Let $f$ be an entire function and suppose that $\exists$ constants $M,R>0$ and $n\in \Bbb N$ such that $|f(z)|<M|z|^n$ for $|z|>R$. Show that $f$ is a polynomial of degree $\le n$. To show that $f$ is a polynomial of degree $\le n$,we show that $f^{(n+1)}(z)=0$ has uncountably many zeros in $\Bbb C$. Since $f$ is […]

Is it true that every entire function is a sum of an entire function bounded on every horizontal strip (horizontal strip is a set of the form $H_y:=\{x+iy : x \in \mathbb R \}$ ) and an entire function bounded on every vertical strip (vertical strip is a set of the form $V_x:=\{x+iy:y\in \mathbb R […]

This is a problem from Complex Variable (Conway’s book) 2nd ed. (Section 4.4) 9. Show that if $f: \mathbb{C}\to\mathbb{C}$ is a continuous function such that $f$ is analytic off $[-1,1]$ then $f$ is an entire function. I already have a solution by Morera’s theorem that split this problem in 5 cases. I think this solution […]

Function $f(z)$ is an entire function such that $$|f(z)| \le |z^{n}|$$ for $z \in \mathbb{C}$ and some $n \in \mathbb{N}$. Show that the singularities of the function $$\frac {f(z)}{z^{n}}$$ are removable. What can be implied about the function $f(z)$ if moreover $f(1) = i$? Draw a far-reaching conclusion. My attempt: If the singularities of $\frac […]

I’m trying to show that $f$ is a holomorphic (around the origin) map satisfying $|f(\frac{1}{n})|\leq \frac{1}{2^n}\forall n\in\mathbb N$ then $f$ is identically zero. Usually, if some function equals an analytic function on a dense set, we can invoke uniqueness of analytic continuation. But in this case, we don’t have equality, just a bound. I tried […]

Let $f$ be an entire function. Suppose that for each $a\in \Bbb R$ there exists at least one coefficient $c_n$ in $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ which is zero. Then: $f^{n}(0)=0$ for infinitely many $n\ge 0$. $f^{2n}(0)=0$ for every $n\ge 0$. $f^{2n+1}(0)=0$ for every $n\ge 0$. $\exists k\ge 0$ such that $f^{n}(0)=0$ for all $n\ge k$. My try: […]

Suppose $f$ is entire and $$\iint_\mathbb{C}|f(z)|^2dxdy < \infty$$Prove that $f\equiv 0.$ So far I have: Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then $$\Big|\iint_\mathbb{C}f^2(z)dA\Big| \leq \iint_\mathbb{C}|f(z)|^2dxdy < \infty$$ We can parameterize \begin{align*} \iint_\mathbb{C}f^2(z)dA &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;d\theta\;rdr \\ &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;ire^{i\theta}[-i\frac{1}{r}e^{-i\theta}]d\theta\;rdr \\ &= […]

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