Find the number of zeros \(z\) with \(\Re(z) > 0\) for the following function: \[ f(z) \da z^3-z+1 .\]
Solution
Let \[ M(z)=z^3+1, \qquad m(z)=-z, \] so that \(f=M+m\). Apply Rouché on the boundary of a large right half-disk.
On the semicircular part \(|z|=R\) with \(R\ge2\), \[ |M(z)|\ge R^3-1>R=|m(z)|. \]
On the imaginary-axis part, write \(z=it\). Then \[ |M(it)|^2=|1-it^3|^2=1+t^6, \qquad |m(it)|^2=t^2. \] If \(t^2\le1\), then \(1+t^6>t^2\); if \(t^2\ge1\), then \(t^6\ge t^2\), so again \(1+t^6>t^2\). Thus \[ |M(it)|>|m(it)| \] on the entire imaginary axis.
Hence \(f\) and \(M\) have the same number of zeros in the right half-disk. The roots of \[ M(z)=z^3+1 \] are \[ e^{i\pi/3},\qquad -1,\qquad e^{5i\pi/3}. \] Exactly two lie in the open right half-plane. Taking \(R\) large enough to contain all zeros of \(f\), we conclude that \(f\) has exactly two zeros with \(\Re z>0\).