It's not $|z_{\inf,i}<1|$ it's $|z_{\inf,i}|<1 $ which means that, as @Mohammad Mohammadi said in the comments, your pole on a complex plane must be inside the unit circle.
Analytically, you get the magnitude with
$$ r = |z| = \sqrt{a^2+b^2} $$
If $r<1$, your pole is stable.
Note that $z_1 = b + ja$ and $z_2 = b-ja$ have the same magnitude, which means that if one is stable, the other is stable as well (in the complex plane, they are symmetrical to the horizontal axis).
Illustration

Matlab code for the illustration
%%
clear
close all
clc;
% STABLE
a = 0.73;
b = 0.24;
P1 = a + b*1i; % Pole 1
P2 = a - b*1i; % Pole 2
r = sqrt(a^2+b^2); % Magnitude
% UNSTABLE
a2 = 1.23;
b2 = 0.84;
r2 = sqrt(a2^2+b2^2);
P12 = a2 + b2*1i; % Pole 1
P22 = a2 - b2*1i; % Pole 2
zplane([0 0; 0 0], [P1 P12; P2 P22]); % Plot the poles on a complex plane
legend('', '', ...
['z = ' num2str(a) ' \pm ' num2str(b) 'i (r = ' num2str(r) ')'],...
['z = ' num2str(a2) ' \pm ' num2str(b2) 'i (r = ' num2str(r2) ')'], ...
'Unit Circle (r=1)')