1
$\begingroup$

That is all the given information.

I should find all the entire functions that are one to one, I have no clue on how to do that.

And there is also another one, that goes like this:

Find all entire functions that are:

$$\forall z \in \mathbb C^* \;\;\;\;\; \lvert f(z)\rvert \le \lvert \frac{\sin z}{z} \rvert$$

$$\mathbb{C}^* \text{ is complex plane without } (0,0)$$

  • 1
    Please ask only one question per post, and [provide context](http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question#9960). – 2017-01-10

2 Answers 2

0

For the first one, there are four cases to consider: constant functions, polynomials of degree $1$, polynomials of degree $>1$, and non-polynomials. The first two cases are obvious. For the third case the derivative is a non-constant polynomial and thus has zeroes, and holomorphic functions are not one-to-one near where their derivatives vanishes. For the last case, $f(\frac{1}{z})$ has an essential singularity and thus takes on almost every value in $\mathbb{C}$ repeatedly as you go to the origin (Picard's theorem). So only the degree $1$ polynomials are injective.

  • 0
    Thanks a lot, that did the trick :) – 2017-01-10
  • 0
    That is not what the Casorati Weierstrass theorem says: all it says is that if $f$ has an essential singularity at $p$ (which may be either in $\Bbb{C}$ or equal to $\infty$), then the range of $f$ when restricted to any neighborhood of $p$ is dense in $\Bbb{C}$. To use the Casorati Weierstrass theorem to prove that entire functions with essential singularities cannot be one to one, you have to use the open mapping theorem as an intermediate step. – 2017-01-10
  • 0
    Right I was thinking of Picard's theorem. – 2017-01-10
0

Hint: Observe that if a function $f$ is entire and one to one, then $f$ cannot have an essential singularity at $\infty$ by the great Picard's theorem. This means the Taylor series expension of $f$ at $0$ has only finitely many terms, and thus $f$ is a polynomial. Now note that a polynomial is one-to-one if and only if the degree is $1$.