{"id":1841,"date":"2024-06-10T23:34:17","date_gmt":"2024-06-10T23:34:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1841"},"modified":"2024-11-27T17:31:33","modified_gmt":"2024-11-27T17:31:33","slug":"complex-numbers-and-operations-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/complex-numbers-and-operations-learn-it-4\/","title":{"raw":"Complex Numbers and Operations: Learn It 4","rendered":"Complex Numbers and Operations: Learn It 4"},"content":{"raw":"

Arithmetic on Complex Numbers Cont.<\/h2>\r\n

Dividing Complex Numbers<\/h3>\r\nDivision of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate<\/strong> of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].\r\n\r\n
\r\n

complex conjugate<\/h3>\r\nThe complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].\r\n\r\n \r\n\r\nImportantly, complex conjugate pairs have a special property. Their product is always real.\r\n

[latex]\\begin{align}(a+bi)(a-bi)&=a^2-abi+abi-b^2i^2\\\\[2mm]&=a^2-b^2(-1)\\\\[2mm]&=a^2+b^2\\end{align}[\/latex]<\/p>\r\n\r\n<\/section>

Find the complex conjugate of each number.\r\n
    \r\n \t
  1. [latex]2+i\\sqrt{5}[\/latex]<\/li>\r\n \t
  2. [latex]-\\frac{1}{2}i[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"349660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"349660\"]\r\n
      \r\n \t
    1. The number is already in the form [latex]a+bi[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]2-i\\sqrt{5}[\/latex].<\/li>\r\n \t
    2. We can rewrite this number in the form [latex]a+bi[\/latex] as [latex]0-\\frac{1}{2}i[\/latex]. The complex conjugate is [latex]a-bi[\/latex], or [latex]0+\\frac{1}{2}i[\/latex]. This can be written simply as [latex]\\frac{1}{2}i[\/latex].<\/li>\r\n<\/ol>\r\nAnalysis of the Solution<\/strong>\r\n\r\nAlthough we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by [latex]i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
      How To: Given two complex numbers, divide one by the other.<\/strong>\r\n
        \r\n \t
      1. Write the division problem as a fraction.<\/li>\r\n \t
      2. Determine the complex conjugate of the denominator.<\/li>\r\n \t
      3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.<\/li>\r\n \t
      4. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>
        Suppose we want to divide [latex]c+di[\/latex] by [latex]a+bi[\/latex], where neither [latex]a[\/latex] nor [latex]b[\/latex] equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.\r\n

        [latex]\\dfrac{c+di}{a+bi}[\/latex] where [latex]a\\ne 0[\/latex] and [latex]b\\ne 0[\/latex].<\/p>\r\nMultiply the numerator and denominator by the complex conjugate of the denominator.\r\n

        [latex]\\dfrac{\\left(c+di\\right)}{\\left(a+bi\\right)}\\cdot \\dfrac{\\left(a-bi\\right)}{\\left(a-bi\\right)}=\\dfrac{\\left(c+di\\right)\\left(a-bi\\right)}{\\left(a+bi\\right)\\left(a-bi\\right)}[\/latex]<\/p>\r\nApply the distributive property.\r\n

        [latex]=\\dfrac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[\/latex]<\/p>\r\nSimplify, remembering that [latex]{i}^{2}=-1[\/latex].\r\n

        [latex]\\begin{align}&=\\dfrac{ca-cbi+adi-bd\\left(-1\\right)}{{a}^{2}-abi+abi-{b}^{2}\\left(-1\\right)} \\\\[2mm] &=\\dfrac{\\left(ca+bd\\right)+\\left(ad-cb\\right)i}{{a}^{2}+{b}^{2}}\\end{align}[\/latex]<\/p>\r\n\r\n<\/section>

        Divide [latex]\\left(2+5i\\right)[\/latex] by [latex]\\left(4-i\\right)[\/latex].[reveal-answer q=\"932961\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932961\"]We begin by writing the problem as a fraction.\r\n

        [latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}[\/latex]<\/p>\r\nThen we multiply the numerator and denominator by the complex conjugate of the denominator.\r\n

        [latex]\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}[\/latex]<\/p>\r\nTo multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).\r\n

        [latex]\\begin{align}\\dfrac{\\left(2+5i\\right)}{\\left(4-i\\right)}\\cdot \\dfrac{\\left(4+i\\right)}{\\left(4+i\\right)}&=\\dfrac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\\\\[2mm] &=\\dfrac{8+2i+20i+5\\left(-1\\right)}{16+4i - 4i-\\left(-1\\right)} && \\text{Because } {i}^{2}=-1 \\\\[2mm] &=\\frac{3+22i}{17} \\\\[2mm] &=\\dfrac{3}{17}+\\frac{22}{17}i && \\text{Separate real and imaginary parts}.\\end{align}[\/latex]<\/p>\r\nNote that this expresses the quotient in standard form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm2_question hide_question_numbers=1]23266[\/ohm2_question]<\/section>","rendered":"

        Arithmetic on Complex Numbers Cont.<\/h2>\n

        Dividing Complex Numbers<\/h3>\n

        Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate<\/strong> of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].<\/p>\n

        \n

        complex conjugate<\/h3>\n

        The complex conjugate of [latex]a+bi[\/latex] is [latex]a-bi[\/latex].<\/p>\n

         <\/p>\n

        Importantly, complex conjugate pairs have a special property. Their product is always real.<\/p>\n

        [latex]\\begin{align}(a+bi)(a-bi)&=a^2-abi+abi-b^2i^2\\\\[2mm]&=a^2-b^2(-1)\\\\[2mm]&=a^2+b^2\\end{align}[\/latex]<\/p>\n<\/section>\n

        Find the complex conjugate of each number.<\/p>\n
          \n
        1. [latex]2+i\\sqrt{5}[\/latex]<\/li>\n
        2. [latex]-\\frac{1}{2}i[\/latex]<\/li>\n<\/ol>\n