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A FIRST COURSE IN LINEAR ALGEBRA An Open Text by Ken Kuttler Complex Numbers - Roots of Complex Numbers Lecture Notes by Karen Seyffarth∗ Adapted by LYRYX SERVICE COURSE SOLUTION ∗ Attribution-NonCommercial-ShareAlike (CC BY-NC-SA) This license lets others remix, tweak, and build upon your work non-commercially, as long as they credit you and license their new creations under the identical terms. Complex Numbers - Roots of Complex Numbers Page 1/14 Roots of Complex Numbers Definition Let z and q be complex numbers, and let n be a positive integer. Then z is called an nth root of q if z n = q. De Moivre’s Theorem and its implication n If θ is any angle and n is a positive integer, e iθ = e inθ . This implies that for any real number r > 0 and any positive integer n, (re iθ )n = r n e inθ . This leads to the following result. Corollary Let q be a nonzero complex number and n a positive integer. Then z n = q has exactly n complex solutions, i.e., q has exactly n complex nth roots. Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 2/14 Example For any positive real number a, z 2 = a has two complex (in this case, real) √ √ solution, z = a and z = − a. This is equivalent to the statement that a has two complex (in this case, real) square roots. One particular example: 25 has two square roots, 5 and −5, and these are the two solutions to z 2 = 25. But we all knew that. A more interesting example is that −1 has no real square roots, but suddenly it has two (complex) square roots, i and −i. These are the two (complex) solutions to z 2 = 1. Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 3/14 Cube Roots Example To find the (three) cube roots of i, we solve the equation z 3 = i. To do so, we express both z and i in polar form: convert i to polar form, and write z = re iθ , giving us (re iθ )3 = e πi/2 . Thus r 3 e 3iθ = 1e πi/2 , implying that r 3 = 1 and 3θ = π 2. Since r is a real number, r 3 = 1 implies that r = 1. The statement 3θ = π2 is not completely correct. The problem that arises is that the argument for i, π2 is not unique. Instead, we could have written i = e 5πi/2 or i = e 9πi/2 or i = e −3πi/2 . In fact, there are infinitely many choices for the argument of i. The important thing to notice is that any two different arguments differ by a multiple of 2π, and thus we may write π 3θ = + 2π`, ` ∈ Z. 2 (Z denotes the set of integers: {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}). Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 4/14 Example (continued) Dividing both sides of 3θ = π 2 + 2π` by 3: θ= π 2 + π`, 6 3 where ` is any integer. The Corollary to De Moivre’s Theorem tells us that there are only three different cube roots. These are obtained by using ` = 0, ` = 1, and ` = 2, resulting in three values of θ: π 5π 9π 3π , , and = . 6 6 6 2 Thus the cube roots of i are e πi/6 , e 5πi/6 , and e 3πi/2 . We now convert these to Cartesian form. Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 5/14 Example (continued) √ πi/6 = e πi/6 = e 3πi/2 = e 3 1 + i, 2√ 2 3 1 − + i, 2 2 − i. You can check your work by computing the cube of each of these. This process is summarized in the following procedure. Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 6/14 Finding Roots of a Complex Number Let w be a complex number. We wish to find the nth roots of w , that is all z such that z n = w . There are n distinct nth roots and they can be found as follows:. 