Just wanted to elaborate the first part given by @above hints We are gonna use Gauss-Lucas Theorem which states that for any complex polynomial $P(z)$, the roots of the derivative $P'(z)$ lies in the convex hull of the roots of $P(z)$ , so basically it means that the roots of $P'(z)$ lies in the smallest convex subset of the complex plane containing all the roots of $P(z)$ \[ P(z) = z^{20} + 63z + 22 \]\[ P'(z) = 20z^{19}+63 =0 \Longleftrightarrow z^{19} + \pi + \epsilon =0 \]

I'll give a proof for Gauss Lucas Thm Here. Gauss-Lucas Thm wrote: For any complex polynomial $P(z)$, the roots of the derivative $P'(z)$ lies in the convex hull of the roots of $P(z).$ Proof. let $P(z)=\lambda \prod_{k=1}^n(z-z_k)^{\alpha_k},$ then if $P'(Z)=0,$ $$\frac{P'(Z)}{P(Z)}=\sum_{k=1}^n\frac{\alpha_k}{Z-z_k}=0.$$$$\Longrightarrow\sum_{k=1}^n\frac{\alpha_k(\overline Z-\overline{z_k})}{|Z-z_k|^2}=0 $$$$\Longrightarrow \sum_{k=1}^n\frac{\alpha_k}{|Z-z_k|^2}Z=\sum_{k=1}^n\frac{\alpha_kz_k}{|Z-z_k|^2} $$Therefore $Z$ can be written as a convex linear combination of $z_1,\ldots ,z_n,$ done $.\Box$ Part 1. By Gauss-Lucas Thm the convex hull of all complex roots of $z^{20}+63z+22=0$ is subset of the convex hull of all complex roots of $20z^{19}+63=0,$ which is a regular 19-gon with length $(63/20)^{1/19}.$ Therefore the radius of incircle of the 19-gon is $$\left(\frac{63}{20}\right)^{1/19}\cos\frac{\pi}{19}>1$$after some "simple" estimations, so the total area is greater than $\pi.\Box$