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

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

Then this function is entire and

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

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

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

 

[blue_raver22]

for your question. Your solution is on the right track, but there are a few mistakes that need to be corrected.

First, when you say "take the circle $|w-w_0|=r$," you should specify that this is a circle in the complex plane, not in the right half plane. The complex plane is the entire domain of f(z), so we want to choose a circle that encompasses the entire domain.

Second, in your definition of g(z), you have written $g(z) = \dfrac{1}{f(z) - w_0}$, but it should be $g(z) = \dfrac{1}{f(z)} - w_0$. This is because we want to show that $g(z)$ is bounded, not $f(z)-w_0$.

Third, in your statement of Liouville's theorem, you wrote "any bounded entire function must be constant." This is not quite correct. The correct statement is "any entire function that is bounded on the entire complex plane must be constant." This is an important distinction, because there are bounded entire functions that are not constant on the entire complex plane. For example, $f(z) = \sin(z)$ is bounded on the entire complex plane, but it is not constant.

Finally, when you say "g(z) is constant, f(z) is constant Contradiction," this is not a complete sentence and it is not clear what you are trying to say here. I believe you are trying to say that since g(z) is constant, f(z) must be constant, and this is a contradiction because we assumed that f(z) is not constant. This is correct, but it would be clearer if you wrote it out in a full sentence.

Overall, your solution is on the right track, but it needs to be clarified and corrected in a few places. I hope this helps!