1. Express both z and w in polar form z = re iθ , w = se iφ . Then z n = w becomes: (re iθ )n = r n e inθ = se iφ We need to solve for r and θ. 2. Solve the following two equations: rn = s e inθ = e iφ Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers (1) Page 7/14 Continued 3. The solutions to r n = s are given by r = 4. The solutions to e inθ = e iφ √ n s. are given by: nθ = φ + 2π`, for ` = 0, 1, 2, · · · , n − 1 or θ= φ 2 + π`, for ` = 0, 1, 2, · · · , n − 1 n n 5. Using the solutions r , θ to the equations given in (1) construct the nth roots of the form z = re iθ . Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 8/14 Problem √ Find all complex numbers z such that z 4 = 2( 3i − 1), and express each in the form a + bi. Solution √ √ 1. Convert 2( 3i − 1) = −2 + 2 3i to polar form: q √ √ |z 4 | = (−2)2 + (2 3)2 = 16 = 4. √ If φ is an argument for −2 + 2 3i, then √ √ 1 2 3 2π −2 3 cos φ = = − and sin φ = = , so φ = . 4 2 4 2 3 Thus z 4 = 4e 2πi/3 . Let z = re iθ . 2. The equation becomes r 4 e i4θ = 4e 2πi/3 , so we need to solve e Complex Numbers - Roots of Complex Numbers r4 = i4θ = e 2πi/3 4 Roots of Complex Numbers Page 9/14 Solution (continued) √ 2 3. Since r 4 = 4, r 2 = ±2. But r is real, √ and so r = 2, implying r = ± 2. However r ≥ 0, and therefore r = 2. 4. The solutions to e i4θ = e 2πi/3 are given by 4θ = 2 π + 2π`, ` = 0, 1, 2, 3. 3 Therefore, θ= 2π 2π` π π` π(3` + 1) + = + = , for ` = 0, 1, 2, 3. 12 4 6 2 6 Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 10/14 Solution (continued) √ 5. Thus r = 2 and θ = 3`+1 π, ` = 0, 1, 2, 3. Converting to 6 Cartesian form: √ √ √ √ √ ` = 0 : z = 2e πi/6 = 2( ( 2 3 + √12 i) = 26√+ 22√i √ √ 3 2 6 ` = 1 : z = 2e 2πi/3 = 2(− 1√ 2 + 2 i) = − √2 + √2 i √ 7πi/6 √ 6 ` = 2 : z = 2e = 2(− 23√− 12 i) = − − 2i √ 2 √ 2 √ 5πi/3 √ 1 ` = 3 : z = 2e = 2( 2 − 23 i) = 22 − 26 i √ Therefore, the fourth roots of 2( 3i − 1) are: √ √ √ √ √ √ √ √ 6 2 2 6 6 2 2 6 + i, − + i, − − i, − i. 2 2 2 2 2 2 2 2 Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 11/14 Roots of Unity Definition A complex number z is a root of unity if there exists a positive integer n so that z n = 1. Problem Find the sixth roots of unity, i.e., all solutions to z 6 = 1. Solution Write z = re iθ and convert 1 to polar form to get (re iθ )6 = e i0 , and so r 6 e 6θi = e i0 . Equating the absolute values and arguments, r 6 = 1 and 6θ = 0 + 2π`, ` = 0, 1, 2, 3, 4, 5. Since r is real, r = 1. The six arguments for the solutions are 2π` π` θ= = , ` = 0, 1, 2, 3, 4, 5. 6 3 Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 12/14 Solution (continued) The six arguments for the solutions are θ= π` 2π` = , ` = 0, 1, 2, 3, 4, 5. 6 3 Converting these to Cartesian form: ` 0 1 2 3 4 5 θ 0 π 3 2π 3 π 4π 3 5π 3 z e 0i = 1 √ e πi/3 = 12 + 23 i√ e 2πi/3 = − 12 + 23 i e πi = −1 √ e 4πi/3 = − 12 −√ 23 i e 5πi/3 = 12 − 23 i If you graph these six point in the complex plane, you’ll see that they result in six equally spaced points on the unit circle, one of them being (1, 0). Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 13/14 Roots of Unity For any integer n ≥ 1, the (complex) solutions to z n = 1 are z = e 2π`i/n for ` = 0, 1, 2, . . . , n − 1. Furthermore, the nth roots of unity correspond to n equally spaced points on the unit circle, one of them being (1, 0). Complex Numbers - Roots of Complex Numbers Roots of Complex Numbers Page 14/14