[proofplan] We use the cited explicit branch formula for the [Stieltjes transform](/page/Stieltjes%20Transform) of the [Marchenko-Pastur distribution](/page/Marchenko-Pastur%20Distribution) and verify directly that it satisfies the stated algebraic equation. The square-root branch is chosen by its behaviour at infinity, which determines the decay $z m_\gamma(z)\to -1$. Substitution into the quadratic gives the equation, solving the quadratic in reverse proves the equivalence with the displayed fixed-point form, and a final square-root branch argument proves uniqueness among analytic solutions with the same sign and infinity conditions. [/proofplan]

[step:Choose the analytic square-root branch determined by infinity]Define the polynomial map \begin{align*} \Delta_\gamma: \mathbb{C}\to\mathbb{C},\quad w\mapsto (w+\gamma-1)^2-4\gamma w. \end{align*} Let $s_\gamma$ denote the analytic branch of $\sqrt{\Delta_\gamma}$ on $\mathbb{C}\setminus[(1-\sqrt{\gamma})^2,(1+\sqrt{\gamma})^2]$ satisfying $s_\gamma(z)/z\to 1$ as $|z|\to\infty$ in the complement of this interval. By the explicit formula for the [Stieltjes transform](/page/Stieltjes%20Transform) of the [Marchenko-Pastur distribution](/page/Marchenko-Pastur%20Distribution), \begin{align*} m_\gamma: \mathbb{C}\setminus\mathbb{R}\to\mathbb{C},\quad z\mapsto \frac{-(z+\gamma-1)+s_\gamma(z)}{2\gamma z}. \end{align*} This cited formula identifies $m_\gamma$ with the Stieltjes transform of the Marchenko-Pastur probability measure $\mu_\gamma$ of aspect ratio $\gamma$ under the convention \begin{align*} m_\gamma(z)=\int_{\mathbb{R}}\frac{1}{x-z}\,d\mu_\gamma(x). \end{align*}[/step]

[guided]The theorem concerns a branch of an algebraic function, so the first task is to make the branch choice explicit. Define \begin{align*} \Delta_\gamma: \mathbb{C}\to\mathbb{C},\quad w\mapsto (w+\gamma-1)^2-4\gamma w. \end{align*} The two zeros of $\Delta_\gamma$ are $(1-\sqrt{\gamma})^2$ and $(1+\sqrt{\gamma})^2$, so the natural cut for the square root is the Marchenko-Pastur support interval. We choose the analytic branch $s_\gamma$ of $\sqrt{\Delta_\gamma}$ on $\mathbb{C}\setminus[(1-\sqrt{\gamma})^2,(1+\sqrt{\gamma})^2]$ by imposing $s_\gamma(z)/z\to 1$ as $|z|\to\infty$. This condition selects one of the two possible square roots. With this branch fixed, we invoke the explicit formula for the [Stieltjes transform](/page/Stieltjes%20Transform) of the [Marchenko-Pastur distribution](/page/Marchenko-Pastur%20Distribution). It states that the Stieltjes transform of the Marchenko-Pastur probability measure $\mu_\gamma$ of aspect ratio $\gamma$ is the map \begin{align*} m_\gamma: \mathbb{C}\setminus\mathbb{R}\to\mathbb{C},\quad z\mapsto \frac{-(z+\gamma-1)+s_\gamma(z)}{2\gamma z}. \end{align*} Here the Stieltjes transform convention is \begin{align*} m_\gamma(z)=\int_{\mathbb{R}}\frac{1}{x-z}\,d\mu_\gamma(x). \end{align*} This is the only non-algebraic input in the proof: after the formula is cited, the rest of the argument verifies the equation, the branch condition, and uniqueness directly. Since $\mu_\gamma$ is a probability measure, this convention gives $z m_\gamma(z)\to -1$ at infinity, matching the normalisation in the theorem.[/guided]

