1. ## [SOLVED] Elliptic functions

Consider the function $w(z)=\int_1^z\frac{dt}{t}$ where the integral is along any rectifiable path from $1$ to $z$. Clearly this is not a single-valued function of $z$, as the value of the integral will depend on the winding number of the path around $0$. Thus $w(z)$ is defined up to an integral multiple of $2\pi i$. Now consider the inverse function $z(w)$; we have $z(w+2\pi i)=z$ and therefore $z$ is a periodic function. (It's the exponential!)

Now consider the function $w(z)=\int_1^z\left(\frac{1}{t}+\frac{i}{t-i}\right)dt$. Now $w$ is defined up to $2\pi i m + 2\pi i n$ where $m,n \in \mathbb{Z}$. Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.

2. Originally Posted by Bruno J.
Consider the function $w(z)=\int_1^z\frac{dt}{t}$ where the integral is along any rectifiable path from $1$ to $z$. Clearly this is not a single-valued function of $z$, as the value of the integral will depend on the winding number of the path around $0$. Thus $w(z)$ is defined up to an integral multiple of $2\pi i$. Now consider the inverse function $z(w)$; we have $z(w+2\pi i)=z$ and therefore $z$ is a periodic function. (It's the exponential!)

Now consider the function $w(z)=\int_1^z\left(\frac{1}{t}+\frac{i}{t-i}\right)dt$. Now $w$ is defined up to $2\pi i m + 2\pi i n$ where $m,n \in \mathbb{Z}$. Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.

I may be off orbit by long miles, but let's give it a try:

Spoiler:
Indeed we get $z(w+2\pi i m+2\pi in)=z(w)$ , so this function is doubly periodic (...really? Read on), and thus, what's lacking to consider it an elliptic function? Well, the periods must be a basis for the real dimensional linear space $\mathbb{C}$ , and in this case we get $\frac{2\pi i}{2\pi i}=1\in\mathbb{R}$ , which screws the whole thing up.
The complete, long and deep explanation may be well beyond what I'd be willing to explain by this means: We need a free abelian group in $\mathbb{C}$ which is also a maximal order there and etc.
Making it short, though, we can simply say: that the above ratio is real and not complex non-real means the function $z(w)$ has actually just one single period (!) which, automatically, disqualifies it from being an elliptic function.

Tonio

3. Oh, I'm sorry, my post should have read

Now is defined up to $2\pi in + 2\pi m$ where .
That's why I used $\frac{i}{t-i}$ instead of $\frac{1}{t-i}$ : so that the residue at $i$ (or $2\pi i$ times the residue, if you prefer) would be real.

So the two "periods" are indeed linearly independent over $\mathbb{R}$.

4. Originally Posted by Bruno J.
Oh, I'm sorry, my post should have read

That's why I used $\frac{i}{t-i}$ instead of $\frac{1}{t-i}$ : so that the residue at $i$ (or $2\pi i$ times the residue, if you prefer) would be real.

So the two "periods" are indeed linearly independent over $\mathbb{R}$.

Spoiler:
Mind you, I did notice the $\frac{i}{t-i}$ thing and thought it is a little odd the second assumed period is pure imaginary, but of course I didn't check it (as I am not checking it, either ) .

Well, then the two periods are a $\mathbb{R}-$basis for the complex, so then the only reason why that function wouldn't be an elliptic one I can think of right now is that the function isn't meromorphic on a fundamental parallelogram. Now, either the function is there holomorphic and thus bounded and thus constant (but STILL would be consider elliptic, imo), or else it has a singularity that it is not a pole....oh, I think I see now! Zero is there...hmmm.
Anyway, it's late here so I shall check this closer tomorrow, perhaps.

Tonio

5. Here's a hint : the universal cover of the twice punctured plane is the disc...

6. Here's the solution I know. I suspect there is a less technical one.

The function $z \mapsto w$ is well defined $\mod 2\pi i \mathbb{Z} \oplus 2\pi \mathbb{Z}$, i.e. it is well defined on the elliptic curve $E=\mathbb{C}/(2\pi i \mathbb{Z} \oplus 2\pi \mathbb{Z})$. Now suppose there were an analytic inverse function $g : E \rightarrow \mathbb{C}-\{0,i\}$. Lifting $g$ to a map between the universal coverings (which are, respectively, $\mathbb{C}$ for $E$ and the unit disc $D$ for the twice-punctured plane $\mathbb{C}-\{0,i\}$), we obtain an analytic map $\tilde g : \mathbb{C} \rightarrow D$, which, by Liouville's theorem, must be constant...