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What are imaginary and complex numbers? Graph it Do Now: Solve for x: x2 + 1 = 0 2 x 1 2 x 1 ? What number when multiplied by itself gives us a negative one? No such real number parabola does not intersect x-axis NO REAL ROOTS Imaginary Numbers 1 is not a real number, then 1 is a non-real or If imaginary number. Definition: A pure imaginary number is any number that can be expressed in the form bi, where b is a real number such that b ≠ 0, and i is the imaginary unit. 1 ab a b i 1 1) 5 25 15i 5 1 5i b = 5 25 25 25 (25 25 1 1 7 7 77 1 1 7i 7i ii 77 b 7 In general, for any real number b, where b > 0: 2 b b 2 1 bi 1 i Powers of i 1 If i2 2 i2 = –1 1 2 i2 = –1 i3 = – 1, then = ? i3 = i2 • i = –1( 1 ) = –i i4 = i2 • i2 = (–1)(–1) = 1 i5 = i4 • i = 1( 1) = i i6 = i4 • i2 = (1)(–1) = –1 i7 = i6 • i = -1( 1 ) = –i i8 = i6 • i2 = (–1)(–1) = 1 What is i82 in simplest form? 82 ÷ 4 = 20 remainder 2 i82 equivalent to i2 = –1 i0 = 1 i1 = i i2 = –1 i3 = –i i4 = 1 i5 = i i6 = –1 i7 = –i i8 = 1 i9 = i i10 = –1 i11 = –i i12 = 1 1 i Properties of i 16 9 Addition: 16 1 9 1 4i + 3i = 7i Subtraction: 25 16 25 1 16 1 5i – 4i = i Multiplication: 36 4 36 1 4 1 (6i)(2i) = 12i2 = –12 note : Division: 36 4 144 16 4 16 4 16 1 4i 2 4 1 2i Complex Numbers Definition: A complex number is any number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit. a + bi real numbers pure imaginary number Any number can be expressed as a complex number: 7 + 0i = 7 a + bi 0 + 2i = 2i The Number System 5 76 -i Complex Numbers Real Numbers i i3 -i Irrational Numbers i9 i 2 + 3i 47 Rational Numbers i Integers i i75 Whole Numbers Counting Numbers 1/2 – 12i -6 – 3i i -i47 i Graphing Complex Numbers (x, y) Complex Number Plane a + bi pure imaginaries 5i (4 + 5i) 4i 3i (0 + 3i) 2i i -5 -4 -3 -2 -1 (–5 – 2i) -i (0 + 0i) 0 1 2 -2i -5i -6i 3 4 5 (3 – 2i) -3i -4i reals (0 – 4i) 6 Vectors Vector - a directed line segment that represents directed force notation: OP pure imaginaries 5i 3 4i 32 42 P (3 + 4i) 4i 3i 25 5 2i OP i reals O -4 -3 -2 -1 0 1 2 3 4 5 6 The-5length of vectors -i is found by using the Pythagorean Theorem & is always positive. -2i -3i The length of a vector representing a complex number is the absolute value of -4i the complex number a bi represented-5iby the equation -6i 2 2 2 2 a bi a b a b Model Problems Express in terms of i and simplify: 100 = 10i 16 = 4/5i 1 300 5i 3 2 25 Write each given power of i in simplest terms: i49 = i i54 = -1 i300 = 1 i2001 = i Add: 4 18 50 4i 9 2 i 25 2 12i 2 5i 2 17i 2 Multiply: 4 5 80 4i 5 4i 5 16i 2 5 2 4i 5 i 16 5 16 5 80 Simplify: 72 32 3 8 i 36 2 i 16 2 3i 4 2 4i 2 6i 2 4i 2 6i 2 Model Problems Which number is included in the shaded region? 1) 2) 3) 4) (-1.5 + 3.5i) (1.5 – 3.5i) (3.5) (1) (4.5i) 5i 4i yi (4) 3i 2i i -5 -4 -3 -2 -1 -i (3) 0 1 2 -2i -3i -4i -5i -6i (2) 3 4 x 5 6 How do we add and subtract complex numbers? Do Now: Simplify: 3 45 125 2 20 3 9 5 25 5 2 4 5 9 5 5 54 5 10 5 Adding Complex Numbers (2 + 3i) + (5 + i) = (2 + 5) + (3i + i) = 7 + 4i In general, addition of complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i Combine the real parts and the imaginary parts separately. Find the sum of convert to complex numbers combine reals and imaginary parts separately (5 36) and (3 16) (5 36) (3 16) (5 i 36) (3 i 16) (5 6i) (3 4i) (5 3) (6i 4i) 8 2i Subtracting Complex Numbers What is the additive inverse of 2 + 3i? -(2 + 3i) or -2 – 3i Subtraction is the addition of an additive inverse (1 + 3i) – (3 + 2i) = (1 + 3i) + (-3 – 2i) = -2 + i In general, subtraction of complex numbers: (a + bi) – (c + di) = (a – c) + (b – d)i Subtract 6 2i 3 from 5 3i 3 5 3i 3 6 2i 3 change to addition problem 5 3i 3 6 2i 3 combine reals and 5 (6) (3i 3) 2i 3 imaginary parts separately 1 i 3 Adding Complex Numbers Graphically (2 + 3i) + (3 + 0i) = (2 + 3) + (3i + 0i) = = 5 + 3i yi 5i 4i vector: 2 + 3i (2 + 3i) 3i vector: 3 + 0i i vector: 5 + 3i -5 (5 + 3i) 2i -4 -3 -2 -1 -i -2i -3i -4i -5i -6i (3 + 0i) 0 1 2 3 4 5 x 6 Adding Vectors Vector - a directed line segment that represents directed force notation: OS S P resultant force O R The vectors that represent the applied forces form two adjacent sides of a parallelogram, and the vector that represents the resultant force is the diagonal of this parallelogram. Subtracting Complex Numbers Graphically (1 + 3i) – (3 + 2i) = (1 + 3i) + (-3 – 2i) = yi -2 + i 5i 4i (1 + 3i) 3i 2i (-2 + i) (3 + 2i) i x -5 -4 -3 -2 -1 (-3 – 2i) -i 0 1 2 3 4 5 -2i -3i The vector representing-4ithe additive inverse is the image of the vector reflected through the -5i origin. Or the image under a rotation about -6i the origin of 1800. 6 Model Problems Add/Subtract and simplify: (10 + 3i) + (5 + 8i) = 15 + 11i (4 – 2i) + (-3 + 2i) =1 2 3 i 1 4 6 1 i 4 i 1 2i 2 6 4 6 4 5 3i 6 4 80 3 20 4 6i 5 Express the difference of 1 in form a + bi 3 4 5 9 2 i 80 2 162