Past exam of the metric spaces and complex analysis course of the University of Oxford / 2015 / 3

One standard form of the Cauchy integral formula is this: if $f$ is holomorphic on an open set containing a contour that is a simple closed curve $\Gamma$ and its interior, and $\Gamma$ has positive orientation of a contour, then for every point $w$ inside it,The hypothesis includes the contour itself; singularities on the contour are excluded. In particular, the formula applies on a circle whose closed disk is contained in the domain of holomorphy.

Put $S={\sum }_{j=0}^{n-1}|a_j|$ and choose $R>\max (1,2S/|a_n|)$. For $|z|>R$, the triangle inequality givesThe reverse triangle inequality therefore yieldsIf $P$ had no root, the function $h=1/P$ would be an entire function. The bound just proved makes $h(z)\to 0$ as $|z|\to \infty$. On the remaining closed disk, continuity and the extreme value theorem make $h$ bounded. Thus $h$ is a bounded entire function, so Liouville's theorem makes it constant. Its limit at infinity forces that constant to be zero, impossible for a reciprocal.

Consequently $P$ has a root in $\mathbb{C}$. This proves the existence part of the fundamental theorem of algebra. The invoked Liouville theorem follows from the Cauchy integral formula: the Cauchy estimate $|h^{\prime }(w)|\le M/r$ on arbitrarily large circles gives $h^{\prime }=0$ when $h$ is bounded by $M$.

List the roots with their multiplicities. By the factor theorem, $Q(z)=a{\prod }_{k=1}^n(z-{\alpha }_k)$ with $a\ne 0$. The product rule, followed by division where $Q(z)\ne 0$, gives its logarithmic derivative:Let $Q^{\prime }(\beta )=0$. If $Q(\beta )=0$ too, then $\beta$ is already one of the roots, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all roots have positive imaginary parts but $\mathrm{Im}\beta \le 0$, thenfor every $k$. Their sum cannot be zero. Thus $\mathrm{Im}\beta >0$.

For the unit disk conclusion, the same sum gives an especially transparent convex combination. Taking complex conjugates and setting $w_k=|\beta -{\alpha }_k{|}^{-2}>0$ givesTherefore, if every $|{\alpha }_k|<1$,The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the Gauss-Lucas theorem: every critical point of a nonconstant polynomial lies in the convex hull of its roots. The common-root case, handled before division, is essential when multiplicities occur.