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1) Find all entire functions that are uniformly continuous on $\mathbb{C}$.

2) Find all entire functions $f(z)$ such that such that for every integer $n \geq 1$,

$\oint_{\partial\mathbb{D}} f(z)\bar{z}^ndz = 0,$ where $\mathbb{D}$ is the unit disk.

I'm a bit shaky on the first one, but I think it's that an entire function has an infinite radius of convergence, so is everywhere normally convergent. So if each term in it's power series is uniformly continuous on $\mathbb{C}$, then the function will be uniformly continuous on $\mathbb{C}$. Am I on the right track?

For the second, I'm not sure how to use the Cauchy Integral Formula since $f(z)\bar{z}^n$ isn't holomorphic.

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    @KannaguchiO. Having |f(z+\delta)-f(z)|<\epsilon for all small $\delta$ does **not** imply |f'(z)|<\epsilon. The definition of derivative involves **divided** difference.2012-10-01

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Definition of uniform continuity: for any $\epsilon$ there is a $\delta$ such that $|x-y| < \delta \implies\ |f(x)−f(y)|<\epsilon$ for any choice of $x, y$.

A trascendental entire function cannot be uniformly continuous, having an essential singularity at $\infty$ (so you can find arithmetical sequences $z_i$ with exponentially growing values of $f(z_i)$).

So we are left with polynomial, which are not uniformly continuous even in the real numbers, except for the linear case $z \to az+b$ which is uniformly continuous almost by definition.

For the second question, the line integral is exactly the Cauchy integral over the unit cicle, as $z^{-1} = \bar{z}$. So the definition is equivalent to saying that every derivative (starting from $n=0$ which is the actual value of the function) is zero, so the only solution is the constant $f:z \to 0$

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    In fact, @LVK's comment on the question is a plainly better route. Just take $g(z) = f(z+\delta/2)-f(z)$. This is entire, and bounded by $\epsilon$, so it's a constant. So the function is linear.2012-10-01