Exercise 16 from chapter 3 of Stein & Shakarchi's complex analysis

Question

Suppose $f$ and $g$ are holomorphic in a region containing the disc $|z| \le 1$.
Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere else in $|z| \le 1$.
Let $f_\epsilon (z) = f(z)+\epsilon g(z)$.
Show that if $\epsilon$ is sufficiently small, then
(a) $f_\epsilon (z)$ has a unique zero in $|z| \le 1$, and
(b) if $z_\epsilon$ is this zero, the mapping $\epsilon \mapsto z_\epsilon$ is continuous.

I already solved (a) by applying Rouche's theorem, but (b) is such a nuisance to me.
I first tried the classical $\epsilon - \delta$ method, but it didn't work. However, I couldn't find any other ways to prove the continuity.
Since $f$ has a simple zero at $z=0$, I found that $f(z)=zh(z)$ for some $h(z)$ that is holomorphic and non-zero in the unit disc and $z_\epsilon h(z_\epsilon) = -\epsilon g(z_\epsilon)$.
Am I on the right track? I don't know how to proceed from this.

Answer 1

$f$ has no zeros on the boundary of the unit circle $\Bbb D$, therefore $$ \varepsilon_0 = \frac{ \min \{ |f(z)| : |z| = 1 \}}{1 + \max \{ |g(z)| : |z| = 1 \} } \, . $$ is strictly positive. Then for $0 \le \varepsilon \le \varepsilon_0$ and $|z| = 1$, $$ | (f(z) + \varepsilon g(z)) - f(z) | = \varepsilon |g(z)| < |f(z)| $$ and it follows from Rouché's theorem that $f(z) + \varepsilon g(z)$ and $f(z)$ have the same number of zeros in $\Bbb D$. Since $f$ has exactly one (simple) zero, it follows that $f(z) + \varepsilon g(z)$ also has exactly one zero $z_\varepsilon$ in the unit disk.

The solution $z_\varepsilon$ of $f(z) + \varepsilon g(z) = 0$ can be represented as $$ z_\varepsilon = \frac{1}{2 \pi i} \int_{\partial \Bbb D} \frac{z (f'(z) + \varepsilon g'(z))}{f(z) + \varepsilon g(z)} \, dz $$

(Proof: For fixed $\varepsilon$, $f(z) + \varepsilon g(z) = (z - z_\varepsilon)h(z)$ where $h$ is holomorphic and not zero in $\Bbb D$. Then $$ \frac{z (f'(z) + \varepsilon g'(z))}{f(z) + \varepsilon g(z)} = z \frac{h'(z)}{h(z)} + 1 + \frac{z_\varepsilon}{z_\varepsilon - z} $$ and that has exactly one (simple) pole in $\Bbb D$, with residue $z_\varepsilon$.)

Since the integrand is continuous as a function of $(z, \varepsilon) \in \partial \Bbb D \times [0, \varepsilon_0]$ it follows that the integral is a continuous function of $\varepsilon $.

(It is even an analytic function of $\varepsilon $ if we consider complex $\varepsilon $ with $|\varepsilon| < \varepsilon_0 $.)

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An alternative proof would be to apply the implicit function theorem to $F(\varepsilon, z) = f(z) + \varepsilon g(z)$, viewed as a function from $\Bbb R \times \Bbb R^2 \to \Bbb R^2$. Writing $f(z) = u(x, y) + i v(x,y)$, the derivative of $F$ with respect to $(x, y)$ at $(0, (0, 0))$ is $$ \begin{pmatrix} u_x(0, 0) & u_y(0, 0) \\ v_x(0, 0) & v_y(0, 0) \end{pmatrix} = \begin{pmatrix} u_x(0, 0) & u_y(0, 0) \\ -u_y(0, 0) & u_x(0, 0) \end{pmatrix} $$ which is invertible because its determinant $$ u_x(0, 0)^2 + u_y(0, 0)^2 = |f'(0)|^2 $$ is not zero.