Use Liouville's theorem to show that two functions are equal

[docnet]

Homework Statement

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Relevant Equations

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$lim_{|z|\rightarrow \infty}\frac{f}{g}=1\neq \frac{\infty}{\infty}$

so $lim_{|z|\rightarrow \infty}f\neq \infty$ and $lim_{|z|\rightarrow \infty}g\neq \infty$.

Because f(z) and g(z) are bounded and entire, f(z) and g(z) are constant functions by Liouville's theorem
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f(z) and g(z) are constant so $\frac{f(z)}{g(z)}=1$, which implies that $f(z)=g(z)$

 

[Office_Shredder]

If $f(z)=g(z)=z$, then f and g satisfy the requirements of the question, but they are not bounded.

 

[Mark44]

Homework Statement:: .
Relevant Equations:: .

View attachment 292736
$lim_{|z|\rightarrow \infty}\frac{f}{g}=1\neq \frac{\infty}{\infty}$

The last inequality makes no sense. You have 1 being unequal to something that is undefined. You could just as well have said that $1 \neq \frac 4 2$, but then you can't come along in a subsequent equation and write that $lim_{|z|\rightarrow \infty}f\neq 4$ and $lim_{|z|\rightarrow \infty}g\neq 2$.

so $lim_{|z|\rightarrow \infty}f\neq \infty$ and $lim_{|z|\rightarrow \infty}g\neq \infty$.

Because f(z) and g(z) are bounded and entire, f(z) and g(z) are constant functions by Louisville.

f(z) and g(z) are constant so $f{(z)}{g(z)}=1$, which implies that $f(z)=g(z)$

Nit: All three distinct versions of the namesake of the theorem that were written in the OP mangle Joseph Liouville's name.

 

[FactChecker]

Consider the function $h(z)=f(z)/g(z)$. Where is it analytic? Is it bounded? What can you conclude?

 

[PeroK]

If $f(z)=g(z)=z$, then f and g satisfy the requirements of the question, but they are not bounded.

Actually, not quite, as $g(0) = 0$. For example: $f(z) = z + 1$ and $g(z) = z$ would apparently contradict the result.

 

[PeroK]

Homework Statement:: .
Relevant Equations:: .

View attachment 292736
$lim_{|z|\rightarrow \infty}\frac{f}{g}=1\neq \frac{\infty}{\infty}$

so $lim_{|z|\rightarrow \infty}f\neq \infty$ and $lim_{|z|\rightarrow \infty}g\neq \infty$.

Because f(z) and g(z) are bounded and entire, f(z) and g(z) are constant functions by Liouville's theorem
.

f(z) and g(z) are constant so $f{(z)}{g(z)}=1$, which implies that $f(z)=g(z)$

If you want to use theorems from complex analysis, then there is no obligation on being able to prove them or do complex analysis yourself.

If, however, you want to learn (how to do) complex analysis, then you need to rethink your approach. You can prove anything in mathematics (whether it is true or not) in three easy steps; as long as you don't care whether the steps are valid or not. Ultimately, you are not learning by bashing out these simple erroneous proofs without any attempt to catch your own errors. Pure mathematics requires you to check your own work rigorously. You need to develop that mindset somehow.

 

[docnet]

We are given that $h=\frac{f}{g}$ and $|g| \neq0$.

$g\cdot h=f$. we know that f and g are entire. Since the entire functions form a C-algebra $h$ is entire.

That $\frac{f}{g}$ is entire can also be shown by expanding $f$ and $g$ into their power series: link. or by computing the C.R. equations.

entirity implies continuity, so f and g are continuous over the complex plane. The complex plane is compact. and continuous functions over compact domains are bounded. so f and g are bounded.

since f and g are bounded and entire, they are constant by the Liouville's Theorem.

Since f and g are constant, $lim_{|z|\rightarrow \infty} \frac{f}{g}=1$ leads to $\frac{f}{g}=1$ and hence f=g

 

[FactChecker]

We are given that $h=\frac{f}{g}$ and $|g| \neq0$.

$g\cdot h=f$. we know that f and g are entire. Since the entire functions form a C-algebra $h$ is entire.

Right.

That $\frac{f}{g}$ is entire can also be shown by expanding $f$ and $g$ into their power series: link. or by computing the C.R. equations

Right.

entirity implies continuity, so f and g are continuous over the complex plane. The complex plane is compact. and continuous functions over compact domains are bounded. so f and g are bounded.

Wrong. Consider $f(z) \equiv g(z) \equiv z$.

since f and g are bounded and entire, they are constant by the Liouville's Theorem.

Since f and g are constant,

This is not proven because of an earlier error.

$lim_{|z|\rightarrow \infty} \frac{f}{g}=1$ leads to $\frac{f}{g}=1$ and hence f=g

Why didn't you just start with this? Step one is to apply Liouville's Theorem directly to $f(z)/g(z)$.

 

[docnet]

Wrong. Consider $f(z) \equiv g(z) \equiv z$.

This is not proven because of an earlier error.

Why didn't you just start with this? Step one is to apply Liouville's Theorem directly to $f(z)/g(z)$.

ahh! this would be wrong for $f(z) \equiv g(z) \equiv z$. i think i see what you mean. there was no need to prove the boundedness of f and g separately .. is this what you are hinting?We are given that $h=\frac{f}{g}$ and $|g| \neq0$.

$|g| \neq0$ means $h=\frac{f}{g}$ is defined over the whole complex plane. $g\cdot h=f$. we know that f and g are entire. Since the entire functions form a C-algebra $h$ is entire.

entirity implies continuity, so $\frac{f}{g}$ is continuous over the complex plane. continuous functions over compact subsets of the complex plane are bounded, so $\frac{f}{g}$ is bounded over compact subsets of the complex plane. This and $lim_{|z|\rightarrow \infty}\frac{f}{g}=1$ together lead to $\frac{f}{g}$ being bounded over the whole complex plane.

since $\frac{f}{g}$ is bounded and entire, it is constant by the Liouville's Theorem.

since $\frac{f}{g}$ is constant, $lim_{|z|\rightarrow \infty} \frac{f}{g}=1$ leads to $\frac{f}{g}=1$ and hence f=g.

 

[FactChecker]

ahh! this would be wrong for $f(z) \equiv g(z) \equiv z$. i think i see what you mean. there was no need to prove the boundedness of f and g separately .. is this what you are hinting?

Exactly.

We are given that $h=\frac{f}{g}$ and $|g| \neq0$.

$|g| \neq0$ means $h=\frac{f}{g}$ is defined over the whole complex plane. $g\cdot h=f$. we know that f and g are entire. Since the entire functions form a C-algebra $h$ is entire.

entirity implies continuity, so $\frac{f}{g}$ is continuous over the complex plane. continuous functions over compact subsets of the complex plane are bounded, so $\frac{f}{g}$ is bounded over compact subsets of the complex plane. This and $lim_{|z|\rightarrow \infty}\frac{f}{g}=1$ together lead to $\frac{f}{g}$ being bounded over the whole complex plane.

since $\frac{f}{g}$ is bounded and entire, it is constant by the Liouville's Theorem.

since $\frac{f}{g}$ is constant, $lim_{|z|\rightarrow \infty} \frac{f}{g}=1$ leads to $\frac{f}{g}=1$ and hence f=g.

Exactly.

 

[martinbn]

In the example $f(z)=g(z)=z$, to avoid that the function has a zero, take $f(z)=g(z)=e^z$.