Analysis Problem of the Day 66

Today’s problem appeared as Problem 12 on the UCLA Spring 2024 Analysis Qual:

Problem 12. Let $P(w,z) = w^3(2-w)^3 + z.$

a) Show that for $z \in \mathbb{D}, P(\cdot,z)$ has precisely three roots for $w \in \mathbb{D}.$

b) Let $w_1(z),w_2(z),w_3(z)$ be the roots in a). Show that $w_1(z)+w_2(z)+w_3(z)$ is holomorphic on $\mathbb{D}.$

Solution: a) We are asked to count roots, so it is natural to use a theorem relating the number of roots, say, Rouche’s theorem. Indeed, for $|z|<1$ and $|w|=1,$ $|z|<|w^3(2-w)^3|=1,$ so $w^3(2-w)^3$ and $w^3(2-w)^3+z$ have the same number of roots on $\mathbb{D}.$ But the former has three roots at $0$ and three roots at $2,$ so $P(\cdot,z)$ has precisely three roots on $\mathbb{D}$ for $z \in \mathbb{D}.$

b) We use the generalized argument principle. Indeed, it states that for $g(w)=w,$ one can write the above function as

$w_1(z)+w_2(z)+w_3(z)= \frac{1}{2\pi i}\int_{|w|=1}w \frac{\partial_w P(w,z)}{P(w,z)} dw.$

It suffices to show that the integral is holomorphic in $z.$ It is clearly continuous in $z$ since $P$ is, and

$\int_\Delta \int_{|w|=1} w \frac{\partial_w P(w,z)}{P(w,z)}dw dz = \int_{|w|=1} w \partial_w P(w,z) \int_{\Delta} \frac{1}{w^3(2-w)^3+z} dz dw.$

Since $|w|=1$ and $|z|<1,$ $w^3(2-w)^3+z$ does not vanish for $z \in \Delta,$ i.e. the integrand is holomorphic and the contour integral thus vanishes by Cauchy’s theorem. It follows by Morera’s theorem that the original integral is continuous in $z$ and its contour integrals vanish over triangles in the unit disk, so by Morera’s theorem it defines a holomorphic function on the unit disk, and we are done.

Remark: This argument generalizes to show that if $P(\cdot,z): V \subseteq \mathbb{C}^2 \to \mathbb{C}$ is a holomorphic function with roots $w_1(z),w_2(z),...,w_n(z)$ and $g$ is holomorphic, then $\sum_{i=1}^n g(w_i(z))$ is holomorphic for $z \in U$ whenever $P(w,z) \not = 0$ for $w \in \partial V, z \in U.$

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