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Imaginary and Complex Numbers 0 The imaginary number i is defined as 1 0 i2 = -1 0 An expression in the form a + bi is called a complex number and has: 0 Real part “a” 0 Imaginary part “bi” 0 When adding/subtracting complex numbers, you can only combine the like parts – real with real and imaginary with imaginary 0 When multiplying binomials containing imaginary parts, using FOIL 0 A fraction containing an imaginary part in the denominator must be rationalized Imaginary and Complex Numbers 0 Simplify 4 0 We can rewrite this as 1 4 0 This will simplify to 2i 0 Simplify 18 0 We can rewrite this as 1 9 2 0 This simplifies as 3i 2 0 Simplify (3 + 2i) + (5 – 7i) 0 Combine like terms: 8 – 5i Imaginary and Complex Numbers 0 Simplify (2 + 3i)(3 – 2i) 0 6 – 4i + 9i – 6i2 0 6 + 5i – 6i2 Remember that i2 = -1 0 6 + 5i – 6(-1) 0 6 + 5i + 6 0 12 + 5i Imaginary and Complex Numbers 0 Simplify 3 2i 0 3 2i i 5i 5i i 0 3i 2i2 Multiply by a form of 1 to get rid of the imaginary number in the denominator 5i2 0 3i 2 1 5 1 2 3i 0 2 3i or 5 5 Imaginary and Complex Numbers 0 Simplify 5 2i 3 4i 0 Because the denominator is a binomial, we cannot 0 0 0 0 simply multiply by i/i to rationalize the denominator We need to multiply by the conjugate of the denominator The conjugate is the binomial that will create a difference of squares when multiplied by the denominator To find the conjugate, simply change the sign of the second term of the denominator For this problem, the conjugate would be 3 – 4i Imaginary and Complex Numbers 0 Simplify 5 2i Multiply by a form of 1 using the conjugate to get rid of the imaginary number in the denominator 0 5 2i 3 34i 4i 3 4i 3 4i 0 15 20i 6i 8i2 Because the conjugate creates a difference of 9 12i 12i 16i2 squares, the middle terms cancel out 0 15 14i 8 1 9 16 1 15 14i 8 0 9 16 0 23 14i 25 Solving Equations 0 Solve: -3x2 = 75 0 x2 = -25 0 x2 25 0 x = ±5i 0 Solve: 2x2 – 4x + 9 = 0 0 b b2 4ac x 2a 0 x 0 4 x 4 4 2 9 22 4 16 72 4 x 4 56 4 x 4 1 4 14 4 2 4 2i 14 x 4 x 2 i 14 2 Quadratics and Complex Numbers 0 Determine if 2x2 – 3x + 5 = 0 has real or imaginary roots. 0 We can use the discriminate (b2 – 4ac) from the quadratic formula to determine if the roots are real or imaginary without having to actually solve the equation. 0 If the discriminate is POSITIVE, the roots will be REAL numbers 0 If the discriminate is NEGATIVE, the roots will be IMAGINARY/COMPLEX numbers Quadratics and Complex Numbers 0 Determine if 2x2 – 3x + 5 = 0 has real or imaginary roots. 0 The discriminate is b2 – 4ac 0 (-3)2 – 4(2)(5) 0 9 – 40 0 -31 0 As the discriminate is negative, this quadratic will have IMAGINARY/COMPLEX roots Quadratic Functions 0 Look at the function below and discuss its zeros. y x We know that all quadratic functions must have two zeros, and that the zeros are the points where y = 0. Generally, this is where the graph crosses the x-axis. The given graph does not cross the x-axis at all, but must still have two zeros. This means that the zeros of the given graph must be imaginary. **NOTE: IMAGINARY ZEROS AND ROOTS MUST ALWAYS OCCUR IN PAIRS.**