Finding n-th Roots

Finding n-th Roots To solve linear differential equations with constant coefficients, we need to be able to find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of zn = α, where α is a complex number, i.e., finding the n-th roots of α. Polar repre­ sentation will be a big help in this. Let’s begin with a special case: the n-th roots of unity: the solutions to zn = 1. To solve this equation, we use polar representation for both sides, setting z = reiθ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes r n einθ = 1 · e(2kπi) , k = 0, ±1, ±2, · · · . Equating the absolute values and the polar angles of the two sides gives r n = 1, nθ = 2kπ, k = 0, ±1, ±2, · · · , from which we conclude that r = 1, θ= 2kπ , n k = 0, 1, · · · , n − 1. (1) In the above, we get only the value r = 1, since r must be real and non­ negative. We don’t need any integer values of k other than 0, · · · , n − 1, since they would not produce a complex number different from the above n numbers. That is, if we add an, an integer multiple of n, to k, we get the same complex number: θ� = 2(k + an)π = θ + 2aπ; n and � eiθ = eiθ , since e2aπi = (e2πi ) a = 1. We conclude from (1) therefore that the n-th roots of 1 are the numbers e2kπi/n , k = 0, · · · , n − 1. (2) This shows there are n complex n-th roots of unity. They all lie on the unit circle in the complex plane, since they have absolute value 1; they are evenly spaced around the unit circle, starting with the root z = 1; the angle between two consecutive roots is 2π/n. These facts are illustrated for the case n = 6 in the figure below y x Fig. 1. The six solutions to the equation z6 = 1 lie on a unit circle in the complex plane. From (2), we get another notation for the roots of unity (ζ is the Greek letter “zeta”): the n-th roots of 1 are 1, ζ, ζ 2 , · · · , ζ n−1 , where ζ = e2πi/n . (3) We now generalize the above to find the n-th roots of an arbitrary com­ plex number w. We begin by writing w in polar form: w = reiθ ; θ = Argw, 0 ≤ θ < 2π, i.e., θ is the principal value of the polar angle of w. Then the same reasoning as we used above shows that if z is an n-th root of w, then √ zn = w = reiθ so z = n rei(θ +2kπ )/n , k = 0, 1, · · · , n − 1. (4) Comparing this with (3), we see that these n roots can be written in the suggestive form √ √ n w = z0 , z0 ζ, z0 ζ 2 , · · · , z0 ζ n−1 , where z0 = n reiθ/n . (5) As a check, we see that all of the n complex numbers in (5) satisfy zn = w : ( z0 ζ i ) n = = z0n ζ ni = z0n · 1i , w, since ζ n = 1, by (3); by the definition (5) of z0 and (4). Example. Find in Cartesian form all values of √ a) 3 1 b) √ 4 i Solution. a) According to (3), the cube roots of 1 are 1, ω, and ω 2 , where √ 1 3 2πi/3 ω=e = cos(2π/3) + i sin(2π/3) = − + i 2 2 √ 1 3 ω 2 = e−2πi/3 = cos(−2π/3) + i sin(−2π/3) = − − i . 2 2 The greek letter ω (“omega”) is traditionally used for this cube root. Note that for the polar angle of ω 2 we used −2π/3 rather than the equiva­ lent angle 4π/3, in order to take advantage of the identities sin(− x ) = − sin( x ). cos(− x ) = cos( x ) Note that ω 2 = ω. Another way to do this problem would be to draw the position of ω 2 and ω on the unit circle and use geometry to figure out their coordinates. √ √ 4 b) To find i, we can use (5). We know that 4 1 = 1, i, −1, −i (either by drawing the unit circle picture or by using (3)). Therefore by (5), we get √ 4 i = z0 , z0 i, −z0 , −z0 i, where z0 = eπi/8 = cos(π/8) + i sin(π/8); = a + ib, −b + ia, − a − ib, b − ia where z0 = a + ib = cos(π/8) + i sin(π/8). Example. Solve the equation x6 − 2x3 + 2 = 0. Solution. Treating this as a quadratic equation in x3 , we solve the quadratic by using the quadratic formula; the two roots are 1 + i and 1 − i (check this!), so the roots of the original equation satisfy either x3 = 1 + i or x3 = 1 − i. This reduces the problem to finding the cube roots of the two complex numbers 1 ± i. We begin by writing them in polar form: √ √ 1 + i = 2eπi/4 , 1 − i = 2e−πi/4 . (Once again, note the use of the negative polar angle for 1 − i, which is more convenient for calculations.) The three cube roots of the first of these are (by (4)), √ 6 √ 6 2eπi/12 = √ 6 √ 6 2 (cos(π/12) + i sin(π/12)) π 2π 3π + = ; 12 3 4 √ √ π 2π 7π 6 6 2e−7πi/12 = 2 (cos(7π/12) − i sin(7π/12)) , since − =− . 12 3 12 � � √ −1 + i −1 + i 6 √ The second cube root can also be written as 2 = √ . 3 2 2 This gives three of the cube roots. The other three are the cube roots of 1 − i, which may be found by replacing i by −i everywhere (i.e., taking the complex conjugate). 2e3πi/4 = 2 (cos(3π/4) + i sin(3π/4)) , since The cube roots can also be described according to (5) as √ √ 6 6 z1 , z1 ω, z1 ω 2 and z2 , z2 ω, z2 ω 2 where z1 = 2eπi/12 , z2 = 2e−πi/12 .