Complex Analysis: Solve Injective Function f(z)=az+b

[jjou]

[SOLVED] Complex Analysis

PROBLEM
Let a function f be entire and injective. Show that f(z)=az+b for some complex numbers a,b where a is not 0. Hint: Apply Casorati-Weierstrass Theorem to f(1/z).

THEOREM
Casorati-Weierstrass Theorem: Let f be holomorphic on a disk $$D=D_r(z_0)\{z_0}$$ and have an essential singularity at $$z_0$$. Then f(D) is dense in the complex plane.

ATTEMPT
I think this is the general idea for the proof:

f is entire thus f(1/z) is holomorphic except at z=0. We want to show that f(1/z) has a removable singularity at 0.

f(1/z) does not have an essential singularity at 0: Suppose f(1/z) has an essential singularity at 0. For any positive integer p, we can find a positive integer q such that, if |w|>1/p, then |f(w)|>q. Thus for any |z|<p, |f(1/z)|>q which is a contradiction. (By CW, $$f(D_p(0)\{0})$$ should be dense in $$\mathbb{C}$$.) Thus f(1/z) does not have an essential singularity at 0.

If it is a removable singularity then f(1/z) is bounded in some disk D' of 0 and thus f(z) is bounded outside of some disk D*. Since D* is compact, f(z) must also be bounded for z in D*. Thus we conclude f(z) is bounded on all of C. We know f cannot be a log, exponential, or inverse function (since those functions are not bounded). Furthermore, f cannot be a trig function (not injective). Thus f must be a polynomial. Then, to satisfy the boundedness condition, the power series expansion of f has finitely many terms.

So $$f(z)=a_0+a_1z+...+a_nz^n$$. First, n>0, otherwise f(z) would be constant and not injective. Furthermore, if n>1, then f(z) would also fail to be injective. Thus n must be exactly 1, ie $$f(z)=a_0+a_1z$$.

QUESTION
I'm having a hard time dealing with the possibility that f(1/z) has a pole singularity at 0. Either I want to show that f(1/z) cannot have a pole at 0 (method 1) or that if f(1/z) has a pole at 0 then f(z) is a polynomial of finite degree (method 2). In both scenarios, I'm having some trouble...

1) If f(1/z) has a pole at 0, then f(z) has a pole at $$\infty$$. In other words, $$|f(z)|\rightarrow\infty$$ as $$|z|\rightarrow\infty$$. Intuitively, I think this might contradict the injectivity of f, but I'm not sure how to show it.

2) If f(1/z) has a pole at 0, then in some disk D centered at 0, $$f(1/z)=\sum_{k=-n}^{\infty}a_kz^k$$. So f(z) for z outside of a certain disk can be represented as
$$f(z)=a_{-n}z^n+...+a_{-1}z+a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+...$$.
I'm not too clear on how to define the degree of a polynomial with negative exponents... Would I say this has degree n (since that is the largest positive exponent) or "infinite degree" (since it has an infinite tail of terms with negative exponents)?

I'm not sure where to go from there or if either method is correct. Any suggestions would be greatly appreciated! :)

 

[jjou]

Ahhh... I just realized that, if f is entire and bounded, by Liouville's Theorem, f must be constant - thus it cannot be injective. So my argument for a removable singularity at 0 does not work. :(

Please help!

 

[jjou]

Okay, I'm redoing my argument since now I have that 0 is neither an essential singularity nor a removable singularity for f(1/z). Thus 0 must be a pole for f(1/z). Then we follow method 2 as described above.

So, for z outside of some disc D, we have
$$f(z)=a_{-n}z^n+...+a_{-1}z+a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+...$$

However, since f is entire, for any point z, we can represent f by a power series:
$$f(z)=b_0+b_1z+b_2z^2+...$$.

Then for any z outside of D,
$$f(z)-f(z)=...+b_{n+2}z^{n+2}+b_{n+1}z^{n+1}+(a_{-n}-b_n)z^n+...+(a_{-1}-b_1)z+(a_0-b_0)+\frac{a_1}{z}+\frac{a_2}{z^2}+...=0$$

Can I then make the inference that $$f(z)=b_0+b_1z+b_2z^2+...b_nz^n$$?

 

[jjou]

Got it. :)