Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 14E (Further Analysis)

The form of Rouché's theorem used here is as follows. Suppose $f$ and $g$ are holomorphic on a neighborhood of the closure of a bounded Jordan domain, and $|g-f|<|f|$ on its positively oriented boundary. Then $f$ and $g$ have the same number of zeros inside, counted with multiplicity.

To prove this form, put $f_t=f+t(g-f)$ for $0\le t\le 1$. The strict inequality prevents $f_t$ from vanishing on the boundary. The argument principle counts its interior zeros by $(2\pi i{)}^{-1}\oint f_t^{\prime }/f_t$. This integral varies continuously with $t$: the integrand is continuous and its denominator stays uniformly away from zero on the compact boundary times $[0,1]$. Its values are integers, so the integral is constant. Its values at $t=0,1$ are the respective zero counts. This proves Rouché's theorem with the multiplicities and boundary hypotheses required below.

On $|z|=3$, the perturbation has modulus $|icz+8|\le 3c+8<11<27=|z^3|$. Thus Rouché's theorem gives three roots inside $|z|<3$, counted with multiplicity.

For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt{3}$, $r_3=1-i\sqrt{3}$. Their pairwise distances are $2\sqrt{3}$. On the circle $|z-r_j|=1$,The three unit discs are disjoint, and Rouché localization of polynomial roots gives one root in each, counted with multiplicity. The disc around $1+i\sqrt{3}$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt{3}$ lies in the fourth quadrant; and that around $-2$ lies in $\mathrm{Re}z<-1$. Consequently there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane. The same count also shows that each root is simple.

At $c=0$ the left-hand root is $-2$, and the derivative of the polynomial with respect to $z$ is $3(-2{)}^2=12\ne 0$. The holomorphic implicit function theorem states that if $P(z,c)$ is holomorphic near $(z_0,c_0)$, $P(z_0,c_0)=0$ and $P_z(z_0,c_0)\ne 0$, there is a unique holomorphic function $z=\gamma (c)$ near $c_0$, with $\gamma (c_0)=z_0$ and $P(\gamma (c),c)=0$. It therefore supplies an analytic root branch $\gamma (c)$ near zero. Differentiate its equation:It follows that ${\gamma }^{\prime }(0)=i/6$ andFor sufficiently small positive $c$ its imaginary part is $c/6+O(c^2)>0$. By the unique-root localization from the preceding part, this is the root in the left half-plane. Thus that root lies above the real axis for all sufficiently small positive $c$.