{"id":2283,"date":"2025-08-12T17:52:52","date_gmt":"2025-08-12T17:52:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2283"},"modified":"2025-08-13T17:05:29","modified_gmt":"2025-08-13T17:05:29","slug":"polar-form-of-complex-numbers-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-form-of-complex-numbers-learn-it-5\/","title":{"raw":"Polar Form of Complex Numbers: Learn It 5","rendered":"Polar Form of Complex Numbers: Learn It 5"},"content":{"raw":"

Finding Powers and Roots of Complex Numbers in Polar Form<\/h2>\r\nFinding powers of complex numbers is greatly simplified using De Moivre\u2019s Theorem. It states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the argument by [latex]n[\/latex]. It is the standard method used in modern mathematics.\r\n\r\n
\r\n

De Moivre\u2019s Theorem<\/h3>\r\nIf [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex] is a complex number, then\r\n

[latex]\\begin{align}&{z}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &{z}^{n}={r}^{n}\\text{cis}\\left(n\\theta \\right)\\end{align}[\/latex]<\/p>\r\nwhere [latex]n[\/latex] is a positive integer.\r\n\r\n<\/section>

Evaluate the expression [latex]{\\left(1+i\\right)}^{5}[\/latex] using De Moivre\u2019s Theorem.[reveal-answer q=\"728625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728625\"]\r\n\r\nSince De Moivre\u2019s Theorem applies to complex numbers written in polar form, we must first write [latex]\\left(1+i\\right)[\/latex] in polar form. Let us find [latex]r[\/latex].\r\n

[latex]\\begin{align}&r=\\sqrt{{x}^{2}+{y}^{2}} \\\\ &r=\\sqrt{{\\left(1\\right)}^{2}+{\\left(1\\right)}^{2}} \\\\ &r=\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nThen we find [latex]\\theta [\/latex]. Using the formula [latex]\\tan \\theta =\\frac{y}{x}[\/latex] gives\r\n

[latex]\\begin{align}&\\tan \\theta =\\frac{1}{1} \\\\ &\\tan \\theta =1 \\\\ &\\theta =\\frac{\\pi }{4} \\end{align}[\/latex]<\/p>\r\nUse De Moivre\u2019s Theorem to evaluate the expression.\r\n

[latex]\\begin{align}&{\\left(a+bi\\right)}^{n}={r}^{n}\\left[\\cos \\left(n\\theta \\right)+i\\sin \\left(n\\theta \\right)\\right]\\\\ &{\\left(1+i\\right)}^{5}={\\left(\\sqrt{2}\\right)}^{5}\\left[\\cos \\left(5\\cdot \\frac{\\pi }{4}\\right)+i\\sin \\left(5\\cdot \\frac{\\pi }{4}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[\\cos \\left(\\frac{5\\pi }{4}\\right)+i\\sin \\left(\\frac{5\\pi }{4}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=4\\sqrt{2}\\left[-\\frac{\\sqrt{2}}{2}+i\\left(-\\frac{\\sqrt{2}}{2}\\right)\\right] \\\\ &{\\left(1+i\\right)}^{5}=-4 - 4i \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Finding Roots of Complex Numbers in Polar Form<\/h2>\r\nTo find the n<\/em>th root of a complex number in polar form, we use the [latex]n\\text{th}[\/latex] Root Theorem or De Moivre\u2019s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding [latex]n\\text{th}[\/latex] roots of complex numbers in polar form.\r\n\r\n
\r\n

the nth root theorem<\/h3>\r\nTo find the [latex]n\\text{th}[\/latex] root of a complex number in polar form, use the formula given as\r\n

[latex]\\begin{align}{z}^{\\frac{1}{n}}={r}^{\\frac{1}{n}}\\left[\\cos \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)+i\\sin \\left(\\frac{\\theta }{n}+\\frac{2k\\pi }{n}\\right)\\right]\\end{align}[\/latex]<\/p>\r\nwhere [latex]k=0,1,2,3,...,n - 1[\/latex]. We add [latex]\\frac{2k\\pi }{n}[\/latex] to [latex]\\frac{\\theta }{n}[\/latex] in order to obtain the periodic roots.\r\n\r\n<\/section>

