# Complex Variables. Chapter 1. Complex Numbers Section 1.2. Basic Algebraic Properties Proofs of Theorems. December 16, 2016

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1 Complex Variables Chapter 1. Complex Numbers Section 1.2. Basic Algebraic Properties Proofs of Theorems December 16, 2016 () Complex Variables December 16, / 12

3 Theorem Theorem For any a 1, z 2, z 3 C we have the following. 1. Commutivity of addition and multiplication: z 1 + z 2 = z 2 + z 1 and z 1 z 2 = z 2 z Associativity of addition and multiplication: (z 1 + z 2 ) + z 3 = z 1 + (z 2 + z 3 ) and (z 1 z 2 )z 3 = z 1 (z 2 z 3 ). 3. Distribution of multiplication over addition: z 1 (z 2 + z 3 ) = z 1 z 2 + z 1 z There is an additive identity 0 = 0 + i0 such that 0 + z = z for all z C. There is a multiplicative identity 1 = 1 + i0 such that z1 = z for all z C. Also, z0 = 0 for all z C. 5. For each z C there is z C such that z + z = 0. z is the additive inverse of z (denoted z). If z 0, then there is z C such that z z = 1. z is the multiplicative inverse of z. () Complex Variables December 16, / 12

4 Theorem (continued 1) Proof. We have abandoned the ordered pair notation, so let z k = x k + iy k for k = 1, 2, 3 and let z = x + iy, where x k, y k, x, y R 1. (Commutivity) For addition, we have z 1 + z 2 = (z 1 + iy 1 ) + (x 2 + iy 2 ) = (x 1 + x 2 ) + i(y 1 + y 2 ) by the definition of + in C = (x 2 + x 1 ) + i(y 2 + y 1 ) since + is commutative in R = (x 2 + iy 2 ) + (x 1 + iy 1 ) by the definition of + in C = z 2 + z 1. () Complex Variables December 16, / 12

5 Theorem (continued 2) Proof. For multiplication we have z 1 z 2 = (x 1 + iy 1 )(x 2 + iy 2 ) = (x 1 x 2 + y 1 y 2 ) + i(y 1 x 2 + x 1 y 2 ) by the definition of in C = (x 2 x 1 y 2 y 1 ) + i(x 1 y 2 + y 1 x 2 ) since and + are commutative in R = (x 2 + iy 2 )(x 1 + iy 1 ) by the definition of in C = z 2 z (Associativity) For addition we have (z 1 + z 2 ) + z 3 = ((x 1 + iy 1 ) + (x 2 + iy 2 )) + (x 3 + iy 3 ) = ((x 1 + x 2 ) + i(y 1 + y 2 )) + (x 3 + iy 3 ) by the definition of + in C () Complex Variables December 16, / 12

6 Theorem (continued 2) Proof. For multiplication we have z 1 z 2 = (x 1 + iy 1 )(x 2 + iy 2 ) = (x 1 x 2 + y 1 y 2 ) + i(y 1 x 2 + x 1 y 2 ) by the definition of in C = (x 2 x 1 y 2 y 1 ) + i(x 1 y 2 + y 1 x 2 ) since and + are commutative in R = (x 2 + iy 2 )(x 1 + iy 1 ) by the definition of in C = z 2 z (Associativity) For addition we have (z 1 + z 2 ) + z 3 = ((x 1 + iy 1 ) + (x 2 + iy 2 )) + (x 3 + iy 3 ) = ((x 1 + x 2 ) + i(y 1 + y 2 )) + (x 3 + iy 3 ) by the definition of + in C () Complex Variables December 16, / 12

7 Theorem (continued 3) (z 1 + z 2 ) + z 3 = ((x 1 + x 2 ) + x 3 ) + i((y 1 + y 2 ) + y 3 ) by the definition of + in C = (x 1 + (x 2 + x 3 )) + i(y 1 + (y 2 + y 3 )) since + is associative in R = (x 1 + iy 1 ) + ((x 2 + x 3 ) + i(y 2 + y 3 )) by the definition of + in C = z 1 + (z 2 + z 3 ) For multiplication we have (z 1 z 2 )z 3 = ((x 1 + iy 1 )(x 2 + iy 2 ))(x 3 + iy 3 ) = ((x 1 x 2 y 1 y 2 ) + i(y 1 x 2 + x 1 y 2 ))(x 3 + iy 3 ) by the definition of in C () Complex Variables December 16, / 12

