This is a question from a previous complex analysis qualifying exam. I'm working problems to study for my own upcoming qual. I'd like to know if my solution below is correct and complete or not. Thanks!


Find all analytic bijections $f: \mathbb{C} \to \mathbb{C}$. Justify that there are no other analytic bijections besides those you found.

Attempted Solution:

(1) Since we're looking for analytic functions on the whole complex plane, we are considering only functions which are entire.

(2) A constant function is neither injective nor surjective, so clearly it is not a bijection. By Liouville's Theorem, any bounded entire function is constant. Thus, such a bijection must have a singularity at infinity since it is analytic everywhere else.

(3) The singularity at infinity must be either a pole or an essential singularity. If there is an essential singularity at infinity, then by the Great Picard theorem, in any neighborhood of infinity, the function attains every value in $C$ (with at most one exception), so these functions would be "infinity-to-1". Thus, not bijections.

(4) Finally, we are left with entire functions that have a pole at infinity. Any such function is a polynomial (since it can be written as a power series), but polynomials of degree $n$ are $n$-to-1 functions. Thus, the only functions that are 1-to-1 are 1st degree polynomials or linear functions of the form $f(z) = \alpha z + \beta$ with $\alpha \neq 0$.

Update to include suggestion by @Greg Martin:

(5) We have shown that if an entire function is a bijection, then it is linear. It remains only to show that if a function is linear, then it is an analytic bijection. We know that any polynomial is analytic.

We have that a function $f(z)=\alpha z + \beta, \; \alpha\neq 0$ is injective if $f(z)=f(w) \implies z=w$. This is straightforward as $$ \alpha z + \beta = \alpha w + \beta \implies z = w $$ by simply subtracting off $\beta$ and dividing both sides by $\alpha \neq 0$.

Then we have that $f$ is surjective if, for every $w \in \mathbb{C}$, there exists a $z \in \mathbb{C}$ such that $f(z)=w$. One can see that $$\alpha z + \beta = w \quad \implies z = \frac{w-\beta}{\alpha}, \; \alpha\neq 0. $$

Thus, for any $w$ there exists such a $z$, and $f(z)$ is an analytic bijection.

  • 1
    $\begingroup$ You've proved "if analytic bijection, then linear", but you should also prove (the easy direction) "if linear, then analytic bijection". Other than that, it looks good! $\endgroup$ Commented Jul 10, 2022 at 0:07
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    $\begingroup$ "... of the form $f(z)=\alpha z+\beta$" with $\alpha\ne 0$ $\endgroup$
    – jjagmath
    Commented Jul 10, 2022 at 0:10
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    $\begingroup$ You certainly do not need Great Picard. That's hugely overkill. Use Casorati-Weierstrass, whose proof you certainly should know. $\endgroup$ Commented Jul 10, 2022 at 0:11
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    $\begingroup$ The proof you gave is the standard one. $\endgroup$ Commented Jul 10, 2022 at 0:12
  • $\begingroup$ Alternatively, you can show any automorphism of the Riemann Sphere is a Mobius transformation via either via the Galois Correspondence or directly (a standard direct argument found in most textbooks requiring nothing but the existence of Laurent Series can be found here). $\endgroup$ Commented Jul 10, 2022 at 0:20


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