Pre-Calc Exam Notes 140

Pre-Calc Exam Notes 140 - 140 Chapter 6 • Additional...

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Unformatted text preview: 140 Chapter 6 • Additional Topics §6.3 The fifth item is a special case of the multiplication formula: (a + bi ) (a − bi ) = ((a)(a) − ( b)(− b)) + ((a)(− b) + ( b)(a)) i = (a2 + b2 ) + (−ab + ba) i = (a2 + b2 ) + 0 i = a2 + b 2 The sixth item comes from using the previous items: a + bi c − di a + bi = · c + di c + di c − di (ac − b(− d )) + (a(− d ) + bc) i = c2 + d 2 (ac + bd ) + ( bc − ad ) i = c2 + d 2 The conjugate a + bi of a complex number a + bi is defined as a + bi = a − bi . Notice that (a + bi ) + (a + bi ) = 2a is a real number, (a + bi ) − (a + bi ) = 2 bi is an imaginary number if b = 0, and (a + bi )(a + bi ) = a2 + b2 is a real number. So for a complex number z = a + bi , z z = a2 + b2 and thus we can define the modulus of z to be z z = a2 + b2 , which we denote by | z |. Example 6.9 Let z1 = −2 + 3 i and z2 = 3 + 4 i . Find z1 + z2 , z1 − z2 , z1 z2 , z1 / z2 , | z1 |, and | z2 |. Solution: Using our rules and definitions, we have: z1 + z2 = (−2 + 3 i ) + (3 + 4 i ) = 1 + 7i z1 − z2 = (−2 + 3 i ) − (3 + 4 i ) = −5 − i z1 z2 = (−2 + 3 i ) (3 + 4 i ) = ((−2)(3) − (3)(4)) + ((−2)(4) + (3)(3)) i = −18 + i −2 + 3 i z1 = z2 3 + 4i (−2)(3) + (3)(4) + ((3)(3) − (−2)(4)) i = 32 + 42 17 6 + i = 25 25 | z1 | = = | z2 | = =5 (−2)2 + 32 13 32 + 42 ...
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