8 Theorem (continued 3) (z 1 + z 2 ) + z 3 = ((x 1 + x 2 ) + x 3 ) + i((y 1 + y 2 ) + y 3 ) by the definition of + in C = (x 1 + (x 2 + x 3 )) + i(y 1 + (y 2 + y 3 )) since + is associative in R = (x 1 + iy 1 ) + ((x 2 + x 3 ) + i(y 2 + y 3 )) by the definition of + in C = z 1 + (z 2 + z 3 ) For multiplication we have (z 1 z 2 )z 3 = ((x 1 + iy 1 )(x 2 + iy 2 ))(x 3 + iy 3 ) = ((x 1 x 2 y 1 y 2 ) + i(y 1 x 2 + x 1 y 2 ))(x 3 + iy 3 ) by the definition of in C () Complex Variables December 16, / 12

9 Theorem (continued 4) (z 1 z 2 )z 3 = ((x 1 x 2 y 1 y 2 )x 3 (y 1 x 2 x 1 y 2 )y 3 ) + i((y 1 x 2 + x 1 y 2 )x 3 + (x 1 x 2 y 1 y 2 )y 3 ) by the definition of in C = (x 1 (x 2 x 3 y 2 y 3 ) y 1 (y 2 x 3 + x 2 y 3 )) +i(y 1 (x 2 x 3 y 2 y 3 ) + x 1 (y 2 x 3 + x 2 y 3 )) by distribution, commutivity, and associativity in R = (x 1 + iy 1 )((x 2 x 3 y 2 y 3 ) + i(y 2 x 3 + x 2 y 3 )) by the definition of in C = (x 2 + iy 1 )((x 2 + iy 2 )(x 3 + iy 3 )) by the definition of in C = z 1 (z 2 z 3 ). () Complex Variables December 16, / 12

10 Theorem (continued 5) 3. (Distribution) For distribution we have z 1 (z 2 + z 3 ) = (z 1 + iy 1 )((x 2 + iy 2 ) + (x 3 + iy 3 )) = (x 1 + iy 1 )((x 2 + x 3 ) + i(y 2 + y 3 )) by the definition of + in C = (x 1 (x 2 + x 3 ) y 1 (y 2 + y 3 )) + i(y 1 (x 2 + x 3 ) + x 1 (y 2 + y 3 )) by the definition of in C = (x 1 x 2 + x 1 x 3 y 1 y 2 y 1 y 3 ) + i(y 1 x 2 + y 1 x 3 + x 1 y 2 + x 1 y 3 ) by distribution in R = ((x 1 x 2 y 1 y 2 ) + i(y 1 x 2 + x 1 y 2 )) + ((x 1 x 3 y 1 y 3 ) +i(y 1 x 3 + x 1 y 3 ) by the definition of + in C = (x 1 + iy 1 )(x 2 + iy 2 ) + (x 1 + iy 1 )(x 3 + iy 3 ) by the definition of in C = z 1 z 2 + z 1 z 3. () Complex Variables December 16, / 12

11 Theorem (continued 6) 4. (Identities) We easily have 0 + z = (0 + i0) + (x + iy) = (0 + x) + i(0 + y) by the definition of + in C = x + iy since 0 is the additive identity in R = z and 1z = (1 + i0) + (x + iy) = ((1)(x) (0)(y)) + i((0)(x) + (1)(y)) by the definition of in C = x + iy since 1 is the multiplicative identity in R = z. () Complex Variables December 16, / 12

12 Theorem (continued 6) 4. (Identities) We easily have 0 + z = (0 + i0) + (x + iy) = (0 + x) + i(0 + y) by the definition of + in C = x + iy since 0 is the additive identity in R = z and 1z = (1 + i0) + (x + iy) = ((1)(x) (0)(y)) + i((0)(x) + (1)(y)) by the definition of in C = x + iy since 1 is the multiplicative identity in R = z. () Complex Variables December 16, / 12

