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Near an isolated essential singularity, a holomorphic function has dense image in the complex plane. If it avoided a disk about $w$, then $1/(f-w)$ would be bounded and extend across the singularity by the Riemann removable singularity theorem. If its extended value is nonzero, $f$ extends holomorphically; if zero, its finite-order zero makes $f$ a pole. Either contradicts an essential singularity.

No such conformal bijection exists. A prescribed point is a polar point for planar Brownian motion: a planar Brownian motion started away from zero hits zero with probability zero. For example, inside an annulus the probability of reaching radius $\epsilon$ before radius $R$ isby the harmonic function $\log |w|$ and the optional stopping theorem. For fixed $R>|z^{\prime }|$ this tends to zero as $\epsilon \downarrow 0$. Any finite-time visit to zero would occur before leaving some sufficiently large disc; the countable union over integer $R$ still has probability zero.

The exit time from $\mathbb{C}\setminus \{0\}$ is therefore infinite almost surely, contradicting part (b) if this were the image of a bounded domain. Equivalently, a hypothetical inverse $\psi :\mathbb{C}\setminus \{0\}\to D$ would be bounded and holomorphic. The Riemann removable singularity theorem extends it over zero, and the Liouville theorem makes it constant, contradicting bijectivity.

A modular form of integer weight $k$ for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane satisfying $f{|}_k\gamma =f$ for $\gamma \in \Gamma$ and holomorphic at every cusp of a modular group. Explicitly, for each $\sigma \in S{L}_2(\mathbb{Z})$ choose a positive integer $h_{\sigma }$ with $\sigma T^{h_{\sigma }}{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ is periodic with period $h_{\sigma }$, and its expansion in $q_{h_{\sigma }}=e^{2\pi iz/h_{\sigma }}$ must have no negative powers. Equivalently it must be bounded as $\mathrm{Im}z\to \infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in \Gamma$, nonzero forms require even weight.

Put $d=[S{L}_2(\mathbb{Z}):\Gamma ]$ and let $f\ne 0$. Form the product over left cosetsChanging a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one modular form of weight $kd$. It is holomorphic on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.

Let $h$ be the least positive integer with $T^h\in \Gamma$, and write $f={\sum }_{n\ge 0}{b}_n{q}_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\le j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\mathrm{Im}z})$, so its level-one order satisfies $v_{\infty }(F)\ge r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.

The permitted valence formula for the modular group bounds $v_{\infty }(F)\le kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\le kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto (b_0,\dots ,b_R)$ is injective: a nonzero form in its kernel would have $r\ge R+1$. This proves the dimension bound for modular forms on a finite-index subgroup:

For the dilation assertion, the original PDF states $N_{1}D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thensince $N_{1}D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the slash operator for modular forms gives $g{|}_k\gamma =g$. It is holomorphic on the upper half-plane.

To verify every cusp, take $\sigma \in S{L}_2(\mathbb{Z})$ and choose $\tau \in S{L}_2(\mathbb{Z})$ taking infinity to the rational cusp $A\sigma \infty$. Then $B={\tau }^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequentlyis a nonzero constant factor times $f{|}_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f{|}_k\tau$ is bounded. Thus $g{|}_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the Riemann removable singularity theorem. Hence the oldform by argument dilation satisfies

Finally ${\Gamma }_0(3)$ has index four: reduction modulo three gives a transitive $S{L}_2({\mathbb{F}}_3)$ action on the four points of ${\mathbb{P}}^1({\mathbb{F}}_3)$, and ${\Gamma }_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $S{L}_2({\mathbb{F}}_3)$ and lift integrally. The dimension bound gives $\dim M_4({\Gamma }_0(3))\le 2$.

The normalized Eisenstein series $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their Fourier expansion of a normalized Eisenstein series starts asA linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the weight-four modular forms on Gamma 0 3 description

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence

Write the Taylor series of the entire function asThe Cauchy estimate on the circle $|z|=R$ givesIf $k>n/2$, letting $R\to \infty$ gives $c_k=0$. Since $n$ is odd,

For the second question, suppose such an $f$ existed. It never vanishes, so $g=1/f$ is analytic on $\mathbb{C}\setminus \{0\}$ andThe Riemann removable singularity theorem extends $g$ analytically across zero with $g(0)=0$. Applying the result just proved with $n=1$, $a=1$, and $b=0$ makes $g$ a polynomial of degree at most zero. It must then be identically zero, contradicting $g=1/f$ away from zero. Therefore no such function exists.

The domains are not conformally equivalent. Suppose that $F:\mathbb{C}\setminus \{0\}\to \{w:0<|w|<1\}$ were a conformal equivalence. Since $F$ is bounded near zero, the Riemann removable singularity theorem extends it holomorphically across zero. The extension is a bounded entire function, so the Liouville theorem makes it constant, contradicting bijectivity. This is the punctured plane is not conformally equivalent to the punctured unit disc obstruction.