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Let a group $G$ of Möbius transformations of the unit disk have an orbit satisfying the Blaschke condition, counted once per distinct orbit point. For its Blaschke product $B$, invariance of the pseudohyperbolic distance and permutation of the orbit give $|B(Tz)|=|B(z)|$. The ratio $B(Tz)/B(z)$ extends across its zeros and has modulus one, so the open mapping theorem makes it a constant $\chi (T)$. Composition gives a group homomorphism $\chi :G\to \{\lambda :|\lambda |=1\}$. It is essential to compare moduli of the products: two arbitrary bounded holomorphic functions with identical zeros need not have constant quotient.

Write the Laurent series on the punctured disc as $f(z)={\sum }_{j=-\infty }^{\infty }{a}_j{z}^j$. A removable singularity means that $f$ extends holomorphically to the origin; exactly all negative-index coefficients vanish. A pole of order $k\ge 1$ means that $z^{k}f(z)$ extends with nonzero value at zero; exactly $a_{-k}\ne 0$ and every coefficient below $-k$ vanishes. An essential singularity is neither of these cases; exactly infinitely many negative-index coefficients are nonzero.

Here is a proof of the required Casorati-Weierstrass theorem consequence. If some disc of values around $w$ were missed in a punctured neighbourhood, then $|f(z)-w|\ge \epsilon >0$ there. The holomorphic function $q(z)=1/(f(z)-w)$ would be bounded, and hence have a removable singularity. If its extension had $q(0)\ne 0$, then $f=w+1/q$ would extend holomorphically. If $q(0)=0$, its zero has a finite positive order, since $q$ is not identically zero; thus $f$ would have a pole. Both contradict an essential singularity. Therefore every sufficiently small punctured neighbourhood has values arbitrarily close to every $w$. Choosing $0<|z_n|<\min (r/2,1/n)$ with $|f(z_n)-w|<1/n$ provesThe open mapping theorem states that a nonconstant holomorphic function on a domain maps open sets to open sets. If $f$ were injective with an essential singularity, choose a small open disc $U$ away from zero and a point $z_0\in U$. Its open image contains a disc about $w=f(z_0)$. The sequence just constructed has $z_n\to 0$ outside $U$, but eventually $f(z_n)\in f(U)$. Thus some $y_n\in U$ has $f(y_n)=f(z_n)$ although $y_n\ne z_n$, contradicting injectivity. An injective holomorphic function cannot have an isolated essential singularity.

For an injective entire function $g$, apply this to $f(z)=g(1/z)$. If its singularity at zero were removable, $g$ would be bounded outside a large disc, hence bounded everywhere and constant by Liouville's theorem. This contradicts injectivity. It must therefore be a pole, say of order $k$, so $|g(z)|\le C|z{|}^k$ for large $|z|$. For any Taylor coefficient of index $j>k$, the Cauchy coefficient formula gives $|a_j|\le C{R}^{k-j}$ on arbitrarily large circles; hence $a_j=0$. Thus $g$ is a nonconstant polynomial. Its derivative never vanishes by the allowed injectivity fact. If its degree exceeded one, the fundamental theorem of algebra would give a zero of this nonconstant derivative. Therefore injective entire functions are affine:

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.