Consider the polynomial $ P (z, a_0, a_1, a_2, ..., a_m) = a_0 + a_1z + ... + a_mz ^ m $ and suppose that it has a simple root $ z_0 $. Show that given $ \varepsilon> 0 $ there exists a $ \delta> 0 $ such that for every polynomial $ P (z, b_0, b_1, ..., b_m) $ with $ | b_i-a_i | <\varepsilon $ $ i = 0,1, ..., m $ has a single simple root $ r (b_0, b_1, ..., b_m) \in B (z_0, \delta) $. the function $ r $ thus defined is of class $ \mathbb {C} ^ {\infty} $.
$z_0$ is a simple root, this is has multiplicity 1, we have $P(z, a_0, a_1, a_2, ..., a_m)=a_0 + a_1z_0 + ... + a_mz_0 ^ m=0$ $(z-z_0)g(z)=0$ with $g(z) \neq 0$.
The claim in your question is marred by typos. It should read
This says that "the roots of a polynomial equation are continuous functions of the coefficients of the polynomial". I have put that into quotes, since there are precisions needed due to multiple roots. In the case at hand we have assumed a simple root $z_0$, and the (complex analytic) implicit function theorem at once gives the result. If you don't have this theorem at your disposal you can argue as follows:
The function $P$ can be rewritten as $${\bf p}: \quad{\mathbb R}^{2m+4}\to{\mathbb R}^2,\qquad(x,y,{\bf c})\mapsto(u,v)$$ in terms of the real variables $x$, $y\in{\mathbb R}$, ${\bf c}\in{\mathbb R}^{2m+2}$. We are given that ${\bf p}(x_0,y_0,{\bf a})=(0,0)$, and that the Jacobian of ${\bf p}$ with respect to $x$, $y$ is $\ne0$ at $(x_0,y_0,{\bf a})$, since the "simple root" assumption guarantees $$J_{{\bf p},(x,y)}(x_0,y_0,{\bf a})=\left|{\partial P\over\partial z}(z_0,{\bf a})\right|^2\ne0\ .$$ Applying the (real) implicit function theorem to this situation and retranslating its claim into the complex environment then gives what you desire.