Suppose that \(f\) is an analytic function in the region \(D\) which contains the point \(a\). Let \[F(z)= z-a-qf(z),\quad \text{where}~ q \ \text{is a complex parameter}.\]
Let \(K\subset D\) be a circle with the center at point \(a\) and also we assume that \(f(z)\not =0\) for \(z\in K\). Prove that the function \(F\) has one and only one zero \(z=w\) on the closed disc \(\bar{K}\) whose boundary is the circle \(K\) if \(\displaystyle{ |q|<\min_{z\in K} \frac{|z-a|}{|f(z)|}.}\)
Let \(G(z)\) be an analytic function on the disk \(\bar{K}\). Apply the residue theorem to prove that \(\displaystyle{ \frac{G(w)}{F'(w)}=\frac{1}{2\pi i}\int_K \frac{G(z)}{F(z)} dz,}\) where \(w\) is the zero from (1).
If \(z\in K\), prove that the function \(\displaystyle{\frac{1}{F(z)}}\) can be represented as a convergent series with respect to \(q\): \(\displaystyle{ \frac{1}{F(z)}=\sum_{n=0}^{\infty} \frac{(qf(z))^n}{(z-a)^{n+1}}.}\)
Solution
Let the circle \(K\) have radius \(R\) and center \(a\). On \(K\) the hypothesis gives \[ |qf(z)|<|z-a|=R. \] Hence, by Rouché’s theorem, the functions \[ z-a \qquad\text{and}\qquad F(z)=(z-a)-qf(z) \] have the same number of zeros in the disk bounded by \(K\), counted with multiplicity. The function \(z-a\) has exactly one zero there. Therefore \(F\) has exactly one zero \(w\) in the disk, counted with multiplicity. The strict boundary inequality also shows that \(F\) has no zero on \(K\). Consequently this unique zero is simple, so \[ F'(w)\ne0. \]
For part (2), \(G/F\) is meromorphic in the disk and has exactly one pole there, the simple pole at \(w\). Its residue is \[ \operatorname{Res}_{z=w}\frac{G(z)}{F(z)} =\frac{G(w)}{F'(w)}. \] The residue theorem therefore gives \[ \boxed{ \frac{G(w)}{F'(w)} =\frac1{2\pi i}\int_K\frac{G(z)}{F(z)}\,dz.} \]
Finally, if \(z\in K\), then \[ \left|\frac{qf(z)}{z-a}\right|<1. \] Thus the geometric series is absolutely convergent and \[ \frac1{F(z)} =\frac1{z-a}\frac1{1-\frac{qf(z)}{z-a}} =\frac1{z-a}\sum_{n=0}^\infty \left(\frac{qf(z)}{z-a}\right)^n. \] Hence \[ \boxed{ \frac1{F(z)} =\sum_{n=0}^\infty\frac{(qf(z))^n}{(z-a)^{n+1}}.} \]