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10.5 Powers and Roots of Complex Numbers

10.5 Powers and Roots of Complex Numbers. Powers of Complex numbers. In the same way,. 10.5 De Moivre’s Theorem. De Moivre’s Theorem If r (cos  + i sin  ) is a complex number, and n is any real number, then In compact form, this is written.

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10.5 Powers and Roots of Complex Numbers

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  1. 10.5 Powers and Roots of Complex Numbers • Powers of Complex numbers In the same way,

  2. 10.5 De Moivre’s Theorem De Moivre’s Theorem If r(cos  + i sin ) is a complex number, and n is any real number, then In compact form, this is written

  3. 10.5 Finding a Power of a Complex Number Example Find and express the result in rectangular form. Solution Convert to trigonometric form. 480º and 120º are coterminal. cos120º = -1/2; sin120º = Rectangular form

  4. 10.5 Roots of Complex Numbers nth Root For a positive integer n, the complex number a+bi is the nth of the complex number x + yi if (a + bi)n = x + yi. • To find three complex cube roots of 8(cos 135º + i sin 135º), for example, look for a complex number, say r(cos  + sin ), that will satisfy

  5. 10.5 Roots of Complex Numbers By De Moivre’s Theorem, becomes Therefore, we must have r3 = 8, or r = 2, and

  6. 10.5 Roots of Complex Numbers Let k take on integer values 0, 1, and 2. It can be shown that for integers k = 3, 4, and 5, these values have repeating solutions. Therefore, all of the cube roots (three of them) can be found by letting k = 0, 1, and 2.

  7. 10.5 Roots of Complex Numbers When k = 0, the root is 2(cos 45º + i sin 45º). When k = 1, the root is 2(cos 165º + i sin 165º). When k = 2, the root is 2(cos 285º + i sin 285º). nth Root Theorem If n is any positive integer, r is a positive real number, and  is in degrees, then the nonzero complex number r(cos  + i sin  ) has exactly n distinct nth roots, given by where

  8. 10.5 Finding Complex Roots Example Find the two square roots of 4i. Write the roots in rectangular form, and check your results directly with a calculator. Solution Firstwrite 4i in trigonometric form as Here, r = 4 and  = /2. The square roots have modulus and arguments as follows.

  9. 10.5 Finding Complex Roots Since there are two roots, let k = 0 and 1. If k = 0, then If k = 1, then Using these values for , the square roots are 2 cis and 2 cis which can be written in rectangular form as

  10. 10.5 Finding Complex Roots Example Find all fourth roots of Write the roots in rectangular form. Solution If k = 0, then  = 30º + 90º·0 = 30º. If k = 1, then  = 30º + 90º·1 = 120º. If k = 2, then  = 30º + 90º·2 = 210º. If k = 3, then  = 30º + 90º·3 = 300º.

  11. 10.5 Finding Complex Roots Using these angles on the previous slide, the fourth roots are 2 cis 30º, 2 cis 120º, 2 cis 210º, and 2 cis 300º. These four roots can be written in rectangular form as

  12. 10.5 Solving an Equation by Finding Complex Roots Example Find all complex number solutions of x5 – 1 = 0. Graph them as vectors in the complex plane. Solution Write the equation as To find the five complex number solutions, write 1 in polar form as The modulus of the fifth roots is

  13. 10.5 Solving an Equation by Finding Complex Roots The arguments are given by Using these arguments, the fifth roots are 1(cos 0º + i sin 0º), k = 0 1(cos 72º + i sin 72º), k = 1 1(cos 144º + i sin 144º), k = 2 1(cos 216º + i sin 216º), k = 3 1(cos 288º + i sin 288º), k = 4

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