Let $f:\mathbb C\to\mathbb C$ be an entire function. Which of the following statements are true and which are false?
- If $f(z)\in\mathbb R$ for all $z\in\mathbb C$, then $f$ is constant.
- If $f(\frac{1}{n})=\frac{i}{n}$ for all $n\in\mathbb N$, then $f(z)=iz$ for all $z\in\mathbb C$.
- If $f$ is a non-constant polynomial, then there is a continuously differentiable path $\gamma: [0,1]\to\mathbb C$ such that $\int_\gamma f(z)dz=2\pi i$.
- The function $1/f$ has no pole in $0$.
- The function $r\mapsto\int_{|z|=r}f(z)dz$ is constant on $(0,\infty)$.
- The series $\sum_{n=0}^\infty\frac{1}{n!}f^{(n)}(1)(z-1)^n$ converges for all $z\in\mathbb C$
Here is what I got:
- True. If $f$ is not constant, then $f(\mathbb C)$ must be a domain which conflicts with $f(\mathbb C)\subseteq\mathbb R$.
- True. On the set $\{1/n|n\in\mathbb N$} the function $f$ coincides with $g(z)=iz$. Since the set does have an accumulation point in the domain of $f$, the identity theorem yields $f=g$.
- I have no idea. How do I start here?
- False. $f(z)=z$.
- True. Since $f$ is holomorphic, the integral is always $0$.
- True. This is just the taylor series of $f$ in $z_0=1$.
What you did is correct.
The assertion 3. is true. For each $w\in\mathbb C$, let $\gamma_w\colon[0,1]\longrightarrow\mathbb C$ be the path $\gamma_w(t)=tw$. Then the map $w\mapsto\int_{\gamma_w}f(z)\,\mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2\pi i$.