Section 7.7 Complex Numbers.

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Presentation transcript:

Section 7.7 Complex Numbers

Objectives Basic Concepts Addition, Subtraction, and Multiplication Powers of i Complex Conjugates and Division

PROPERTIES OF THE IMAGINARY UNIT i THE EXPRESSION

Example Write each square root using the imaginary i. a. b. c. Solution a. b. c.

SUM OR DIFFERENCE OF COMPLEX NUMBERS Let a + bi and c + di be two complex numbers. Then (a + bi) + (c+ di) = (a + c) + (b + d)i Sum and (a + bi) − (c+ di) = (a − c) + (b − d)i. Difference

Example Write each sum or difference in standard form. a. (−8 + 2i) + (5 + 6i) b. 9i – (3 – 2i) Solution a. (−8 + 2i) + (5 + 6i) = (−8 + 5) + (2 + 6)I = −3 + 8i b. 9i – (3 – 2i) = 9i – 3 + 2i = – 3 + (9 + 2)I = – 3 + 11i

Example Write each product in standard form. a. (6 − 3i)(2 + 2i) b. (6 + 7i)(6 – 7i) Solution a. (6 − 3i)(2 + 2i) = (6)(2) + (6)(2i) – (2)(3i) – (3i)(2i) = 12 + 12i – 6i – 6i2 = 12 + 12i – 6i – 6(−1) = 18 + 6i

Example (cont) Write each product in standard form. a. (6 − 3i)(2 + 2i) b. (6 + 7i)(6 – 7i) Solution b. (6 + 7i)(6 – 7i) = (6)(6) − (6)(7i) + (6)(7i) − (7i)(7i) = 36 − 42i + 42i − 49i2 = 36 − 49i2 = 36 − 49(−1) = 85

POWERS OF i The value of in can be found by dividing n (a positive integer) by 4. If the remainder is r, then in = ir. Note that i0 = 1, i1 = i, i2 = −1, and i3 = −i.

Example Evaluate each expression. a. i25 b. i7 c. i44 Solution a. When 25 is divided by 4, the result is 6 with the remainder of 1. Thus i25 = i1 = i. b. When 7 is divided by 4, the result is 1 with the remainder of 3. Thus i7 = i3 = −i. c. When 44 is divided by 4, the result is 11 with the remainder of 0. Thus i44 = i0 = 1.

Example Write each quotient in standard form. a. b. Solution a.

Example (cont) b.