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9

You are right, and your example shows that not every holomorphic function on the open unit disk is the uniform limit of polynomials. However, every function that is continuous on the closed unit disk and holomorphic on the open unit disk is the uniform limit of polynomials, and every holomorphic function on the open unit disk is the locally uniform limit of ...


6

Of course you are right: it is impossible to approximate an unbounded function uniformly with bounded functions.


2

We can identify the group of automorphisms of the upper half plane with $PSL(2,\mathbb{R})$, the group of $2\times 2$ matrices with determinant 1, up to $\pm I$. That is, they are transformations of the form $$z \rightarrow \frac{az + b}{cz + d},$$ where $ad - bc = 1$ and $(a,b,c,d) \sim (\alpha a, \alpha b, \alpha c, \alpha d)$ for $\alpha \neq 0$, $a, b, ...


1

For the function $z^n-1$ we have equality in the statement. Two observations: (1) We can assume that $|a_j|\le1$ for every $j$. Proof: If $|z|=1$ then $|z-a|=|z|\cdot|\bar{z}-\bar{a}|=|1-\bar{a}z|=|a|\cdot\left|z-\frac1{\bar a}\right|$, and $1+|a|=|a|\cdot\left(1+\frac1{|\bar{a}|}\right)$. If we replace the factor $z-a$ in the polynomial by ...


1

The geometric interpretation is the following : any holomorphic function of the Riemann Sphere/complex projective line is constant.


1

Note that $\cos(ix)=\cosh(x)$ and $\sin(ix)=i\sinh(x)$, so that $$f(z) = \sin(x)\cosh(y)+ i(\cos(x)\sinh(y)+c)=\sin(x)\cos(iy)+ \cos(x)\sin(iy)+ic$$ $$=\sin(x+iy)+ic=\sin(z)+ic$$


1

Berenstein/Gay: Complex variables. Imo it satisfies the first four points. Concerning a sane statement of Liouville's theorem: the usual statement is the sane one; if you don't understand this, think harder about it. It also does not treat complex analysis in several variables, for this you should take a look at Hörmander; I am also not sure as far as power ...



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