Evaluate the cube roots of [latex]z=8\\left(\\cos \\left(\\frac{2\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{3}\\right)\\right)[\/latex].[reveal-answer q=\"791675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"791675\"]\r\n\r\nWe have\r\n

[latex]\\begin{align}&{z}^{\\frac{1}{3}}={8}^{\\frac{1}{3}}\\left[\\cos \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2k\\pi }{3}\\right)\\right] \\\\ &{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{2k\\pi }{3}\\right)\\right] \\end{align}[\/latex]<\/p>\r\nThere will be three roots: [latex]k=0,1,2[\/latex]. When [latex]k=0[\/latex], we have\r\n

[latex]{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{2\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}\\right)\\right)[\/latex]<\/p>\r\nWhen [latex]k=1[\/latex], we have\r\n

[latex]\\begin{align}&{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{6\\pi }{9}\\right)\\right] && \\text{ Add }\\frac{2\\left(1\\right)\\pi }{3}\\text{ to each angle.} \\\\ &{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{8\\pi }{9}\\right)+i\\sin \\left(\\frac{8\\pi }{9}\\right)\\right) \\end{align}[\/latex]<\/p>\r\nWhen [latex]k=2[\/latex], we have\r\n

[latex]\\begin{align}&{z}^{\\frac{1}{3}}=2\\left[\\cos \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)+i\\sin \\left(\\frac{2\\pi }{9}+\\frac{12\\pi }{9}\\right)\\right]&& \\text{Add }\\frac{2\\left(2\\right)\\pi }{3}\\text{ to each angle.} \\\\ &{z}^{\\frac{1}{3}}=2\\left(\\cos \\left(\\frac{14\\pi }{9}\\right)+i\\sin \\left(\\frac{14\\pi }{9}\\right)\\right)\\end{align}[\/latex]<\/p>\r\nRemember to find the common denominator to simplify fractions in situations like this one. For [latex]k=1[\/latex], the angle simplification is\r\n

[latex]\\begin{align}\\frac{\\frac{2\\pi }{3}}{3}+\\frac{2\\left(1\\right)\\pi }{3}&=\\frac{2\\pi }{3}\\left(\\frac{1}{3}\\right)+\\frac{2\\left(1\\right)\\pi }{3}\\left(\\frac{3}{3}\\right)\\\\ &=\\frac{2\\pi }{9}+\\frac{6\\pi }{9} \\\\ &=\\frac{8\\pi }{9} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Find the four fourth roots of [latex]16\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex].[reveal-answer q=\"929013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929013\"][latex]{z}_{0}=2\\left(\\cos \\left(30^\\circ \\right)+i\\sin \\left(30^\\circ \\right)\\right)[\/latex][latex]{z}_{1}=2\\left(\\cos \\left(120^\\circ \\right)+i\\sin \\left(120^\\circ \\right)\\right)[\/latex]\r\n\r\n[latex]{z}_{2}=2\\left(\\cos \\left(210^\\circ \\right)+i\\sin \\left(210^\\circ \\right)\\right)[\/latex]\r\n\r\n[latex]{z}_{3}=2\\left(\\cos \\left(300^\\circ \\right)+i\\sin \\left(300^\\circ \\right)\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Finding Powers and Roots of Complex Numbers in Polar Form<\/h2>\n

Finding powers of complex numbers is greatly simplified using De Moivre\u2019s Theorem. It states that, for a positive integer [latex]n,{z}^{n}[\/latex] is found by raising the modulus to the [latex]n\\text{th}[\/latex] power and multiplying the argument by [latex]n[\/latex]. It is the standard method used in modern mathematics.<\/p>\n

\n

De Moivre\u2019s Theorem<\/h3>\n

If [latex]z=r\\left(\\cos \\theta +i\\sin \\theta \\right)[\/latex] is a complex number, then<\/p>\n

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

where [latex]n[\/latex] is a positive integer.<\/p>\n<\/section>\n

Evaluate the expression [latex]{\\left(1+i\\right)}^{5}[\/latex] using De Moivre\u2019s Theorem.<\/p>\n