Números complejos con Python.

Algebra_140314.ipynb

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      "# Algebra Lineal\n",
      "\n",
      "Ismael Venegas Castell\u00f3: <[email protected]>\n",
      "## N\u00fameros complejos.\n",
      "\n",
      "###14-III-14\n",
      "\n",
      "#### Acerca de numpy: [Numpy](http://es.wikipedia.org/wiki/NumPy) "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Primero importamos el m\u00f3dulo \"numpy\" (\"python n\u00famerico\", del ingles 'numerical python')\n",
      "# bajo el alias \"np\":\n",
      "import numpy as np"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 17
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Ahora utilizamos la funci\u00f3n \"np.sqrt\" (\"raiz cuadrada\", del ingles 'square root')\n",
      "# del m\u00f3dulo \"np\" utilizando la notaci\u00f3n \"modulo.funci\u00f3n\" ,\n",
      "# usando el n\u00famero complejo/imaginario -1 (en Python \"j\" denota el n\u00famero imaginario \"i\"):\n",
      "np.sqrt(complex(-1))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
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       "output_type": "pyout",
       "prompt_number": 18,
       "text": [
        "1j"
       ]
      }
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     "prompt_number": 18
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     "source": [
      "|Potencia|Desarrollo|Equivalencia|\n",
      "|--------|----------|------------|\n",
      "|$i^1$|$\\sqrt{-1}$|$i$|\n",
      "|$i^2$|$(\\sqrt{-1})(\\sqrt{-1})=(\\sqrt{-1})^2$|$-1$|\n",
      "|$i^3$|$i^2\u00b7i=(-1)(\\sqrt{-1})$|$-i$|\n",
      "|$i^4$|$i^2\u00b7i^2=(-1)(-1)$|$1$|\n",
      "|$i^5$|$i^4\u00b7i=(1)(i)$|$i$|\n",
      "|$i^6$|$i^4\u00b7i^2=(1)(-1)$|$-1$|"
     ]
    },
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     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Para encontrar $i$ elevado a cualquier potencia, hay que dividir la potencia entre 4, ya que $i^4=1$. El residuo va a ser la potencia a la cual se eleve $i$ y se multiplique por 1, ejemplo:\n",
      "\n",
      "$i^8$\n",
      "\n",
      "$\\frac{8}{4} = (i^2)^4 i^0 = (1)(1)=1$ \n",
      "\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# La funci\u00f3n \"divmod\" regresa una tupla con el cociente y dividendo de una divisi\u00f3n:\n",
      "cociente, residuo = divmod(8, 4) # \"divmod\" nos da el dividendo y el modulo (residuo) de 8/4.\n",
      "print('cociente: {0}\\n'\n",
      "      ' residuo: {1}'.format(cociente, residuo))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "cociente: 2\n",
        " residuo: 0\n"
       ]
      }
     ],
     "prompt_number": 19
    },
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     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "### Representaci\u00f3n binomica o cartesiana.\n",
      "\n",
      "$a+bi$\n",
      "\n",
      "Donde $a$ es la parte real y $bi$ es la parte imaginaria.\n",
      "Suponiendo que:\n",
      "\n",
      "* $2x^2-2x+5=0$\n",
      "\n",
      "Usando la formula general:\n",
      "\n",
      "$x = \\frac{-b \\pm \\sqrt{b\u00b2-4ac}}{2a}$"
     ]
    },
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     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Despues definimos una funci\u00f3n que compute la formula general,\n",
      "# \"fgen\" regresa una tupla con el valor de (x1, x2),\n",
      "# tenemos que usar la funcion \"complex\" para tener un\n",
      "# numero complejo en caso de que la raiz sea negativa:\n",
      "def fgen(a, b, c):\n",
      "    x1 = (-b + np.sqrt(complex(b**2 - 4*a*c))) / (2*a)\n",
      "    x2 = (-b - np.sqrt(complex(b**2 - 4*a*c))) / (2*a)\n",
      "    return (x1, x2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 20
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     "input": [
      "# Por ultimo llamamos a la funci\u00f3n \"fgen\" con los argumentos (2, -2, 5).\n",
      "# a = 2; b = -2; c = 5:\n",
      "x1, x2 = fgen(2, -2, 5)\n",
      "\n",
      "print('x1 = {0}\\n'\n",
      "      'x2 = {1}'.format(x1, x2))"
     ],
     "language": "python",
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     "outputs": [
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       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "x1 = (0.5+1.5j)\n",
        "x2 = (0.5-1.5j)\n"
       ]
      }
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     "prompt_number": 21
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     "source": [
      "Es decir, el resultado es: \n",
      "\n",
      "$x_1 = \\frac{1}{2} + \\frac{3}{2}i$\n",
      "\n",
      "$x_2 = \\frac{1}{2} - \\frac{3}{2}i$"
     ]
    }
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