# complex functions - how can we prove f(z) = k g(z) for every z?

Suppose f and g are entire and that, for every complex z, |f(z)| ≤ |g(z)|. Show that, also for every z, f(z) = k g(z), where k is a complex constant Ig g is not indentically 0, then |k| ≤ 1.

It's immediate that, if the claim is true, then we must have |k| ≤ 1 if g is not identically 0.

If g is identically 0, the result is trivial and we have nothing to prove. And any k will do.

If g never vanishes, then the function f/g is entire and bounded by 1. Hence, the desired conclusion follows easily from Liouville Theorem.

But what if the set Zg of zeroes of g is not empty and is not all of C either? Then, f/g is not defined on Zg. Hence, It's not entire and has a singularity at every element of Zg. We can't apply Liouville. Can anyone help me to deal with this case? If each element of Zg were a removable singularity, then it would be possible to extend f/g so as to obtain an entire function h such that h(z) = f(z)/g(z) for every z ∉ Zg. In virtue of the continuity of h, we readily see that h would be bounded by 1, which would allow us to apply Liouville to conclude h were constant. This would imply that f(z) = k g(z) for every z ∉ Zg. But since f and g are continuous, for every w ∈ Zg we'd have f(w) = lim (z → w) f(z) = lim (z → w) k g(z) =k g(0) = 0 = k g(w), so that we'd have f(z) = k g(z) for every z. And we'd be done.

I think I'm missing something. Well, another conclusion is, since g is not identically 0, then it's zeroes are isolated points of Zg. Hence, each z in Zg has a neighborhood where h is given by h = f/g and where h is holomorphic and bounded. But this doesn't imply z is a removable singularity, does this?

So, for the case where Zg is not empty and is a proper subset of the plane, I couldn't find a proof. I'd like some hint, please.

Thank you very much.

Oh, actually each z in Zg has a neighborhood V such that in V - {z} h is given by h = f/g, is holomorphic and bounded.

### 2 Answers

- Anonymous9 years agoBest Answer
Short answer: Since f/g is the quotient of entire functions, it's meromorphic, and the zeroes of g are either removable singularities of f/g or poles of f/g. They can't be poles, because at a pole p, lim z->p [f(z)/g(z)] = ∞, which contradicts the fact that |f/g| is bounded in some punctured neighborhood of p.

---

Long answer:

CLAIM. In the situation in the problem, if g(w) = 0, then either lim z->w [f(z) / g(z)] = c for some complex number c, or lim z->w [f(z) / g(z)] = ∞.

PROOF OF CLAIM. Let w be such that g(w) = 0.

Note that we must have f(w) = 0 also, because of the condition |f(w)| ≤ |g(w)|.

Assume that the f and g are not the same function. (If they are, then the result you're trying to prove is obvious; f(z) = 1 g(z).)

Let us try to compute lim z->w [f(z) / g(z)].

Since f(w) = g(w) = 0, then we may apply L'Hospital's rule; by applying this finitely many times (say, n times), we arrive at the first time where the numerator and denominator are not both zero at w, say

lim z->w [f(z) / g(z)] = lim z->w [f^(n)(z) / g^(n)(z)].

(Note that L'Hospital's Rule stops after finitely many steps because otherwise all derivatives of f(z) and g(z) would be zero at w, which would mean that all coefficients in the Taylor series at w of f and g are zero, making f and g both identically zero, which contradicts the assumption that f and g aren't the same function.)

If g^(n)(w) ≠ 0, then lim z->w [f^(n)(z) / g^(n)(z)] = f^(n)(w) / g^(n)(w) (because f^(n) and g^(n) are continuous at w, since they're derivatives of analytic functions), so the singularity is removable.

If g^(n)(w) = 0, then f^(n)(w) ≠ 0; so lim z->w [f^(n)(z) / g^n(z)] = ∞ (because the numerator is bounded near w, and the denominator grows arbitarily small in magnitude).

This completes the proof of the claim.

So all we have to do is rule out the case lim z->w [f(z) / g(z)] = ∞. But this is easy; as you've stated, w has a punctured neighborhood where f/g is bounded. This obviously contradicts lim z->w [f(z) / g(z)] = ∞ (since the limit being ∞ at w *requires* unboundedness in *all* punctured neighborhoods of the point).

- Josh SwansonLv 69 years ago
"But this doesn't imply z is a removable singularity, does this?" It does. See my reference.