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: \[ c_k = \begin{cases} e^2 + i\pi & \text{if } k = 100, \\ \ln(2) + \zeta(3)\,i & \text{if } k = 99, \\ \sqrt{\pi} + e^{i/2} & \text{if } k = 98, \\ \frac{(-1)^k}{\Gamma(k+1)} + \sin(k) \, i & \text{for } 0 \leq k \leq 97, \end{cases} \]
Does this have a nice answer? It seems quite ad hoc.
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.