Let \(f\) be analytic in a domain \(D\). Fix \(z_0 \in D\) and let \(w_0 = f(z_0)\). Suppose \(z_0\) is a zero of finite multiplicity \(m\) for \(f(z) - w_0 = 0\). Show that there exist \(\delta > 0\) and \(\varepsilon > 0\) such that for each \(w\) with \(0 < \abs{w - w_0} < \varepsilon\), the equation \(f(z) - w = 0\) has exactly \(m\) distinct solutions inside the disk \(\abs{z - z_0} < \delta\).
Solution
Because \(z_0\) is a zero of multiplicity \(m\) of \(f-w_0\), write \[ f(z)-w_0 = (z-z_0)^m g(z), \] where \(g\) is analytic near \(z_0\) and \[ g(z_0)\neq0. \]
1. There is \(\delta>0\) such that the closed disk\[ \overline{B}(z_0,\delta) \]lies in \(D\) and, throughout that disk,\[ g(z)\neq0 \]and\[ m g(z)+(z-z_0)g'(z)\neq0. \]
Since \(D\) is open, some closed disk centered at \(z_0\) lies in \(D\). Moreover, \[ g(z_0)\neq0 \] and \[ m g(z_0)+(z_0-z_0)g'(z_0) = m g(z_0) \neq0. \] Both displayed functions are continuous. Shrinking the disk if necessary therefore makes both nonzero on its closure.
2. On the punctured disk\[ 0<\abs{z-z_0}\leq\delta, \]one has\[ f'(z)\neq0. \]
Differentiate the factorization above: \[ f'(z) = (z-z_0)^{m-1} \left( m g(z)+(z-z_0)g'(z) \right). \] For \(z\neq z_0\), the first factor is nonzero, and the second factor is nonzero by step 1.
3. Define\[ \varepsilon = \min_{\abs{z-z_0}=\delta} \abs{f(z)-w_0}. \]Then\[ \varepsilon>0. \]
On the boundary circle, \[ f(z)-w_0 = (z-z_0)^m g(z). \] Step 1 gives \(g(z)\neq0\), so this expression never vanishes on the circle. Its modulus is continuous on the compact circle, hence attains a strictly positive minimum.
4. If\[ 0<\abs{w-w_0}<\varepsilon, \]then \(f(z)-w\) has exactly \(m\) zeros in\[ \abs{z-z_0}<\delta, \]counting multiplicity.
On the boundary circle, \[ \abs{ (f(z)-w)-(f(z)-w_0) } = \abs{w-w_0} < \varepsilon \leq \abs{f(z)-w_0}. \] Rouché’s theorem implies that \(f-w\) and \(f-w_0\) have the same number of zeros in the disk, counting multiplicity.
By step 1, \[ f(z)-w_0=(z-z_0)^m g(z) \] with \(g\) nonzero throughout the disk. Hence \(f-w_0\) has exactly the one zero \(z_0\), of multiplicity \(m\). Thus \(f-w\) has exactly \(m\) zeros counting multiplicity.
5. Every zero of \(f-w\) in the disk from step 4 is simple.
Because \(w\neq w_0\), \[ f(z_0)-w = w_0-w \neq0. \] Thus none of the zeros counted in step 4 is \(z_0\). Every such zero lies in the punctured disk, where step 2 gives \(f'(z)\neq0\). Therefore the derivative of \(f(z)-w\) is nonzero at each zero, so each zero is simple.
6. For every \(w\) satisfying\[ 0<\abs{w-w_0}<\varepsilon, \]the equation\[ f(z)=w \]has exactly \(m\) distinct solutions in\[ \abs{z-z_0}<\delta. \]
Step 4 gives \(m\) solutions counted with multiplicity. Step 5 says that every counted zero has multiplicity one. Hence there are exactly \(m\) distinct solutions.
7. Q.E.D.
Steps 1, 3, and 6 provide the required positive numbers \(\delta\) and \(\varepsilon\) and establish the conclusion.