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Complex Numbers and Equation Solving 1. Simple Equations 2. Compound Equations 3. Systems of Equations 4. Quadratic Equations 5. Determining Quadratic Equations Complex Numbers and Equation Solving In this section we will consider the following: a) simple equations b) compound equations c) systems of equations d) quadratic equations Simple Equations: Goal: isolate the variable on either the left or right side of the equation and to solve for one positive variable. Example 1: This example reviews the basic rules for solving a simple equation 3x 4 5(x 2) 3 3x 4 5x 10 3 1. Remove brackets or parentheses that exist within the equation using the distributive property. Reminder: make sure the value outside the ( ) or [ ] is distributed to all terms within, with a 3x 5x 4 10 3 special focus on any sign changes. 2. Isolate the terms containing the variable remembering that if 2x 17 terms are moved across the equal sign, the sign of the term 1 1 must change. x 17 3. Simplify both sides of the equation. 2 2 17 17 4. Solve for “1” positive variable by multiplying both sides by the reciprocal of the coefficient or in simple terms divide both x , SS 2 2 sides by the coefficient of the variable Example 2: Real coefficient assigned to the variable with real and imaginary numbers 3x 4 5i 8x 7 2i 3x 8x 4 5i 7 2i 11x 3 3i 3 3i 3 3i x , SS 11 11 1. Isolate the terms containing the variable “x” on the left side of the equation and move all real and imaginary numbers to the right side. 2. Simplify both sides of the equation. 3. Solve for one positive “x” Example 3: Imaginary coefficients assigned to the variable with imaginary numbers 4ix 2i 3(3ix 5i) 4ix 2i 9ix 15i 4ix 9ix 2i 15i 5ix 17i x 17 17 , SS 5 5 1. Remove any parentheses. 2. Isolate the terms containing the variable “x” on the left side of the equation and move all imaginary numbers to the right side. 3. Simplify both sides of the equation. 4. Solve for one positive “x”. Dividing both sides of the equation by “-5i” results in a situation where an imaginary number is contained in both numerator and denominator and as a result the value “i” cancels. Example 4: Imaginary coefficients assigned to the variable with real and imaginary numbers 4ix 3 2(ix 5i) 7 4i 4ix 3 2ix 10i 7 4i 1. Remove any parentheses. 2. Isolate terms containing the variable “x” on the left side and the real and imaginary numbers on the 4ix 2ix 3 10i 7 4i right side of the equation. 3. Simplify both sides of the equation 2ix 4 14i 4. When solving for one positive “x” you must divide 4 14i i 4i 14i 2 by an “i” and following the rule of no imaginary x 2 2i i 2i number in the denominator, both numerator and denominator must be multiplied by an “i”. 4i 14 x 2i 7 SS 2i 7 5. Simplify the resulting fraction 2 Example 5: Real and imaginary coefficients assigned to the variable with real numbers 1. Isolate and simplify as in previous examples 2. Remove the common factor “x” from the left side of the equation resulting in a complex number factor (7 - 2i). 7x 2ix 11 3. When solving for one positive “x” you must divide by the complex number factor and following the rule of no x(7 2i) 11 imaginary number in the denominator, the denominator 11 (7 2i) 77 22i and numerator on the right side must be multiplied by x 2 (7 2i) (7 2i) 49 4i the conjugate of the complex number factor 77 22i 77 22i 77 22i (7 + 2i). x SS 49 4(1) 53 53 4. Simplify the resulting fraction 5x 7 2ix 4ix 4 2x 5x 2ix 4ix 2x 7 4 Example 6: Real and imaginary coefficients assigned to the variable with imaginary numbers 4ix 5i 2x 3(2ix 3i) 4x 4ix 5i 2x 6ix 9i 4x 4ix 2x 6ix 4x 5i 9i 2ix 6x 14i x(2i 6) 14i 14i (2i 6) 28i 2 84i x 2 (2i 6) (2i 6) 4i 36 28(1) 84i 28 84i x 4(1) 36 40 7 21i SS 10 1. Remove any parenthesis. 2. Isolate the variables on the left side and the imaginary numbers on the right side of the equation. 3. After simplifying the equation, remove the common factor “x” from the left side of the equation resulting in a complex number factor (-2i - 6). 