MHB Families of holomorphic functions and uniform convergence on compact sets

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The discussion focuses on the sequence of complex functions defined by \( f_n = \tan(nz) \) for \( z \) in the upper half-plane. It establishes that this sequence is uniformly locally bounded, with the bound expressed in terms of the hyperbolic cotangent function, \( \coth(ny) \). A correction is noted regarding the inequality involving \( \coth \), emphasizing that it should reflect the decreasing nature of the function. Additionally, the sequence converges uniformly to the constant function \( i \) on compact subsets of the upper half-plane, with a method suggested for demonstrating this convergence. The analysis provides a solid foundation for understanding the behavior of holomorphic functions in this context.
pantboio
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Consider the sequence $\{f_n\}$ of complex valued functions, where $f_n=tan(nz)$, $n=1,2,3\ldots$ and $z$ is in the upper half plane $Im(z)>0$. I want to show two facts about this sequence:
1) it's uniformly locally bounded: for every $z_0=x_0+iy_0$ in the upper half plane, ther exist $r_0,M_0>0$ such that $|f_n(z)|\leq M_0$ for every $n$ and for every $z$ with $|z-z_0|<r_0$.
2) the sequence converges uniformly to the function identically equal to $i$ on the compact subsets of the upper half plane.

For point 1), i write $tan(nz)$ in terms of exponentials:

$$tan(nz)=i\frac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$$
Thus
$$|tan(nz)|=\frac{|e^{-inz}-e^{inz}|}{|e^{-inz}+e^{inz}|}\leq\frac{|e^{-inz}|+|e^{inz}|}{|e^{-inz}|-|e^{inz}|}=\frac{e^{ny}+e^{-ny}}{e^{ny}-e^{-ny}}=coth(ny)$$

Since $coth(y)$ is monotonically decreasing for $y=Im(z)>0$, we have
$$|tan(nz)|\leq coth(ny)\leq coth(y)\leq coth(y_0+r_0)=:M_0$$

Do you think there's some error in what i wrote?
For the point 2) i think i need a suggestion
 
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pantboio said:
Consider the sequence $\{f_n\}$ of complex valued functions, where $f_n=tan(nz)$, $n=1,2,3\ldots$ and $z$ is in the upper half plane $Im(z)>0$. I want to show two facts about this sequence:
1) it's uniformly locally bounded: for every $z_0=x_0+iy_0$ in the upper half plane, ther exist $r_0,M_0>0$ such that $|f_n(z)|\leq M_0$ for every $n$ and for every $z$ with $|z-z_0|<r_0$.
2) the sequence converges uniformly to the function identically equal to $i$ on the compact subsets of the upper half plane.

For point 1), i write $tan(nz)$ in terms of exponentials:

$$tan(nz)=i\frac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$$
Thus
$$|tan(nz)|=\frac{|e^{-inz}-e^{inz}|}{|e^{-inz}+e^{inz}|}\leq\frac{|e^{-inz}|+|e^{inz}|}{|e^{-inz}|-|e^{inz}|}=\frac{e^{ny}+e^{-ny}}{e^{ny}-e^{-ny}}=coth(ny)$$

Since $coth(y)$ is monotonically decreasing for $y=Im(z)>0$, we have
$$|tan(nz)|\leq coth(ny)\leq coth(y)\leq coth(y_0+r_0)=:M_0$$

Do you think there's some error in what i wrote?
For the point 2) i think i need a suggestion
I think that the only error comes right at the end, where you have $\coth(y)\leqslant \coth(y_0+r_0)$. Since $\coth$ is a decreasing function, that should be $\coth(y)\leqslant \coth(y_0-r_0)$. If you then take $r_0 = y_0/2$, you see that you can take $M_0 = \coth(y_0/2)$.

For 2), start from $\tan(nz)=i\dfrac{e^{-inz}-e^{inz}}{e^{-inz}+e^{inz}}$, to show that $\tan(nz)-i =\dfrac{-2ie^{inz}}{e^{-inz}+e^{inz}} =\dfrac{-2i}{e^{-2inz}+1}$, and deduce that $\bigl|\tan(nz)-i\bigr| \leqslant \dfrac2{e^{2ny}-1}$. Then show that this goes to $0$ uniformly on compact subsets of the upper half plane.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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