MHB Can Analytic Functions Exhibit These Complex Behaviors?

[Markov2]

Denote $D=\{z\in\mathbb C:|z|<1\}$

1) Let $\mathcal U\to\mathbb C$ open and $L\subset\mathbb C$ a line. If $f:\mathcal U\to\mathbb C$ is a continuous function which is analytic on all the points $z\in\mathcal U\cap L^c,$ show that $f$ is analytic on $\mathcal U.$

2) Does exist a function $f:\overline D\to\mathbb C$ analytic, bijective and with analytic inverse?

3) Compute the Laurent series of $f(z)=\log\dfrac{z-1}{z+1}$ around $z=0.$ Show where the series converges to $f.$

4) Let $R>0$ and $\Omega=\{z\in\mathbb C:|z-1|-|z|>R\}.$ Does exist an analytic function $f:\mathbb C\to\Omega$ ?

Attempts:

1) No ideas here, what is the key theorem?

2) I think it doesn't, but I don't see a counterexample.

3) I have a problem here, I don't know if $|z|<1$ or $|z|>1,$ which one should I assume? Because I can write $\displaystyle f(z) = \log \left( {1 - \frac{1}{z}} \right) - \log \left( {1 + \frac{1}{z}} \right) = - \sum\limits_{k = 1}^\infty {\frac{1}{k}{{\left( { - \frac{1}{z}} \right)}^k}} + \sum\limits_{k = 1}^\infty {\frac{1}{k}{{\left( {\frac{1}{z}} \right)}^k}} ,$ which converges for $\dfrac1{|z|}<1,$ so that means indeed that $|z|>1,$ does this make sense?

4) Well first some of algebra, we have $|z-1|-|z|=\sqrt{(x-1)^2+y^2}-\sqrt{x^2+y^2}>R,$ this is very, very messy, is there a way to indentify the curve? Because by having that I could conclude that such function exists or doesn't.

 

[tarantella]

2) Hint: if such a function $f$ exists, its inverse cannot be analytic by Liouville theorem.

 

[Markov2]

Okay so you're looking for a contradiction, but how do you use Liouville to prove it?

Can you please check my attempts or help me with other problems please?

 

[tarantella]

$f^{-1}$ is analytic, since $f^{-1}(z)\in\overline D$ we have $|f^{-1}(z)|\leq 1$ for all $z\in\mathbb C$, so $f^{-1}$ is constant. In particular it cannot be injective, so we got the contradiction.

In fact, there is a theorem of Riemann, which says that for each simply connected open $U$ subset of $\mathbb C$ (i.e. without holes) which is different from $\mathbb C$ we can find $f\colon U\to D$ analytic and bijective.

 

[Markov2]

Very nice, I can get it now.

Can you help me with other problems please?

 

[AlexYoucis]

2) Hint: if such a function $f$ exists, its inverse cannot be analytic by Liouville theorem.

Or you could just use the fact that any such $f$ is a homeomorphism, $\overline{\mathbb{D}}$ is compact, and $\mathbb{C}$ is not.

 

[Markov2]

Can anybody give me a hand for 1) and 4) please? Is my work for 3) correct?

 

[Markov2]

For problem 1) it looks like Morera's Theorem works, but I don't know how to make it, how to do it?

 

[tarantella]

AlexYoucis: in fact I didn't see $f$ was supposed to be defined on $\bar D$; indeed in this case we don't need complex analysis argument but only topological one. But the question is more interesting with $D$ instead of $\bar D$.

Markov: for question 3 you have to specify which branch of logarithm you are using.

 

[Markov2]

For problem 1), it should be $\mathcal U\subset\mathbb C,$ but now I see the problem, isn't it easy? Because $\mathcal U\cap L^c=\mathcal U$ and the conclusion follows.

Could you help me for problem 4)?