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Section 1.5 Complex Numbers Why? (Part I – 4:36) Why? (Part II – 10:16) The Imaginary Unit (9:28) Imaginary Number Recap i 1 i2 1 1 1 i3 i2 i i i4 i2 i2 1 i42 i40i2 1 1 1 i37 i36 i 1 i i The Algebra of Imaginary Numbers 98 1 49 2 7i 2 27 1 9 3 3i 3 2i 3i 5i 7i 8i i 2i 3i 6i2 6 1 6 3i 3 4i 4 5 5 5 i 5i i 2 7 7i 7i i 7i 7 2i 14i The Complex Number a bi Real Part Imaginary Part The Algebra of Complex Numbers 2 3i 7 4i 9 i 2 3 i 6 2i 7 6i 2 3i 5 9i 2i 4 5i 8i 10i2 10 8i 7 i i 7 4 i 3 i 6 i 6 i 4 i 12 4i 3i i2 13 i 36 i2 37 16 8i i2 15 8i 2 The Conjugate 2 3 Conjugate 2 3 72 5 Conjugate 7 2 5 7 2i Conjugate 7 2i 9 3i Conjugate 9 3i 4 5i Conjugate 4 5i 72 7 Conjugate 7 2 7 3 i 3 3 i 4i 4i i 4 2 2 3 i 6 2i 3 1 i 3 i 3 i 3 i 9 1 5 5 6 i 6 i 2 i 12 8i i2 11 8 i 2i 2i 2i 4 1 5 5 3 2i 3 2i 4 5i 12 23i 10i2 2 23 i 4 5i 4 5i 4 5i 41 41 41 5 i 5 i 6 i 30 11i i2 29 11 i 6 i 6i 6i 36 1 37 37 Now try these: 7 6i 4 3i 3 2i 2 8i 3 3i 6 20i 16i2 1 i1 i 3 4i 2 3 4i 3 4i 3 4i 6 8i 25 6 8 i 25 25 3 2i 5 2i 22 20i 3 2i 5 2i 5 2i 5 2i 2 3i 5 i 5 i 15 16i 4i2 29 11 16 i 29 29 2 3i 26 1 3 i 13 26 The Discriminant b 4ac 2 b2 4ac 0 two unique real solutions b2 4ac 0 one real solution...double root b2 4ac 0 two complex answers If b2 4ac is a perfect square, can factor. 2x 2 5x 6 0 5 2 4 2 6 23 x 2 6x 10 0 6 2 6 4 2 5 23 2 2 5 i 23 x 4 4 110 4 x 6 2i 3i 2 Now try these…..(do discriminant first) 3x 2 2x 1 0 2x 2 5x 4 0 2 4 3 1 16 5 2 3x 1 x 1 0 1 x , x 1 3 2x 2 6x 3 0 6 2 4 2 3 12 62 3 3 3 x x 4 2 x 2 4 2 4 7 5 i 7 5 i 7 x 4 4 4 2x 2 5x 4 0 5 4 2 4 57 2 x 5 57 4