[proofplan] We first check that the denominator $cz+d$ never vanishes on the upper half-plane, so the fractional [linear map](/page/Linear%20Map) is well-defined. We then write $z=x+iy$ and compute the imaginary part after multiplying numerator and denominator by $c\bar z+d$. The determinant condition $ad-bc=1$ gives the stated formula, which also shows that the upper half-plane is preserved. Finally, we verify holomorphicity, inverses, and compatibility with multiplication to obtain the asserted $SL_2(\mathbb{Z})$-action by holomorphic automorphisms. [/proofplan] [step:Show the fractional linear expression is defined on $\mathbb{H}$] Let $z \in \mathbb{H}$. Suppose first that $c=0$. Since $\gamma \in SL_2(\mathbb{R})$, we have $ad-bc=ad=1$, so $d \neq 0$, and hence $cz+d=d \neq 0$. Suppose now that $c \neq 0$. If $cz+d=0$, then \begin{align*} z=-\frac{d}{c}. \end{align*} Since $c,d \in \mathbb{R}$, the number $-d/c$ is real, contradicting $z \in \mathbb{H}$. Therefore $cz+d \neq 0$ for every $z \in \mathbb{H}$, and $T_\gamma$ is well-defined on $\mathbb{H}$. [/step] [step:Compute the imaginary part by rationalising the denominator] Let $z=x+iy$, where $x,y \in \mathbb{R}$ and $y>0$. Since $cz+d \neq 0$, we may multiply numerator and denominator by $c\bar z+d$: \begin{align*} T_\gamma(z) &=\frac{az+b}{cz+d} =\frac{(az+b)(c\bar z+d)}{(cz+d)(c\bar z+d)} \\ &=\frac{ac|z|^2+ad z+bc\bar z+bd}{|cz+d|^2}. \end{align*} The denominator $|cz+d|^2$ is a positive real number. The terms $ac|z|^2$ and $bd$ are real, while \begin{align*} \operatorname{Im}(ad z+bc\bar z) &=ad\,\operatorname{Im}(z)+bc\,\operatorname{Im}(\bar z) \\ &=ad\,y-bc\,y \\ &=(ad-bc)y. \end{align*} Because $\gamma \in SL_2(\mathbb{R})$, $ad-bc=1$. Hence \begin{align*} \operatorname{Im}(T_\gamma(z)) =\frac{y}{|cz+d|^2} =\frac{\operatorname{Im}(z)}{|cz+d|^2}. \end{align*} [guided] Let $z=x+iy$ with $x,y \in \mathbb{R}$ and $y>0$. The goal is to isolate the imaginary part of \begin{align*} \frac{az+b}{cz+d}. \end{align*} The denominator is nonzero by the previous step, so multiplying numerator and denominator by the complex conjugate $c\bar z+d$ is legitimate. This gives \begin{align*} T_\gamma(z) &=\frac{az+b}{cz+d} =\frac{(az+b)(c\bar z+d)}{(cz+d)(c\bar z+d)} \\ &=\frac{ac z\bar z+ad z+bc\bar z+bd}{|cz+d|^2} \\ &=\frac{ac|z|^2+ad z+bc\bar z+bd}{|cz+d|^2}. \end{align*} The denominator $|cz+d|^2$ is a positive real number, so the imaginary part of the quotient is the imaginary part of the numerator divided by $|cz+d|^2$. Now $ac|z|^2$ and $bd$ are real because $a,b,c,d \in \mathbb{R}$ and $|z|^2 \in \mathbb{R}$. Thus only $ad z+bc\bar z$ contributes to the imaginary part. Since $\operatorname{Im}(z)=y$ and $\operatorname{Im}(\bar z)=-y$, we compute \begin{align*} \operatorname{Im}(ad z+bc\bar z) &=ad\,\operatorname{Im}(z)+bc\,\operatorname{Im}(\bar z) \\ &=ad\,y-bc\,y \\ &=(ad-bc)y. \end{align*} The determinant condition for $\gamma \in SL_2(\mathbb{R})$ is exactly $ad-bc=1$, so this becomes $y$. Therefore \begin{align*} \operatorname{Im}(T_\gamma(z)) =\frac{y}{|cz+d|^2} =\frac{\operatorname{Im}(z)}{|cz+d|^2}. \end{align*} [/guided] [/step] [step:Deduce that each transformation preserves the upper half-plane] For every $z \in \mathbb{H}$, the formula just proved gives \begin{align*} \operatorname{Im}(T_\gamma(z)) =\frac{\operatorname{Im}(z)}{|cz+d|^2}. \end{align*} Here $\operatorname{Im}(z)>0$ and $|cz+d|^2>0$, so $\operatorname{Im}(T_\gamma(z))>0$. Hence $T_\gamma(z) \in \mathbb{H}$ for every $z \in \mathbb{H}$, and therefore \begin{align*} T_\gamma: \mathbb{H} \to \mathbb{H} \end{align*} is a well-defined map. [/step] [step:Verify holomorphicity and identify the inverse transformation] Since $z \mapsto az+b$ and $z \mapsto cz+d$ are complex polynomial maps and $cz+d$ has no zero on $\mathbb{H}$, the quotient \begin{align*} T_\gamma: \mathbb{H} &\to \mathbb{H} \\ z &\mapsto \frac{az+b}{cz+d} \end{align*} is holomorphic. Its complex derivative is \begin{align*} T_\gamma'(z) &=\frac{a(cz+d)-c(az+b)}{(cz+d)^2} \\ &=\frac{ad-bc}{(cz+d)^2} \\ &=\frac{1}{(cz+d)^2}. \end{align*} The inverse matrix is \begin{align*} \gamma^{-1}= \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \in SL_2(\mathbb{R}). \end{align*} For every $z \in \mathbb{H}$, direct substitution gives \begin{align*} T_{\gamma^{-1}}(T_\gamma(z))=z \quad\text{and}\quad T_\gamma(T_{\gamma^{-1}}(z))=z. \end{align*} Thus $T_\gamma$ is a bijective holomorphic map $\mathbb{H} \to \mathbb{H}$ whose inverse $T_{\gamma^{-1}}$ is also holomorphic. Hence $T_\gamma$ is a holomorphic automorphism of $\mathbb{H}$. [/step] [step:Check that integer matrices give a group action by automorphisms] Let \begin{align*} \eta= \begin{pmatrix} e & f \\ g & h \end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} For $z \in \mathbb{H}$, set $w=T_\eta(z)$. Since $T_\eta(z) \in \mathbb{H}$, the expression $T_\gamma(w)$ is defined. A direct computation gives \begin{align*} T_\gamma(T_\eta(z)) &=\frac{a\frac{ez+f}{gz+h}+b}{c\frac{ez+f}{gz+h}+d} \\ &=\frac{(ae+bg)z+(af+bh)}{(ce+dg)z+(cf+dh)} \\ &=T_{\gamma\eta}(z). \end{align*} Also $T_I(z)=z$ for the identity matrix $I \in SL_2(\mathbb{Z})$. Therefore the assignment $\gamma \mapsto T_\gamma$ is an action of $SL_2(\mathbb{Z})$ on $\mathbb{H}$. Since every $T_\gamma$ with $\gamma \in SL_2(\mathbb{Z})$ is a holomorphic automorphism by the previous step, this action is by holomorphic automorphisms. [/step]