{"id":2282,"date":"2025-08-12T17:52:50","date_gmt":"2025-08-12T17:52:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2282"},"modified":"2025-08-13T17:05:14","modified_gmt":"2025-08-13T17:05:14","slug":"polar-form-of-complex-numbers-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-form-of-complex-numbers-learn-it-4\/","title":{"raw":"Polar Form of Complex Numbers: Learn It 4","rendered":"Polar Form of Complex Numbers: Learn It 4"},"content":{"raw":"

Finding Products and Quotients of Complex Numbers in Polar Form<\/h2>\r\nNow that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.\r\n\r\n
\r\n

products of complex numbers<\/h3>\r\nIf [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the product of these numbers is given as:\r\n

[latex]\\begin{align}{z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\left[\\cos \\left({\\theta }_{1}+{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}+{\\theta }_{2}\\right)\\right] \\\\ {z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\text{cis}\\left({\\theta }_{1}+{\\theta }_{2}\\right) \\end{align}[\/latex]<\/p>\r\nNotice that the product calls for multiplying the moduli and adding the angles.\r\n\r\n<\/section>

Find the product of [latex]{z}_{1}{z}_{2}[\/latex], given [latex]{z}_{1}=4\\left(\\cos \\left(80^\\circ \\right)+i\\sin \\left(80^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(145^\\circ \\right)+i\\sin \\left(145^\\circ \\right)\\right)[\/latex].[reveal-answer q=\"385516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"385516\"]\r\n\r\nFollow the formula\r\n

[latex]\\begin{align}&{z}_{1}{z}_{2}=4\\cdot 2\\left[\\cos \\left(80^\\circ +145^\\circ \\right)+i\\sin \\left(80^\\circ +145^\\circ \\right)\\right] \\\\ &{z}_{1}{z}_{2}=8\\left[\\cos \\left(225^\\circ \\right)+i\\sin \\left(225^\\circ \\right)\\right] \\\\ &{z}_{1}{z}_{2}=8\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ {z}_{1}{z}_{2}=8\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &{z}_{1}{z}_{2}=-4\\sqrt{2}-4i\\sqrt{2} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Finding Quotients of Complex Numbers in Polar Form<\/h2>\r\nThe quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.\r\n\r\n
\r\n

quotients of complex numbers<\/h3>\r\nIf [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the quotient of these numbers is\r\n

[latex]\\begin{align}&\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\left[\\cos \\left({\\theta }_{1}-{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}-{\\theta }_{2}\\right)\\right],{z}_{2}\\ne 0\\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{{r}_{1}}{{r}_{2}}\\text{cis}\\left({\\theta }_{1}-{\\theta }_{2}\\right),{z}_{2}\\ne 0\\end{align}[\/latex]<\/p>\r\nNotice that the moduli are divided, and the angles are subtracted.\r\n\r\n<\/section>

How To: Given two complex numbers in polar form, find the quotient.\r\n<\/strong>\r\n
    \r\n \t
  1. Divide [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex].<\/li>\r\n \t
  2. Find [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\r\n \t
  3. Substitute the results into the formula: [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex]. Replace [latex]r[\/latex] with [latex]\\frac{{r}_{1}}{{r}_{2}}[\/latex], and replace [latex]\\theta [\/latex] with [latex]{\\theta }_{1}-{\\theta }_{2}[\/latex].<\/li>\r\n \t
  4. Calculate the new trigonometric expressions and multiply through by [latex]r[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Find the quotient of [latex]{z}_{1}=2\\left(\\cos \\left(213^\\circ \\right)+i\\sin \\left(213^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=4\\left(\\cos \\left(33^\\circ \\right)+i\\sin \\left(33^\\circ \\right)\\right)[\/latex].[reveal-answer q=\"244676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244676\"]\r\n\r\nUsing the formula, we have\r\n

    [latex]\\begin{align}&\\frac{{z}_{1}}{{z}_{2}}=\\frac{2}{4}\\left[\\cos \\left(213^\\circ -33^\\circ \\right)+i\\sin \\left(213^\\circ -33^\\circ \\right)\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[\\cos \\left(180^\\circ \\right)+i\\sin \\left(180^\\circ \\right)\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=\\frac{1}{2}\\left[-1+0i\\right] \\\\ &\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2}+0i \\\\ &\\frac{{z}_{1}}{{z}_{2}}=-\\frac{1}{2} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    \r\n
    \r\n\r\nFind the product and the quotient of [latex]{z}_{1}=2\\sqrt{3}\\left(\\cos \\left(150^\\circ \\right)+i\\sin \\left(150^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"381268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"381268\"]\r\n\r\n[latex]{z}_{1}{z}_{2}=-4\\sqrt{3};\\frac{{z}_{1}}{{z}_{2}}=-\\frac{\\sqrt{3}}{2}+\\frac{3}{2}i[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
    [ohm_question hide_question_numbers=1]173851[\/ohm_question]<\/section>","rendered":"

    Finding Products and Quotients of Complex Numbers in Polar Form<\/h2>\n

    Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.<\/p>\n

    \n

    products of complex numbers<\/h3>\n

    If [latex]{z}_{1}={r}_{1}\\left(\\cos {\\theta }_{1}+i\\sin {\\theta }_{1}\\right)[\/latex] and [latex]{z}_{2}={r}_{2}\\left(\\cos {\\theta }_{2}+i\\sin {\\theta }_{2}\\right)[\/latex], then the product of these numbers is given as:<\/p>\n

    [latex]\\begin{align}{z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\left[\\cos \\left({\\theta }_{1}+{\\theta }_{2}\\right)+i\\sin \\left({\\theta }_{1}+{\\theta }_{2}\\right)\\right] \\\\ {z}_{1}{z}_{2}&={r}_{1}{r}_{2}\\text{cis}\\left({\\theta }_{1}+{\\theta }_{2}\\right) \\end{align}[\/latex]<\/p>\n

    Notice that the product calls for multiplying the moduli and adding the angles.<\/p>\n<\/section>\n

    Find the product of [latex]{z}_{1}{z}_{2}[\/latex], given [latex]{z}_{1}=4\\left(\\cos \\left(80^\\circ \\right)+i\\sin \\left(80^\\circ \\right)\\right)[\/latex] and [latex]{z}_{2}=2\\left(\\cos \\left(145^\\circ \\right)+i\\sin \\left(145^\\circ \\right)\\right)[\/latex].<\/p>\n