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MATH 57091 - Algebra for High School Teachers Division of Complex Numbers Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1/9 Complex Conjugates The computation of the multiplicative inverse of a complex number will require use of the “complex conjugate” of the number. √ This will √ be analogous to, for example, using the 1“conjugate” 3 − 5 of 3 + 5 to “rationalize” the denominator of 3+√5 : √ √ √ 1 1 3− 5 3− 5 3− 5 3 1√ √ = √ · √ = = = − 5. 9−5 4 4 4 3+ 5 3+ 5 3− 5 This method is based on the “difference of squares” factorization, (a + b)(a − b) = a2 − b 2 . With this in mind, we make the following definition: Definition The (complex) conjugate of the complex number z = a + bi (where a, b ∈ R) is z = a + bi = a − bi. D.L. White (Kent State University) 2/9 Complex Conjugates We have the following properties of the complex conjugate. Theorem 1 If z and w are complex numbers, then i z + w = z + w; ii z · w = z · w. Proof: Exercise. (Let z = a + bi and w = c + di, so z = a − bi and w = c − di, and compute the right hand sides of the equations.) D.L. White (Kent State University) 3/9 Complex Conjugates Theorem 2 If z = a + bi is a complex number, then i z + z = 2a = 2Re(z); ii z − z = 2bi = 2Im(z) · i; iii z · z = a2 + b 2 , a non-negative real number, and z · z = 0 if and only if z = 0 + 0i = 0. Proof: We leave (i) and (ii) as exercises and prove (iii). We have z · z = (a + bi)(a − bi) = a2 − abi + bai − (bi)2 = a2 − abi + abi − b 2 (−1) = a2 + b 2 . Since a and b are real numbers, so is a2 + b 2 . Also, a2 > 0 and b 2 > 0, and so a2 + b 2 > 0, and a2 + b 2 = 0 if and only if a = b = 0; i.e., z = 0. D.L. White (Kent State University) 4/9 Multiplicative Inverses 1 We want to write the multiplicative inverse of a + bi, that is, a+bi , a+bi or the quotient c+di , in the standard form A + Bi, where A, B ∈ R. Of course, if 0 6= r ∈ R, then r has a multiplicative inverse, r −1 or 1r . It is easy to verify, using the definition of multiplication in C, that for a complex number a + bi, we have r ( ar + br i) = (r + 0i)( ar + br i) = a + bi. That is, a + bi = r 1 r · (a + bi) = a r + br i, and so we can divide a complex number by a real number. It follows that in order to write any quotient of complex numbers in the standard form, we only need to “real-ize” the denominator. D.L. White (Kent State University) 5/9 Multiplicative Inverses By Theorem 2(iii), if z = a + bi 6= 0, then zz = a2 + b 2 is a positive real number, and so a − bi a b z = 2 = 2 − 2 i 2 2 zz a +b a +b a + b2 as above. But notice also that z· z zz = = 1, zz zz z hence zz is the multiplicative inverse of z. In particular, every non-zero complex number has a multiplicative inverse. Therefore, we have the following theorem. Theorem The set of complex numbers is a field under the operations defined above. D.L. White (Kent State University) 6/9 Multiplicative Inverses If z = a + bi 6= 0, then we have (a + bi)−1 = 1 a b = 2 − 2 i. 2 a + bi a +b a + b2 Despite the fact that the formula is red, and is inside a red box, it is not worth memorizing. Just learn the method: 1 “real-ize” the denominator of a+bi just as you would “rationalize” the denominator of 3+1√5 . Multiply numerator and denominator by the conjugate of the denominator: 1 1 a − bi a − bi a b = · = 2 = 2 − 2 i. 2 2 a + bi a + bi a − bi a +b a +b a + b2 D.L. White (Kent State University) 7/9 Examples 1 Find the multiplicative inverse of 2 + 3i. (2 + 3i)−1 = = 2 − 3i 2 − 3i 1 · = 2 + 3i 2 − 3i (2 + 3i)(2 − 3i) 2 − 3i 2 3 2 − 3i = = − i. 2 2 2 +3 13 13 13 In standard form, (2 + 3i)−1 = 2 2 13 − 3 13 i (and not 2−3i 13 ). Find the multiplicative inverse of 4 − 7i. (4 − 7i)−1 = = D.L. White (Kent State University) 4 + 7i 4 + 7i 1 · = 4 − 7i 4 + 7i (4 − 7i)(4 + 7i) 42 4 + 7i 4 + 7i 4 7 = = + i. 2 + (−7) 65 65 65 8/9 Division a + bi , by the same method as an inverse: c + di a + bi (a + bi)(c − di) (ac + bd) + (−ad + bc)i = = , c + di (c + di)(c − di) c2 + d2 We compute a quotient, and so ac + bd a + bi ad − bc = 2 − 2 i 2 c + di c +d c + d2 is another formula that is not worth memorizing. For example, 3 + 2i −4 + 5i = (3 + 2i)(−4 − 5i) (−4 + 5i)(−4 − 5i) = [3(−4) − 2(−5)] + [3(−5) + 2(−4)]i (−4)2 + 52 = (−12 + 10) + (−15 − 8)i 2 23 = − − i. 16 + 25 41 41 D.L. White (Kent State University) 9/9