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Math 0031 Section 3.1 Page 1 of 4 3.1 Complex Numbers The Number i The number i is defined such that i2 = −1. √ In this course we also assume that i = −1. Example Express each number in terms of i. (a) √ −36. Solution: (b) √ p √ √ √ −36 = (−1) · 36 = −1 · 36 = i · 6 = 6i −32. Solution: p √ √ √ √ √ √ √ −32 = (−1) · 2 · 16 = −1 · 2 · 16 = i · 2 · 4 = 4i 2 = 4 2 i Complex Numbers A complex number is a number of the form a + bi, where a and b are real numbers. The number a is said to be the real part of a+bi and the number b is said to be the imaginary part of a + bi. If b = 0 then a + bi = a + 0i = a is a real number. So every real number is a complex number. If a = 0 then a + bi = 0 + bi = bi is an imaginary number (or a pure imaginary number). Note: b is a real number and bi is an imaginary number. Sometimes i is also called an imaginary number because i = 1i. Addition and Subtraction (a + bi) + (c + di) = (a + c) + (b + d)i. (a + bi) − (c + di) = (a − c) + (b − d)i. Example Add or subtract each of the following. (a) (2 + 6i) + (7 − 11i) Solution: (b) (2 + 6i) + (7 − 11i) = (2 + 7) + (6 + (−11))i = 9 − 5i (2 + 6i) − (7 − 11i) Solution: (2 + 6i) − (7 − 11i) = (2 − 7) + (6 − (−11))i = −5 + 17i Math 0031 Section 3.1 Page 2 of 4 Multiplication (a + bi) · (c + di) = (ac − bd) + (ad + bc)i. Indeed, (a + bi) · (c + di) = ac + adi + bci + bdi2 = ac + (ad + bc)i − bd = (ac − bd) + (ad + bc)i. √ √ Note: In real numbers 1√= 1. In complex numbers 1 can be both 1 and −1 because 2 2 (−1) = 1 and 1 = 1. So 1 = ±1. Due to this duality we have, for example √ √ √ √ √ √ −4 · −9 = −1 · 4 · −1 · 9 = i · 2 · i · 3 = 6i2 = −6 p √ √ √ and −4 · −9 = (−4)(−9) = 36 = 6 The ambiguity comes from the fact that p √ √ √ √ √ √ √ √ −4 · −9 = −1 · 4 · −1 · 9 = (−1)(−1) · 36 = 1 · 6 = ±1 · 6 = ±6. For the sake of certainty in this course we define that for positive reals a and b the only correct multiplication of two imaginary numbers in the radical notation is √ √ √ √ √ √ √ √ √ √ −a · −b = −1 · a · −1 · b = i · a · i · b = i2 ab = − ab Example Multiply and simplify each of the following. (a) √ −10 · √ −8 Solution: (b) (2 + i)(7 − 3i) Solution: (c) √ √ √ √ √ √ √ √ √ −10 · −8 = −1 · 10 · −1 · 8 = i2 10 · 8 = − 5 · 16 = −4 5 (2 + i)(7 − 3i) = (14 − (−3)) + (−6 + 7)i = 17 + i (2 + 3i)2 Solution: (2 + 3i)2 = (2 + 3i)(2 + 3i) = (4 − 9) + (6 + 6)i = −5 + 12i Powers of i It is easy to calculate the following powers of i i2 = −1. i3 = i2 · i = −1i = −i. i4 = i2 · i2 = (−1)(−1) = 1. i5 = i4 · i = 1 · i = i. and the cycle is closed. Math 0031 Section 3.1 Page 3 of 4 Therefore for positive integers n, m, and k such that k < 4 and n = 4m + k we have in = i4m+k = i4m · ik = (i4 )m · ik = 1m · ik = ik Example Simplify each of the following. (a) i49 Solution: (b) 49 = 4 · 12 + 1. Therefore i49 = i1 = 1 i91 Solution: 91 = 4 · 22 + 3. Therefore i91 = i3 = −i Conjugates and Division Conjugate of a Complex Number The conjugate of a complex number a + bi is a − bi. The numbers a + bi and a − bi are complex conjugates. Example Find the conjugate to the number −12 − 3i. Solution: It is −12 − (−3)i = 12 + 3i Example Find the product of the number 2 − 3i and its conjugate. Solution: The conjugate is 2 + 3i. Then (2 − 3i)(2 + 3i) = (4 − (−9)) + (6 − 6)i = 13 Product of conjugates (a + bi)(a − bi) = a2 + b2 Division (a + bi) ÷ (c + di) = ac + bd bc − ad + 2 i. c2 + d 2 c + d2 Indeed, (a + bi) ÷ (c + di) = a + bi (a + bi)(c − di) (ac + bd) + (bc − ad)i ac + bd bc − ad = = = 2 + 2 i. 2 2 c + di (c + di)(c − di) c +d c + d2 c + d2 Example Divide and simplify each of the following. (a) (2 + i) ÷ (3 + 5i) Math 0031 Solution: (b) Section 3.1 (2 + i) ÷ (3 + 5i) = 6+5 3 − 10 + i= 32 + 52 32 + 52 Page 4 of 4 11 34 − 7 34 i (−2 + 5i) ÷ (7 − 4i) Solution: (−2 + 5i) ÷ (7 − 4i) = −14 − 20 35 − (−2)(−4) + + i = − 34 65 2 2 2 2 7 +4 7 +4 27 65 i