[step:Verify the quadratic equation by substituting the branch formula] For $z\in\mathbb{C}\setminus\mathbb{R}$, set $m=m_\gamma(z)$ and $s=s_\gamma(z)$. The defining formula gives \begin{align*} 2\gamma z m=-(z+\gamma-1)+s. \end{align*} Hence \begin{align*} s=z+\gamma-1+2\gamma z m. \end{align*} Using $s^2=\Delta_\gamma(z)=(z+\gamma-1)^2-4\gamma z$, we obtain \begin{align*} (z+\gamma-1+2\gamma z m)^2=(z+\gamma-1)^2-4\gamma z. \end{align*} Subtracting $(z+\gamma-1)^2$ from both sides and dividing by $4\gamma z$, which is non-zero because $\gamma>0$ and $z\notin\mathbb{R}$, gives \begin{align*} (z+\gamma-1)m+\gamma z m^2=-1. \end{align*} Therefore \begin{align*} \gamma z m_\gamma(z)^2+(z+\gamma-1)m_\gamma(z)+1=0. \end{align*} [/step]

[step:Recover the fixed-point equation from the quadratic] The quadratic identity can be rewritten as \begin{align*} m_\gamma(z)\bigl(\gamma z m_\gamma(z)+z+\gamma-1\bigr)=-1. \end{align*} Thus $m_\gamma(z)\ne 0$. If $1+\gamma m_\gamma(z)=0$, then $m_\gamma(z)=-1/\gamma$, and substituting into the quadratic gives $1/\gamma=0$, impossible because $\gamma>0$. Hence $1+\gamma m_\gamma(z)\ne 0$. Dividing the quadratic identity by $m_\gamma(z)(1+\gamma m_\gamma(z))$ yields \begin{align*} z=-\frac{1}{m_\gamma(z)}+\frac{1}{1+\gamma m_\gamma(z)}. \end{align*} Conversely, multiplying this fixed-point equation by $m_\gamma(z)(1+\gamma m_\gamma(z))$ gives the quadratic identity, so the two displayed equations are equivalent whenever the denominators are non-zero. [/step]

[step:Prove that the sign and infinity conditions select only this analytic branch] Let \begin{align*} h: \mathbb{C}\setminus\mathbb{R}\to\mathbb{C} \end{align*} be an analytic map satisfying the quadratic identity, the sign condition $\operatorname{Im}h(z)$ has the same sign as $\operatorname{Im}z$, and the normalisation $z h(z)\to -1$ as $|z|\to\infty$ in $\mathbb{C}\setminus\mathbb{R}$. Define \begin{align*} r: \mathbb{C}\setminus\mathbb{R}\to\mathbb{C},\quad z\mapsto z+\gamma-1+2\gamma z h(z). \end{align*} The same algebra used in the substitution step gives $r(z)^2=\Delta_\gamma(z)$ for every $z\in\mathbb{C}\setminus\mathbb{R}$. Moreover, $z h(z)\to -1$ implies $h(z)\to 0$, and therefore \begin{align*} \frac{r(z)}{z}=1+\frac{\gamma-1}{z}+2\gamma h(z)\to 1. \end{align*} On each connected half-plane $\mathbb{C}_+=\{z\in\mathbb{C}:\operatorname{Im}z>0\}$ and $\mathbb{C}_-=\{z\in\mathbb{C}:\operatorname{Im}z<0\}$, both $r$ and $s_\gamma$ are analytic square roots of $\Delta_\gamma$. Since $s_\gamma(z)/z\to 1$, the quotient $r/s_\gamma$ is an analytic function with square equal to $1$ and limit $1$ at infinity; hence $r=s_\gamma$ on each half-plane. Solving the defining relation for $h$ gives \begin{align*} h(z)=\frac{-(z+\gamma-1)+s_\gamma(z)}{2\gamma z}=m_\gamma(z). \end{align*} Thus no other analytic solution can satisfy the same sign and infinity conditions. [/step]

[step:Check the Stieltjes branch and the normalisation at infinity] Since $\mu_\gamma$ is a probability measure and \begin{align*} m_\gamma(z)=\int_{\mathbb{R}}\frac{1}{x-z}\,d\mu_\gamma(x), \end{align*} the imaginary part of $(x-z)^{-1}$ has the same sign as $\operatorname{Im} z$ for every $x\in\mathbb{R}$. Therefore $\operatorname{Im}m_\gamma(z)$ has the same sign as $\operatorname{Im}z$. Also $s_\gamma(z)/z\to1$ and \begin{align*} s_\gamma(z)=z-(\gamma+1)+o(1) \end{align*} as $|z|\to\infty$ in the chosen branch, so \begin{align*} z m_\gamma(z)=\frac{-(z+\gamma-1)+s_\gamma(z)}{2\gamma}\to -1. \end{align*} Thus the explicit Stieltjes transform is precisely the branch of the quadratic solution characterised in the theorem. [/step]