Re(f(z))}e^{iIm(f(z))}$$

Then this function is entire and

$$|g(z)|\leq e^{Re(z)} \leq 1$$

By Liouville theorem $g$ is constant hence $g'=0$

$$g'=f'(z) e^{f(z)}= 0$$ hence we have $f'(z)=0$