13 Theorem (continued 6) Also, z0 = (x + iy)(0 + i0) = ((x)(0) (y)(0)) + i((y)(0) + (x)(0)) by the definition of in C = 0 + i0 since r0 = 0 for all r R = (Inverses) For z = x + iy, we take z = ( x) + i( y) and then z + z = (x + iy) + (( x) + i( y)) = (x + ( x)) + i(y + ( y)) by the definition of + in C = 0 + i0 since x and y are the + inverses of x and y, respectively, in R = 0. () Complex Variables December 16, / 12

14 Theorem (continued 6) Also, z0 = (x + iy)(0 + i0) = ((x)(0) (y)(0)) + i((y)(0) + (x)(0)) by the definition of in C = 0 + i0 since r0 = 0 for all r R = (Inverses) For z = x + iy, we take z = ( x) + i( y) and then z + z = (x + iy) + (( x) + i( y)) = (x + ( x)) + i(y + ( y)) by the definition of + in C = 0 + i0 since x and y are the + inverses of x and y, respectively, in R = 0. () Complex Variables December 16, / 12

15 Theorem (continued 6) For z = x + iy 0, take z = x/(x 2 + y 2 ) + i( y)/(x 2 + y 2 ) (see page 4 of the text for motivation). Then zz = (x + iy)(x/(x 2 + y 2 ) + i( y)/(x 2 + y 2 )) = ((x)(x/(x 2 + y 2 ) (y)(( y)/x 2 + y 2 ))) +i((y)(x/(x 2 + y 2 )) + x(( y)/(x 2 + y 2 ))) by the definition of in C = 1 + i0 by the multiplicative and additive inverse properties in R = 1. () Complex Variables December 16, / 12

16 Corollary Corollary Corollary For z 1, z 2, z 3 C if z 1 z 2 = 0 then either z 1 = 0 or z 2 = 0. That is, C has no zero divisors. Proof. Suppose one of z 1 or z 2 in nonzero. WLOG, say z 1 0. Then, by Theorem 1.2.1(5), there is a multiplicative inverse z1 1 C such that z 1 z1 1 = 1. () Complex Variables December 16, / 12

17 Corollary Corollary Corollary For z 1, z 2, z 3 C if z 1 z 2 = 0 then either z 1 = 0 or z 2 = 0. That is, C has no zero divisors. Proof. Suppose one of z 1 or z 2 in nonzero. WLOG, say z 1 0. Then, by Theorem 1.2.1(5), there is a multiplicative inverse z1 1 C such that z 1 z1 1 = 1. So z 2 = z 2 1 by Theorem 1.2.1(4) ( identity) = z 2 (z 1 z2 1 = (z 1 z1 1 2 = (z1 1 1)z 2 by Theorem 1.2.1(1) (commutivity of ) = z 1 1 (z 1z 2 ) by Theorem 1.2.1(2) (associativity of ) (0) by hypothesis = z1 1 = 0 by Theorem 1.2.1(4). So if one of z 1, z 2 is nonzero, then the other is 0 and the result follows. () Complex Variables December 16, / 12

18 Corollary Corollary Corollary For z 1, z 2, z 3 C if z 1 z 2 = 0 then either z 1 = 0 or z 2 = 0. That is, C has no zero divisors. Proof. Suppose one of z 1 or z 2 in nonzero. WLOG, say z 1 0. Then, by Theorem 1.2.1(5), there is a multiplicative inverse z1 1 C such that z 1 z1 1 = 1. So z 2 = z 2 1 by Theorem 1.2.1(4) ( identity) = z 2 (z 1 z2 1 = (z 1 z1 1 2 = (z1 1 1)z 2 by Theorem 1.2.1(1) (commutivity of ) = z 1 1 (z 1z 2 ) by Theorem 1.2.1(2) (associativity of ) (0) by hypothesis = z1 1 = 0 by Theorem 1.2.1(4). So if one of z 1, z 2 is nonzero, then the other is 0 and the result follows. () Complex Variables December 16, / 12

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