Tarea 2 Numeros Complejos
Author
Michell Gallardo
Last Updated
10 лет назад
License
Creative Commons CC BY 4.0
Abstract
ALGEBRA
\documentclass[a4paper]{article}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\title{Tarea 2 Numeros Complejos}
\author{GALLARDO ZATARAIN CECILIA MICHELL 14210440}
\date{25 Agosto, 2014}
\begin{document}
\maketitle
\LARGE Resuelve lo siguiente
Z = 1 + 2i
W = 5 + 3i
V = 4 + i
1)X.W
2)$\bar{W} - \bar{V}$
3)6W+2Z
\section{Ejercicio 1}
1)X.W =
$(1+2i)(5+3i) = 5+31+10+6^2 = 5+13i-6 = -1+13i$
$z =-1+13i$
$ \bar {z} = -1-13i$
$\left | z \right | = \sqrt{ \alpha^2 + \beta^2} = \sqrt{ (-1)^2 + (-13)^2} = \sqrt{170}$
\begin{figure}
\centering
\includegraphics[width=0.3\textwidth]{graf_1}
\end{figure}
$\theta = \Pi- (tan^-1(\frac {13} {-1})) = 1.6475 $
\section{Ejercicio 2}
2)$\bar{w} - \bar{v}$ =
$(5-3i)-(4-i) = 5-3i-4+i = 1-2i$
$z =-1-2i$
$ \bar {z} = -1+2i$
$\left | z \right | = \sqrt{ \alpha^2 + \beta^2} = \sqrt{ (1)^2 + (-2)^2} = \sqrt{5}$
\begin{figure}
\centering
\includegraphics[width=0.3\textwidth]{GRAF2}
\end{figure}
$\theta = 2\Pi- (tan^-1(\frac {-2} {1})) = 5.1760 $
\section{Ejercicio 3}
3)6w+2z =
$6(5+3i)-2(1+2i) = 30+18i+2+4i = 32+22i$
$z =32+22i$
$ \bar {z} = 32+22i$
$\left | z \right | = \sqrt{ \alpha^2 + \beta^2} = \sqrt{ (32)^2 + (22)^2} = \sqrt{1024+484} = \sqrt{1580}$
$\theta = (tan^-1(\frac {22} {32})) = 0.6022 $
\begin{figure}
\centering
\includegraphics[width=0.3\textwidth]{GRAF3}
\end{figure}
\end{document}