4. When solving for one positive “x” you must divide by the complex number factor and following the rule of no imaginary number in the denominator, the denominator and numerator on the right side must be multiplied by (-2i + 6). 5. Simplify the resulting fraction. Some basic items to take into consideration: a) the distributive property 2 b) the value of i 1 c) when reducing a fraction every term is affected by a common factor or a negative sign. In this example “-4 ” Example 7: Real and imaginary coefficients assigned to the variable with real and imaginary numbers 3xi 6 5x 2i 4(2ix 3) 2x 7i 1. Remove any parenthesis. 2. Isolate the variables on the left 3xi 6 5x 2i 8ix 12 2x 7i side and the real and imaginary 3xi 5x 8ix 2x 6 2i 12 7i numbers on the right side of the equation. 5ix 3x 6 5i 3. After simplifying the equation, x(5i 3) 6 5i remove the common factor “x” (6 5i) (5i 3) 30i 18 25i 2 15i from the left side of the x equation resulting in a complex 2 (5i 3) (5i 3) 25i 9 number factor (-5i - 3). 4. When solving for one positive 30i 18 25(1) 15i 45i 43 x “x” you must divide by the 25(1) 9 34 complex number factor and 45i 43 following the rule of no imaginary number in the denominator, the SS denominator and numerator on the right side must be multiplied by 34 (-5i + 3). 5. Simplify the resulting fraction. Some basic items to consider: a) binomial expansion 2 b) the value of i 1 c) when reducing a fraction every term is affected by a common factor or a negative sign. In this example “a negative sign”. Compound Equations By definition - an equation containing two different variables - one with real coefficients assigned and the second having assigned imaginary coefficients. Goal - to create two equations by isolating the real equation from the imaginary and solving each equation for each particular variable following the procedures outlined for simple equations. (Check procedure by referring to identified example) Example 1: 3x 2i 6iy 5x 6 9iy Imaginary Equation Real Equation 3x 5x 6 3x 5x 6 2x 6 6 x 3 2 Check: The real equation formed by equating the real values from the left with the real values on the right side of the equal sign. (Check example 1) 2i 6iy 9iy 6iy 9iy 2i 3iy 2i 2i 2 y 3i 3 3(3) 2i 6i 2 3 5(3) 6 9i 2 3 The imaginary equation formed by equating the imaginary values from the left with the imaginary values on the right side of the equal sign (Check example 3) 9 2i 4i 15 6 6i 9 6i 9 6i Therefore, the complex number on the left side = the complex number on the right Solution Set 3, 2 3 Example 2 Remember to remove 3(3ix 5) 2(2y 4i) 7ix 6i 7y 3 parenthesis 9ix 15 4y 8i 7ix 6i 7y 3 Imaginary Equation Real Equation 9ix 8i 7ix 6i 9ix 7ix 8i 6i 2ix 2i 2i x 1 2i Check The imaginary equation formed by equating the imaginary values from the left with the imaginary values on the right side of the equal sign (Check example 3) 15 4y 7y 3 4y 7y 15 3 3y 18 18 y 6 3 The real equation formed by equating the real values from the left with the real values on the right side of the equal sign. (Check example 1) 3(3i[1] 5) 2(2[6] 4i) 7i[1] 6i 7[6] 3 9i 15 24 8i 7i 6i 42 3 i 39 i 39 Therefore, the complex number on the left side = the complex number on the right Solution Set {(-1, 6)} System of Equations By definition - an equation containing two different variables - each with real and imaginary coefficients assigned. Goal - a) to create two equations by isolating the real equation from the imaginary following procedure of real values equal real values and imaginary values equal imaginary values b) to simplify each equation using procedures of simple equations c) to solve the two equations as a system using the procedures of elimination or substitution. Review procedures by referring to Solving Systems of Equations 5ix 3y 7 2iy 5x 4i Example 1 Imaginary Equation Real Equation 3y 7 5x 5x 3y 7 After each equation has been formed, each is simplified and put into the form of Ax + By = C 6ix 2iy 4i 6ix 2iy 4i 6x 2y 4 Solving the elimination - removing the x and then solving for one positive ‘y” The system of equations 5x 3y 7 6(5x 3y 7) 30x 18y 42 Solving for “y” Substitution