\documentclass[12pt,twoside]{article}
%\RequirePackage[discret]{correct}
\usepackage{amssymb}
\usepackage{amsmath}%matrix
\usepackage{bezier}
\usepackage[cp1251]{inputenc}
\usepackage[english,russian]{babel}
\usepackage[T2A]{fontenc}
\usepackage[dvips]{graphicx}
%\usepackage{cyrcm1}
%\RequirePackage[discret]{correct}

%\textwidth=16cm \textheight=23.7cm
%\topmargin=-0.5cm \oddsidemargin=-0.04cm
\textwidth=16cm \textheight=23.4cm \oddsidemargin=0.1cm
\evensidemargin=0.1cm \topmargin=-0.5cm
%\textwidth=16.9cm \textheight=24.7cm
%\topmargin=-1.2cm \oddsidemargin=-0.35cm

\parindent=8mm
\newcommand{\cc}{\centering}
\newcommand{\ds}{\displaystyle}
%\renewcommand{\dfrac}[2]{{\displaystyle {\frac {#1}{#2}}}}
\newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}}
%\newcommand{\iint}{\mathop{\int \int}}
%\newcommand{\iiint}{\mathop{\int \int \int}}
\newcommand{\res}{\mathop{\rm res\,}}
\newcommand{\Ln}{\mathop{\rm Ln}\nolimits}
\newcommand{\rot}{\mathop{\rm rot}\nolimits}
\renewcommand{\div}{\mathop{\rm div}\nolimits}
\newcommand{\grad}{\mathop{\rm grad}\nolimits}
\renewcommand{\geq}{\geqslant}
\renewcommand{\leq}{\leqslant}
\renewcommand{\Re}{\mathop{\rm Re}\nolimits}
\renewcommand{\Im}{\mathop{\rm Im}\nolimits}
\renewcommand{\sinh}{\mathop{\rm sh}\nolimits}
\renewcommand{\cosh}{\mathop{\rm ch}\nolimits}
\renewcommand{\coth}{\mathop{\rm cth}\nolimits}
\renewcommand{\tanh}{\mathop{\rm th}\nolimits}
\renewcommand{\tan}{\mathop{\rm tg}\nolimits}
\renewcommand{\cot}{\mathop{\rm ctg}\nolimits}
\renewcommand{\arctan}{\mathop{\rm arctg}\nolimits}
\newcommand{\pno}{\par\noindent}
\renewcommand{\i}{\mathbf{i}}
\renewcommand{\j}{\mathbf{j}}
\renewcommand{\k}{\mathbf{k}}
\renewcommand{\matrix}{\begin{matrix}}
\renewcommand{\endmatrix}{\end{matrix}}

\begin{document}

{\fontsize{15.8pt}{20pt}\selectfont

Пусть точка $P$ изображает комплексное число $z$ на комплексной плоскости 
(см. рис. 1). Тогда $P = \langle -1,\; \sqrt{3} \rangle$ ($x = \Re{z} = -1$,
$y = \Im{z} = \sqrt{3}$).

\begin{figure}[h]
\begin{center}
\leavevmode
\includegraphics*{fig_1.1}
\end{center}
\begin{center}
Рис.~1
\end{center}
\end{figure}
                  
Пусть $P_{1}$~--- проекция точки $P$ на ось $Ox$. В треугольнике
$P_{1}OP$ имеем: $OP = |z| = 2$, $OP_{1} = |x| = 1$. Поэтому
$\ds \cos{\angle P_{1}OP} = \frac{OP_{1}}{OP} = \frac{1}{2}$.
Отсюда $\ds \angle P_{1}OP = \arccos{\frac{1}{2}} = \frac{\pi}{3}$.

\end{document}
