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Introduction

Introduction

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Introduction

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  1. Introduction Finding sums and differences of complex numbers is similar to finding sums and differences of expressions involving numeric and algebraic quantities. The real parts of the complex number are similar to the numeric quantities, and the imaginary parts of the complex number are similar to the algebraic quantities. Before finding sums or differences, each complex number should be in the form a + bi. If iis raised to a power n, use the remainder of n ÷ 4 to simplify in. 4.3.2: Adding and Subtracting Complex Numbers

  2. Key Concepts First, find the sum or difference of the real parts of the complex number. Then, to find the sum or difference of the imaginary numbers, add or subtract the coefficients of i. The resulting sum of the real parts and the imaginary parts is the solution. In the following equation, let a, b, c, and d be real numbers. (a + bi) + (c + di) = a + c + bi + di = (a + c) + (b + d)i 4.3.2: Adding and Subtracting Complex Numbers

  3. Key Concepts, continued a + c is the real part of the sum, and (b + d)iis the imaginary part of the sum. When finding the difference, distribute the negative throughout both parts of the second complex number. (a + bi) – (c + di) = a + bi – c – di = (a – c) + (b – d)i a – c is the real part of the difference, and (b – d)iis the imaginary part of the difference. The sum or difference of two complex numbers can be wholly real (having only real parts), wholly imaginary (having only imaginary parts), or complex (having both real and imaginary parts). 4.3.2: Adding and Subtracting Complex Numbers

  4. Common Errors/Misconceptions failing to distribute a negative throughout both the real and imaginary parts of a complex number before simplifying a difference adding the multiples of two powers of iwhen the powers of iare not equal, such as 2i2+ 3i = 5i 4.3.2: Adding and Subtracting Complex Numbers

  5. Guided Practice Example 2 Is (5 + 6i9) – (5 + 3i15) wholly real or wholly imaginary, or does it have both a real and an imaginary part? 4.3.2: Adding and Subtracting Complex Numbers

  6. Guided Practice: Example 2, continued Simplify any expressions containing in. Two expressions, 6i9and 3i15, contain in. Divide each power of i by 4 and use the remainder to simplify in. 9 ÷ 4 = 2 remainder 1, so 9 = 2 • 4 + 1. i9= i2 • 4 • i1= i 15 ÷ 4 = 3 remainder 3, so 15 = 3 • 4 + 3. i15= i3 • 4 • i3= –i 4.3.2: Adding and Subtracting Complex Numbers

  7. Guided Practice: Example 2, continued Replace each occurrence of inin the expressions with the simplified versions, and replace the original expressions in the difference with the simplified expressions. 6i9= 6 • (i) = 6i 3i15= 3 • (–i) = –3i (5 + 6i9) – (5 + 3i15) = (5 + 6i) – [5 + (–3i)] 4.3.2: Adding and Subtracting Complex Numbers

  8. Guided Practice: Example 2, continued Distribute the difference through both parts of the complex number. 4.3.2: Adding and Subtracting Complex Numbers

  9. Guided Practice: Example 2, continued Find the sum or difference of the real parts. 5 – 5 = 0 4.3.2: Adding and Subtracting Complex Numbers

  10. Guided Practice: Example 2, continued Find the sum or difference of the imaginary parts. 6i + 3i = 9i 4.3.2: Adding and Subtracting Complex Numbers

  11. Guided Practice: Example 2, continued Find the sum of the real and imaginary parts. 0 + 9i = 9i 4.3.2: Adding and Subtracting Complex Numbers

  12. Guided Practice: Example 2, continued Use the form of the sum to determine if it is wholly real or wholly imaginary, or if it has both a real and an imaginary part. 9i has only an imaginary part, 9i, so the difference is wholly imaginary. ✔ 4.3.2: Adding and Subtracting Complex Numbers

  13. Guided Practice: Example 2, continued 4.3.2: Adding and Subtracting Complex Numbers

  14. Guided Practice Example 3 Is (12 – I20) + (–18 – 4i18) wholly real or wholly imaginary, or does it have both a real and an imaginary part? 4.3.2: Adding and Subtracting Complex Numbers

  15. Guided Practice: Example 3, continued Simplify any expressions containing in. Two expressions, i20and 4i18, contain in. Divide each power of i by 4 and use the remainder to simplify in. 20 ÷ 4 = 5 remainder 0, so 20 = 5 • 4 + 0. i20= i5 • 4= 1 18 ÷ 4 = 4 remainder 2, so 18 = 4 • 4 + 2. i18= i4 • 4 • i2= –1 4.3.2: Adding and Subtracting Complex Numbers

  16. Guided Practice: Example 3, continued Replace each occurrence of inin the expressions with the simplified versions, and replace the original expressions in the difference with the simplified expressions. i20= 1 4i18= 4 • (–1) = –4 (12 – i20) + (–18 – 4i18) = (12 – 1) + [–18 – (–4)] 4.3.2: Adding and Subtracting Complex Numbers

  17. Guided Practice: Example 3, continued Find the sum or difference of the real parts. 4.3.2: Adding and Subtracting Complex Numbers

  18. Guided Practice: Example 3, continued Find the sum or difference of the imaginary parts. The expression contains no multiples of i, so there are no imaginary parts, and the multiple of iis 0: 0i. 4.3.2: Adding and Subtracting Complex Numbers

  19. Guided Practice: Example 3, continued Find the sum of the real and imaginary parts. –3 + 0i = –3 4.3.2: Adding and Subtracting Complex Numbers

  20. Guided Practice: Example 3, continued Use the form of the sum to determine if it is wholly real or wholly imaginary, or if it has both a real and an imaginary part. –3 has only a real part, –3, so the sum is wholly real. ✔ 4.3.2: Adding and Subtracting Complex Numbers

  21. Guided Practice: Example 3, continued 4.3.2: Adding and Subtracting Complex Numbers

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