Inequality for a complex polynomial

Question

Let $p(z), q(z)$, and $r(z)$ be polynomials with complex coefficients in the complex plane. Suppose that $|p(z)|+|q(z)| \leq|r(z)|$ for every $z$. Show that there exist two complex numbers $a, b$ such that $|a|^2+|b|^2=1$ and $a p(z)+b q(z)=0$ for every $z$.

I tried the following approach but not sure if it is correct or not:

To find the complex numbers $a$ and $b$ such that $|a|^2+|b|^2=1$ and $a p(z)+b q(z)=0$ for every $z$, we can use the given inequality and the properties of complex numbers and their modulus.

Given the inequality $$ |p(z)|+|q(z)| \leq|r(z)| $$ we can consider the functions $f(z)$ and $g(z)$ defined by $$ \begin{aligned} & f(z)=a p(z)+b q(z) \\ & g(z)=a p(z)-b q(z) \end{aligned} $$ We want to find $a$ and $b$ such that $f(z)=0$ for every $z$. From the triangle inequality, we have that $$ |f(z)|+|g(z)|=|a p(z)+b q(z)|+|a p(z)-b q(z)| \leq|r(z)|+|r(z)|=2|r(z)| \text {. } $$ We want to find $a$ and $b$ such that $|f(z)|=0$. This means that we want to minimize $|f(z)|$, subject to the constraint $|a|^2+|b|^2=1$.

To minimize $|f(z)|$, we can write $f(z)$ in terms of $r(z)$ and differentiate with respect to $z$. Setting the derivative equal to zero gives $$ \frac{d}{d z}|f(z)|=\frac{d}{d z}|a p(z)+b q(z)|=0, $$ which yields the condition that $$ a \frac{d}{d z} p(z)+b \frac{d}{d z} q(z)=0 . $$ Thus, we have a system of equations for $a$ and $b$ : $$ \left\{\begin{array}{l} a p(z)+b q(z)=0, \\ a \frac{d}{d z} p(z)+b \frac{d}{d z} q(z)=0 . \end{array}\right. $$ We also have the constraint $|a|^2+|b|^2=1$. We can solve this system of equations for $a$ and $b$ using various methods, such as substitution or elimination. Once we find $a$ and $b$, we can verify that they satisfy $|a|^2+$ $|b|^2=1$.

Answer 1

This does not work. There are various errors, just to name a few:

• $|a p(z)-b q(z)|$ is not equal to $|r(z)|$.
• $\frac{d}{d z}|f(z)|$ is not the same as $\frac{d}{d z} f(z)$. In fact the absolute value of a holomorphic function is not complex differentiable unless the function is constant.
• If you want to minimize $|f(z)|$ as a function of $a$ and $b$ then you would have to differentiate with respect to those variables, not with respect to $z$.

For a correct solution, note that $$ |p(z)| \le |p(z)| + |q(z)| \le |r(z)| \, . $$ It follows that $p(z) = \alpha r(z)$ with some constant $\alpha \in \Bbb C$, see for example Property of Entire Functions. In the same way we can conclude that $q(z) = \beta r(z)$ with some constant $\beta \in \Bbb C$.

If $\alpha=\beta = 0$ then both $p(z)$ and $q(z)$ are identically zero, and we can choose (for example) $a=1$ and $b=0$.

Otherwise, $$ a = \frac{\beta}{\sqrt{|\alpha|^2+|\beta|^2}} \, , \, b = -\frac{\alpha}{\sqrt{|\alpha|^2+|\beta|^2}} $$ satisfy $|a|^2+|b|^2=1$ and $$ a p(z) + bq(z) = \frac{\beta \alpha r(z) - \alpha \beta r(z)}{\sqrt{|\alpha|^2+|\beta|^2}} = 0 \, . $$