- Thread starter
- #1

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 $f:\mathbb C\to\Omega$ ?

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.

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.

Last edited: