Exercise 16 from chapter 3 of Stein & Shakarchi's complex analysis 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.
 A: $f$ 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 $.)

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.