of 11 4 6x 2y 4 5(6x 2y 4) 30x 10y 20 6x - 2(11 4) = -4 8y = 22 for “y” and solving for “x” Check 4 6x - 4 2(11 4) = 4 -4 y = 22 8 11 4 24x 22 16 24x 6 x 6 24 1 4 5ix 3y 7 2iy 5x 4i 5i 1 4 3 11 4 7 2i 11 4 5 1 4 4i 4 5i 1 4 4 3 11 4 4 7 4 2i 11 4 4 5 1 4 4 4i 5i 33 28 22i 5 16i 5i 5 5i 5 Therefore, the complex number on the left side = the complex number on the right 1 , 11 Solution Set 4 4 The next example will use the process of substitution to solve the system of equations 3x 4iy 7 5xi 5y 6i Example 2 Imaginary Equation Real Equation 3x 7 5y 3x 5y 7 After each equation has been formed, each is simplified and put into the form of Ax + By = C 4yi 2xi 6i 5xi 4iy 6i 5x 4y 6 The system of equations 3x 5y 7 3x 5y 7 x 5y 7 3 Solving for “x” 5x 4y 6 Substitution of 5y 7 3 for “x” and solving for one positive “y” 5y 7 5 4y 6 3 5y 7 3 5 3 4y 3 6 3 25y 35 12y 18 13y 53 y 5313 3x - 5 5313 7 Substitution of 53 for “y” 13 and solving for one positive “x” 13 3x -13 5 5313 13 7 39x - 265 = -91 39x = 174 x = 174 39 58 13 Check 3 5813 4i 5313 7 5 58 13 i 5 5313 6i 13 3 5813 13 4i 5313 13 7 13 5 5813 i 13 5 5313 13 6i 174 212i 91 290i 265 78i 265 212i 212i 265 Therefore, the complex number on the left side = the complex number on the right Solution Set 58 , 53 13 13 Quadratic Equations The previous examples contained a visual “i” and it was easy to identify each equation as one that contained imaginary or complex numbers. The quadratic equation with real coefficients hides any reference to the existence of imaginary or complex numbers until one is asked to solve (determine the roots, find the critical zeros, calculate the x-intercepts) the equation. The procedure used most often is the application of the quadratic formula: b b 2 4ac x 2a and in particular, the discriminant b 2 4ac. If the value of the discriminant is less than zero (a negative value) the nature of the roots can be interpreted as two unequal and imaginary roots. A further interpretation is that a graph of this function would not have any xintercepts or critical zeros. One other note is that imaginary roots always occur in pairs and are conjugates on one another. Example 1: Determine the roots of the quadratic equation x 2 3x 11 0 b b 4ac x ; a 1, b 3, c 11 2a Note: 1. Take time when writing out and substituting values into the formula paying particular attention to negative signs. 2 2. Use the concept of i 1 to remove the negative values from under the radical sign 3. Remove all perfect from under the radical. 2 3 (3)2 4(1)(11) x 2(1) 3 9 44 3 35 3 i 35 x 2 2 2 Once the roots have been determined one task remains and that is to check whether the calculated roots are the correct values. Two distinct procedures can be used. a) The first procedure requires the substitution of a root into the original equation, simplification and a result of 0 = 0. b) The second procedure utilizes two parts i) the formulas -b/a and c/a which equals the sum and the product of the roots respectively. ii) the addition of and multiplication of the calculated roots iii) comparison of the results from i) and ii) Procedure One To help simplify the equation, remove 2 3 i 35 3 i 35 the fractions be multiplying each term 2 x 3x 11 0 3 11 0 by a common denominator.Remember 2 2 to use the distributive property on the 2 9 3i 35 3i 35 i 35 9 3i 35 second fraction. 11 0 4 2 9 6i 35 35 9 3i 35 6i 35 26 9 3i 35 11 0 4 4 4 11 4 0 4 2 4 2 Note: 6i 35 26 18 6i 35 44 0 0 0 1. Adding conjugates results in the radical term being Procedure Two eliminated. b 3 c 11 2 2. When multiplying conjugates x 3x 11 0 s1 3, p1 11 a 1 a 1 the middle term is always 3 i 35 3 i 35 6 eliminated. s2 3 3. A radical times itself always 2 2 2 results in an answer of the 2 3 i 35 3 i 35 9 i 35 9 35 44 radicand. p2 11 2 2 4 4 4 35 35 35 s1 s2, p1 p2 3 i 35 3 i 35 Solution Set: , 2 2 35 35 35 i 35 i 35 35i 2 35 i 35 i 35 35i 2 35 Example #2: Determine the roots of the quadratic equation b b 4ac x ; a 3, b 5, c 7 2a - 3x 2 5x 7 0 2 5 (5)2 4(3)(7) x 2(3) x Note: Always pay attention to the substitution of negative values for a, b and c 5 