**1**How to Find the Square Root of a Complex Number Stanley Rabinowitz 12 Vine Brook Road Westford, Massachusetts 01886 USA It is known that every polynomial with complex coefficients has a complex root. This is called “The Fundamental Theorem of Algebra”. In particular, the equation z2= c where c is a complex number, always has a solution. In other words, every complex number has a square root. We could write this square root as√c. But – it would be nice to find an explicit representation for that square root in the form p + qi where p and q are real numbers. It is the purpose of this note to show how to actually find the square root of a given complex number. This method is not new (see for example page 95 of Mostowski and Stark [1]) but appears to be little-known. Let us start with the complex number c = a + bi where a and b are real (b ?= 0) and attempt to find an explicit representation for its square root. Of course, every complex number (other than 0) will have two square roots. If w is one square root, then the other one will be −w. We will find the one whose real part is non-negative. Let us assume that a square root of c is p + qi where p and q are real. Then we have (p + qi)2= a + bi. Equating the real and imaginary parts gives us the two equations p2− q2= a 2pq = b. (1) (2) We must have p ?= 0 since b ?= 0. Solving equation (2) for q gives b q = (3) 2p and we can substitute this value for q into equation (1) to get ?b ?2 = a p2− 2p Reprinted from Mathematics and Informatics Quarterly, 3(1993)54–562 or 4p4− 4ap2− b2= 0. This is a quadratic in p2, so we can solve for p2using the quadratic formula. We get (taking just the positive solution): p2=a +√a2+ b2 2 so that ? ? 1 p = a + a2+ b2. √2 From equation (3), we find b b q = 2p= ?√a2+ b2+ a b ?√a2+ b2+ a √2 ?√a2+ b2− a =sgnb √2 2 √2 ?√a2+ b2− a = ?√a2+ b2− a · 2 √2 ?√a2+ b2− a b = ?(a2+ b2) − a2 √b2 ?? ?√a2+ b2− a |b| b b = = √2 √2 a2+ b2− a . √b2= |b|, so that b/|b| = sgn(b), the sign of b (defined to be +1 if b > 0 and -1 Note that if b < 0). Thus we have our answer: Theorem 1. If a and b are real (b ?= 0), then √a + bi = p + qi where p and q are real and are given by ?? 1 a2+ b2+ a p = √2 and ?? q =sgnb a2+ b2− a . √2

3 In practice, square roots of complex numbers are more easily found by first converting to polar form and then using DeMoivre’s Theorem. Any complex number a + bi can be written as r(cosθ + isinθ) where ? cosθ =a sinθ =b a2+ b2, and (4) r = r, r DeMoivre’s Theorem states that if n is any positive real number, then (a + bi)n= rn(cosnθ + isinnθ). In particular, if n = 1/2, we have ? ? √a + bi =√r cosθ 2+ isinθ (5) . 2 This gives us a straightforward way to calculate√a + bi. This method also gives us an alternate proof of Theorem 1. If we apply the half-angle formulae cosθ 2= ± and sinθ 2= ± to equation (5), we get ?? ? ? 1 + cosθ 2 1 − cosθ 2 ? ? 1 + cosθ 2 1 − cosθ 2 √a + bi =√r ± i where we have arbitrarily chosen the “+” sign for the first radical. Using the value for cosθ from equation (4), we get ?? ?r + a ?√a2+ b2+ a 2 ? ? 1 + a/r 2 1 − a/r 2 √a + bi =√r ± i ?r − a = ± i 2 2 ?√a2+ b2− a = ± i 2 which is equivalent to Theorem 1. As before, the “±” sign should be chosen to be the same as the sign of b.

4 We sometimes need to find the square root of an expression of the form s+√−d where s and d are real numbers and d > 0. We can use Theorem 1 to get an explicit formula for this square root which is of the form p+qi where p and q are real. Since s+√−d = s+i√d, we can let a = s and b =√d in Theorem 1, to get the result: Theorem 2. If s and d are real with d > 0, then ?? ?? ? 1 s2+ d + s + i1 s +√−d = s2+ d − s . √2 √2 Reference [1] A. Mostowski and M. Stark, Introduction to Higher Algebra. Pergamon Press. New York: 1964.