I am interested in an optimization problem of the form $$\min_{\boldsymbol z} \max_j \vert f_j(\boldsymbol z) \vert = \min_{\boldsymbol z} \Vert f_j(\boldsymbol z) \Vert_\infty. $$ Here, the optimization/decision variables are $\boldsymbol z \in \mathbb C^{N} $ and $f_i: \mathbb C^N \to \mathbb C$.

The $f_j$ are essentially polynomials in $\lambda_j \in \mathbb C$ $$f_j(\boldsymbol z ) = \prod_{k=1}^N \big(1 - z_k \lambda_j\big).$$

To provide some more context, I am essentially trying to optimize the common roots of complex polynomials. Write the polynomial $p(\lambda) $ with $p(0) = 1$ as $$p(\lambda) = \prod_{k=1}^N \bigg(1 - \frac{ \lambda}{\tilde z_k} \bigg)$$ where $\tilde z_k$ are the roots of the polynomial. For $z_k := 1/\tilde z_k$ you obtain the representation above.

In principle, one has also to enforce that the roots come in complex-conjugated pairs which would give a linear constraint like $$A \boldsymbol z = \boldsymbol 0.$$

  • $\begingroup$ Of relevance: Application of complex numbers in Linear Programming? $\endgroup$ Jun 17, 2022 at 12:22
  • $\begingroup$ Without additional constraints, I don't see how the maximum could exist. Let $\boldsymbol{z}_0 \in \mathbb C^{N}$ be such that $\vert f(\boldsymbol{z}_0) \vert \leq 1.$ Then for any $\alpha > 1,$ $ \vert f(\alpha \boldsymbol{z}_1) \vert \leq 1,$ where $\boldsymbol{z}_1 = \boldsymbol{z}_0 / \alpha.$ $\endgroup$
    – prubin
    Jun 17, 2022 at 15:17
  • $\begingroup$ @prubin Yeah you are absolutely right. I re-formulated the problem. $\endgroup$
    – Dan Doe
    Jun 18, 2022 at 10:14

1 Answer 1


I sat down and worked a bit on this. Since I am only interested in the magnitude of the polynomial it might be helpful to employ the polar representation, i.e., $1 - z_j \lambda = r_j e^{i\phi_j}$. Then,

\begin{align} &\left \vert \prod_{j=1}^N 1 - z_j \lambda \right\vert = \left \vert \prod_{j=1}^N r_j e^{i\phi_j}\right\vert = \left \vert \left(\prod_{j=1}^N r_j \right) \cdot \exp \left( \sum_{j=1}^N i\phi_j\right)\right\vert \\ =&\left \vert \left(\prod_{j=1}^N r_j \right) \right \vert \cdot \left \vert \exp \left( \sum_{j=1}^N i\phi_j\right)\right\vert = \left \vert \left(\prod_{j=1}^N r_j \right) \right \vert \cdot 1= \prod_{j=1}^N r_j \end{align}

Now let's take a closer look on the $r_j$. Write $z_j = a_j + b_j i $, $\lambda = c+d i$. Then, $$r_j = \sqrt{\big[1 - \text{Re} (z_j \lambda)\big]^2 + \text{Im}^2 (z_j \lambda)}$$ Note that $$\text{Re} (z_j \lambda) = a_jc -b_jd, \text{Im} (z_j \lambda) = a_jd + b_j c $$ one obtains a formulation with no complex variables involved at all! The optimization is now over $\boldsymbol a, \boldsymbol b$. While the problem is of course highly nonlinear and most likely very complicated, my main concern could be lifted.


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