25 84 5 59 5 i 59 6 6 6 Check: 2 5 i 59 5 i 59 3x 2 5x 7 0 3 5 7 0 6 6 25 5i 59 5i 59 i 2 59 5 i 59 25 10i 59 59 5 i 59 3 5 7 0 3 5 7 0 4 6 36 2 30i 59 102 25 5i 59 75 30i 59 177 25 5i 59 7 0 36 36 36 7 36 0 36 6 36 6 30i 59 102 150 30i 59 252 0 0 0 Be careful with sign changes when simplifying this part of the equation Check 2: b 5 5 c 7 7 2 3x 5x 7 0 s1 , p1 a 3 3 a 3 3 5 i 59 5 i 59 10 5 s2 6 6 6 3 s1 s2, p1 p2 5 i 59 5 i 59 25 i 2 59 25 59 84 7 p2 6 6 36 36 36 3 Solution Set 5 i 59 5 i 59 , 6 6 Determining a Quadratic Equation The ability to calculate the roots of a quadratic equation also allows for the mathematical procedure of determining a quadratic equation given the roots of an equation. Two procedures can be used. The include: a) writing the equation as a product of factors and then simplifying the equation b) using the formula x 2 (r1 r2 )x r1 r2 0 where r1 and r2 represent the given roots The following examples will demonstrate the two procedures. Evaluate both procedures and decide which one is best suits your mathematical skills. Example 1: non fractional roots Determine the quadratic equation that has a solution set of 3 i 5,3 i 5 Procedure 1 - writing as factors x - 3 i 5x - 3 i 5 0 x x3 i 5 x3 i 5 3 i 5 3 i 5 0 2 x 3x ix 5 3x ix 5 9 i 5 0 x 6x 9 5 0 2 2 2 x 6x 14 0 2 Procedure 2 - using the formula x 2 (r1 r2 )x r1 r2 0 x 2 3 i 5 3 i 5 x 3 i 5 3 i 5 0 x 2 6x 9 i 2 5 0 x 2 6x 9 5 0 x 2 6x 14 0 When the roots are not fractions the amount of work required does not appear to be that much different but additional work is still required when using binomial expansion in the first procedure as identified by the red boxes. Example 2: fractional roots Determine the quadratic equation that has a solution set of 2 i 7 2 i 7 , 5 5 Procedure 1 - writing as factors 2 i 7 2 i 7 2 i 7 2 i 7 2 i 7 2 i 7 2 x x 0 x x x 0 5 5 5 5 5 5 2x ix 7 2x ix 7 4 i 7 2x ix 7 2x ix 7 4 7 x 0 x2 0 5 5 25 5 5 25 4x 11 4x 11 x 2 0 25 x 2 25 25 25 0 25x 2 20x 11 0 5 25 5 25 Procedure 2 - using the formula 2 2 2 i 7 2 i 7 2 i 7 2 i 7 x 2 (r1 r2 )x r1 r2 0 x 2 x 0 5 5 5 5 4 4 i 2 7 4 11 2 x x 0 x x 0 5 25 5 25 2 4 11 25 x 2 25 x 25 25 0 25x 2 20x 11 0 5 25 This example demonstrates that the first procedure requires a lot more work especially when the binomial expansion involves fractions as shown by the red boxes. The last part of both procedures requires the same amount of calculation Example 3: the roots written as a combined expression Determine the quadratic equation that has a solution set of Procedure 1 - writing as factors 4 i 11 6 To determine the equation it is necessary to identify the given 2 roots 4 i 11 4 i 11 , 6 6 4 i 11 4 i 11 4 i 11 4 i 11 4 i 11 4 i 11 2 x x 0 x x x 0 6 6 6 6 6 6 4x ix 11 4x ix 11 16 i 121 4x ix 11 4x ix 11 16 121 x 0 x2 0 6 6 36 6 6 36 8x 137 8x 137 x2 0 36 x 2 36 36 36 0 36x 2 48x 137 0 6 36 6 36 Procedure 2 - using the formula 2 2 4 i 11 4 i 7 4 i 11 4 i 11 x 2 (r1 r2 )x r1 r2 0 x 2 x 0 6 6 6 6 8 16 i 2 121 8 137 2 x x 0 x x 0 6 36 6 36 2 8 137 36 x 2 36 x 36 36 0 36x 2 48x 137 0 6 36 If not indicated, the choice of which procedure (factors or formula) is left up to you. Make your choice based on mathematical skill and chance for best success in determining the equation A special application of imaginary numbers A quick focus on factoring: x 2 4 (x 2)(x 2) 9x 2 25 (3x 5)(3x 5) Each of these examples represent a difference of squares and the factors are a set conjugates formed by taking the square root of each of the values in the question. 2 2 2 Until now questions of the form x 4, x 25, 9x 49 have not been factorable. The existence of imaginary numbers permits the factoring of an equation because of this which in term allows basic fact 2 2 16 16i 4i 4i positive perfect squares to 2i 2i 4i 4 2 be re-written as a product 2 25 25i 5i 5i 8i 8i 64i 64 as indicated in these 2 144 144i 2 11i 11i 11i 11i 121i 121 examples 2 2 To factor questions of the form x y Step 1: re-write the question so that appears as a difference of squares (this is accomplished by using the above fact) Include binomial multiplkicatiom