# Existence of holomorphic functions such that...

1. Does there exist a function holomorphic on the unit disk such that $$f(\frac{1}{n}) = \frac{ (-1)^n}{n^2}$$, $$n \in \mathbb{N}, n \geq 2$$

2. The same but now the condition is $$f(\frac{1}{n}) = \frac{2 + n}{3 + n}$$.

I don't have the solution to check myself and I have just begun to study this topic, so I have no idea if my thoughts regarding given above problems are correct.

Here are my attempts of solution:

According to identity theorem (which is satisfied because the limit point of $$1/n$$ belongs to the disk) all functions which coincides on {$$1/n$$} are equal. For the first problem consider the function $$f(z) = (-1)^{1/z} z^2$$

Looks like the condition on function from the problem is satisfied, this function is not holomorphic, does it mean that such a function doesn't exist?

For the second problem it looks like $$f(z) = \frac{2z + 1}{3z + 1}$$ is good and the answer is yes because it is obviously holomorphic on the unit disc. But I have a feeling that something is wrong about it.

I would appreciate correct solution or pointing out my mistakes.

• How do you define $(-1)^z$? Aug 5, 2017 at 21:44
• I guess there is no another way but $e^{zln(-1)}$. The thing is, I am not quite sure if it's OK for $-1$, however complex logarithm does exist for negative numbers Aug 5, 2017 at 21:47
• For the first problem: look at the case where n even, what does that imply about the function? After you've done that, see if the condition f(1)=-1 is satisfied. Aug 5, 2017 at 21:50
• For 1. consider $g(z):=z^2$ and $h(z):=-z^2$ on $D_1:=\{1/n| n\text{ odd }\}$ and $D_2:=\{1/n| n\text{ even }\}$. Aug 5, 2017 at 21:50
• @Vladislav And how do you choose the logarithm? The number $-1$ has infinitely many of them. Aug 5, 2017 at 21:52

1. Suppose that such a function exists and let $g(z)=z^2$. Then the equality $f\left(\frac1n\right)=\frac{(-1)^n}{n^2}=g\left(\frac1n\right)$ holds for all even natural numbers $n$. Therefore, by the identity theorem, $f=g$, which is impossible, since $f\left(\frac13\right)\neq g\left(\frac13\right)$.
2. Suppose that such a function exists and let $g(z)=\frac{2z+1}{3z+1}$ ($z\in D(0,1)\setminus\left\{-\frac13\right\}$). Then, by the identity theorem, $f=g$. But this is impossible, because the limit $\lim_{z\to-\frac13}f(z)$ exists, whereas the limit $\lim_{z\to-\frac13}g(z)$ doesn't.