I created the image below from various bits of the book "The Cauchy-Schwarz Master Class". In particular, it's from a question in chapter 6.
Could someone explain to me why $\theta_j$ is not more than $2\psi$ or how one arrive at the inequality just before (14.54) in the image.
(Edited on 29-11-2020)
I would like to draw your attention to an answer to this question:
GM-AM-QM inequalities for complex numbers?!
It seems the constant $2\psi$ also appeared in the answer but the person just said it works.
The complex AGM inequality states that if $z_k, k=1,..,n$ are (nonzero) complex numbers for which the argument spread is at most $2\psi, \psi <\pi/2$, or in other words, choosing say the principal argument in $(-\pi, \pi]$ or $[0, 2\pi)$ but having a fixed choice for all $z_k$, we have $|\arg(z_k)-\arg (z_l)| \le 2\psi <\pi$,
then $(|z_1..z_n|)^{1/n}\cos \psi \leq |\frac{z_1+..z_n}{n}|$
and the proof follows from the usual AGM inequality for positive numbers, noting that by a rotation which changes neither LHS, nor RHS (so send the lowest most argument of the $z_k$ taken in $(-\pi, \pi]$ say to $-\psi$, then all the others are at most $\psi$ by the spread hypothesis), we can assume that the arguments are all in $[-\psi, \psi]$ so $\Re z_k \geq |z_k|\cos \psi >0$ hence:
$(|z_1..z_n|)^{1/n}\cos \psi \leq (\Re (z_1)..\Re (z_n))^{1/n} \leq \frac{\Re z_1+..\Re z_n} {n} = \frac{\Re (z_1+..z_n)}{n} \le |\frac{z_1+..z_n}{n}|$ (note that if we let $\psi=\pi/2$ we still can apply the argument above, but the inequality is vacuous)
Apply this here with $z_k=1/(z-r_k)$ noting that the hypothesis shows that the arguments of $z-r_k$ hence of $z_k$ have spread at most $2\psi(z)$ for some $\psi(z) < \pi/2$ - considering $w_k=z-r_k$ their arguments are contained in the angle drawn by definition, so their difference is at most $2\psi$ as the picture says (the angle in Wilf's inequality depends on $z$ of course and can get very close to $\pi/2$ if $z$ gets very close to a side of the polygon) and get the bound in the line above (14.54).
But now since by substitution LHS is $|a_n/P(z)|^{1/n}\cos \psi$ as $P(z)=a_n(z-r_1)..(z-r_n)$ and RHS is $|\frac{P'(z)}{nP(z)}|$ by the usual formula for $(P'/P)(z)$ as the sum of the reciprocals of $1/(z-r_k)$, (14.54) follows by rewritting and moving the $\cos \psi$ in the denominator of RHS