[webglfan]

what are the zeros of the following polynomial:

    \[
    P(z) = \sum_{k=0}^{100} c_k z^k
    \]

    where the coefficients \( c_k \) are defined as:

    \[
    c_k = 
    \begin{cases}
    e^2 + i\pi & \text{if } k = 100, \\
    \ln(2) + \zeta(3)\,i & \text{if } k = 99, \\
    \sqrt{\pi} + e^{i/2} & \text{if } k = 98, \\
    \frac{(-1)^k}{\Gamma(k+1)} + \sin(k) \, i & \text{for } 0 \leq k \leq 97,
    \end{cases}
    \]

[drodgers]

o3 handles this perfectly by writing a solver using numpy: https://chatgpt.com/share/680aab8e-cf9c-8012-9f48-301ef62948...

    The polynomial has no closed-form factorisation, so the only practical way to get its zeros is numerical root-finding.
    I have supplied an interactive table (“Zeros of P(z)”) just above that lists all 100 roots to full machine precision (real part, imaginary part, and magnitude). You can sort or filter it as needed.
    
     Reliability notes
     • Computed with numpy.roots, which first companion-matrixes then uses QR; typical error is ≈10-12 ulp for coefficients of this size.
     • Condition number is modest (coefficients range from O(1) down to 1/97!), so the results should be accurate to at least 10 significant figures.
     • All roots are simple (pairwise distinct to >10 σ): no evidence of multiplicities.
    
    If you need higher precision (e.g. 30+ digits) let me know and I can rerun the solve with mpmath’s arbitrary-precision eigen-solver.

[Chinjut]

Does this have a nice answer? It seems quite ad hoc.

[webglfan]

Not to my knowledge. I asked Deepseek: "create me a random polynomial of degree 100 using complex numbers as coefficients. It must have at least 3 different transcendental numbers." Then I messed with some of the exponents.