MATH 117 The Roots of Complex Numbers Imaginary numbers were discovered while attempting to evaluate the square roots of negative numbers within the context of attempting to solve the depressed cubic equation. This discovery led to the initial definition of the imaginary number i = −1. Square roots of other negative numbers then could be defined such as −9 = 3i. We can now solve equations that previously had no real solutions. For example if x2+ 9 = 0, then x2= −9 and x = ± −9 = ± 3i. The polynomial x2+ 9 then factors as (x + 3 i)(x − 3 i). What about the equation x4+ 9 = 0? How many solutions will there be? In general, what are the solutions to xn = a? This general equation requires that we compute the nth root of the real number a. Because n could be even or odd and a could be positive or negative, there seems to be four possible cases. However we need complex numbers to compute an only when n is even and a is negative. On the other hand, xn = a has more solutions than just x = an. There are actually n distinct complex solutions to this equation. How do we find all such solutions? Can we similarly solve xn = z, where z is a complex number? The Polar Form Solution To solve the equation xn = z, first recall the forms in which the complex number z may be written: z = a + b i = r eiθ = r (cosθ+ i sinθ), where r is a real number with r ≥ 0, and θ is an angle such that 0 ≤ θ < 2π. It is especially important that we initially write θ as an angle between 0 and 2π. By adding multiples of 2π, we obtain angles that are co-terminal to θ and therefore have the same cosine and sine. In other words we obtain the same complex number z = r eiθ = r ei (θ+ 2 k π), for all integers k, in particular for k = 0, 1, 2, . . . , n − 1. Now for 0 ≤ θ < 2π and 0 ≤ k ≤ n − 1, 0 ≤ θ + 2 k π < 2π + 2 k π = 2 π (k + 1) ≤ 2 nπ. Thus for k = 0, 1, 2, . . . , n − 1, we have 0 ≤ θ+ 2k πn < 2 π. As k varies, there are n distinct angles described here which are equally spaced around the unit circle between 0 and 2π.Because the radius r is a non-negative real number, the value rn is defined. So consider the n distinct complex numbers zk = rncosθ+ 2 k πn    + i sinθ+ 2k πn        = rneiθ+ 2 k πn      , for k = 0, 1, 2, . . . , n − 1. Then zk( )n = rneiθ+ 2 k πn             n = r ei (θ+ 2 k π) = z. Thus there are n distinct complex solutions to the equation xn = z given by the values of zk for k = 0, 1, 2, . . . , n − 1. These solutions will be equally spaced around a circle of radius rn. Example 1. Find all complex solutions to the equation x5 = 16+ 16 3 i. Solution. We first write 16+ 16 3 i in polar form. Here r = 162+ (16 3 )2 = 32. Also tan−1(b / a ) = tan−1( 3) = π/3. So θ = π/3 since the point (16, 16 3) is in the 1st quadrant. So z = 32 cos(π / 3) + i sin(π / 3)( ). The five solutions to x5 = z are zk= 325cosπ / 3 + 2 k π5    + i sinπ / 3 + 2 k π5        = 2 cosπ15+2 k π5    + i sinπ15+2k π5        = 2 cos 12+ 72 k( )+ i sin 12+ 72 k( )( ), for k = 0, 1, 2, 3, 4. We have written the final form in degrees so as to better visualize the result. To compute the rectangular form of the solutions, we simply let k vary from 0 to 4 and evaluate. The five solutions are equally spaced around a circle of radius 2. z0= 2 cos(12) +i sin(12)( ) ≈ 1.956295 + 0. 4158234 i. z1= 2 cos(84) + i sin(84)( ) ≈ 0.209 + 1. 989 iz2= 2 cos(156) + i sin(156)( ) ≈ −1.827 + 0. 81347 i z3= 2 cos(228) + i sin(228)( ) ≈ −1.33826 − 1. 4863 i z4= 2 cos(300) + i sin(300)( ) ≈ 1− 1. 732 i Example 2. Find all complex solutions to the equation x4 = –256. Solution. Because –256 is the point (–256, 0), the polar coordinates are r = 256 and θ = π. The four solutions to the equation are zk= 2564cosπ + 2 k π4    + i sinπ + 2 k π4        = 4 cosπ4+k π2    + i sinπ4+k π2        = 4 cos 45+ 90 k( )+ i sin 45+ 90k( )( ), for k = 0, 1, 2, 3. The rectangular form of the solutions are z0= 4 cos(45) + i sin(45)( )= 2 2 + 2 2 i z1= 4 cos(135) + i sin(135)( )= −2 2 + 2 2 i z2= 4 cos(225) + i sin(225)( )= −2 2 − 2 2 i z3= 4 cos(315) + i sin(315)( )= 2 2 − 2 2 i Notice that the pairs ( z0, z3) and ( z1, z2) are conjugates having the form a ± b i. When a polynomial equation has real coefficients, then it is always the case that the complex solutions occur in conjugate pairs. Thus consider the following four cases: (i) x6 = 100 (ii) x6 = –100 (iii) x7 = 100 (iv) x7 = –100 Case (i): There are two real solutions ± 1006 and four complex solutions occurring in two complex conjugate pairs.Case (ii): There are no real solutions, but six complex solutions occurring in three complex conjugate pairs. Case (iii): There is one real solution 1007 and six complex solutions occurring in three complex conjugate pairs. Case (iv): There is one real solution – 1007 and six complex solutions occurring in three complex conjugate pairs. These cases can be generalized for all (positive) even and odd exponents. Exercises Find all complex solutions to the equations (a) x6 = 15625 i (b) x5 = –2.48832 (c) x4 = −648 + 648 3