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i) $$ \begin{align} \text{Let }f:D(0,1)\to \mathbb{C} \text{ holomorphic ,$\\$ Show that } f(\frac{1}{n})\ne \frac{1}{n+1} \end{align} $$ for all natural numbers,except maybe for some finite cases.

I considered $\alpha_n=\frac{1}{n}$ and $\frac{1}{n+1}=\frac{\alpha_n}{\alpha_n +1}$ . Also $g(z)=f(z)- \frac{z}{z+1}$ and we can see that as $\alpha_n\to0, g(\alpha_n)=f(\alpha_n) +\frac{\alpha_n}{\alpha_n+1}=0$ , hence $f(z)=\frac{z}{z+1}$ by the identity theorem and this where I get stuck.

ii) $$ \begin{align} \\ \\ \text{Let }f:D(0,1)\to \mathbb{C} \text{ holomorphic }\\ f(1-\frac{1}{n})=0 , \forall n= 1,2,.. \end{align} $$

Can such a holomorphic $f$ exist?

I'm thinking that this would hold for $f\equiv0$

but taking $ 1-\frac{1}{n} \to 1 $ confuses me

helplessKirk
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1 Answers1

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(ii) As you can see, main condition of identity theorem $\left(1\in D\right)$ does not hold. This can help us assume that not only $z=0$ is demanded function. If we remember Picard theorem, we conclude that $f$ can have essential singularity at $1$ and condition will hold. Indeed, $$f(z)=\sin \frac{\pi}{1-z} $$ satisfies conditions: $$\sin \frac{\pi}{1-z}=0 \Leftrightarrow \frac{\pi}{1-z}=\pi k, k\in \mathbb{Z} \Leftrightarrow$$ $$\frac{1}{1-z}=k \Leftrightarrow 1-z=\frac{1}{k}, k\in \mathbb{Z}-0 \Leftrightarrow z = 1-\frac{1}{k}, k\in \mathbb{Z}$$ As $\mathbb{N} \subset \mathbb{Z}$, we have found non-zero corresponding function.

cool
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