f is a holomorphic function at every z in C. We know that lim f(z)=0 when z->+00. Show that f(z)=0 for every Z in C.

Can you please give me some kind of hint ?? i would be sooo grateful...(Rofl)

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- May 14th 2013, 03:50 AMEleniproof a complex function is zero
f is a holomorphic function at every z in C. We know that lim f(z)=0 when z->+00. Show that f(z)=0 for every Z in C.

Can you please give me some kind of hint ?? i would be sooo grateful...(Rofl) - May 15th 2013, 12:44 PMGJARe: proof a complex function is zero
Hi Eleni,

Since $\displaystyle f(z)$ is holomorphic at each point of $\displaystyle \mathbb{C},$ $\displaystyle f$ is an entire function. The idea here is to use Liouville's Theorem. Try finding a way to use the information that $\displaystyle f(z)\rightarrow 0$ to prove that $\displaystyle f$ is bounded. Once you have $\displaystyle f$ bounded and entire Liouville's Theorem will tell you that $\displaystyle f$ is constant.

Does this get things on the right track? Good luck! - May 15th 2013, 04:30 PMhollywoodRe: proof a complex function is zero
GJA - I don't see a way to easily prove that f is bounded. You can't just use the fact that f(z) goes to zero - there are functions that go to zero but are not bounded.

I think you can expand f(z) in a power series and use Cauchy's Integral Formula to prove all the coefficients are zero except for the constant coefficient. So f(z) is a constant which must be zero. That's the way the proof of Liouville's Theorem goes.

I think it might be possible to argue directly from Cauchy's Integral Formula. If so, that would be a much better solution.

- Hollywood - May 16th 2013, 07:41 AMGJARe: proof a complex function is zero
If $\displaystyle f(z)\rightarrow 0$ as $\displaystyle |z|\rightarrow\infty,$ then there is a compact disc centered at the origin such that $\displaystyle |f(z)|<1$ for $\displaystyle z$ outside of this disc. Since this disc is compact and $\displaystyle f$ is entire (so continuous), $\displaystyle f$ is bounded on the disc. The bound for $\displaystyle f$ is then the maximum of 1 and the bound on the disc.

- May 16th 2013, 08:13 AMhollywoodRe: proof a complex function is zero
Sure enough. Thanks.

Eleni: GJA's post shows that f is bounded, and you know f is entire. So by Liouville's Theorem, f is constant. Since its limit is zero, then, it must be identically zero.

- Hollywood