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Where Are the Quadratic's Complex Roots?

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Where Are the Quadratic's Complex Roots? Volume 8, Number 1, 2015 WHERE ARE THE QUADRATIC’S COMPLEX ROOTS ? Ágnes Orsolya Páll-Szabó Abstract: A picture worth more than a thousand words – in mathematics too. Many students fail in learning mathematics because, in some cases, teachers do not offer the necessary visualization. Nowadays technology overcomes this problem: computer aided instruction is one of the most efficients methods in teaching mathematics. In this article we try to discuss some methods of visualizing complex roots, with examples. The first method presented is the auxiliary polynomial method, followed by the the tangent line method. One of the most interesting but less known method is Lill’s circle. Another method discussed in the article is Domain Coloring via the Color Wheel. Key words: quadratic equations; complex roots; visualizing roots; Mathematics Education 1.Introduction According ro Arcavi`s definition “visualization is the ability, the process and the product of creation, interpretation, use of and reflection upon pictures, images, diagrams, in our minds, on paper or with technological tools, with the purpose of depicting and communicating information, thinking about and developing previously unknown ideas and advancing understandings”.[1] Visualization is very important for every teacher, especially for mathematics one, because they must often present abstract concepts and symbols which are difficult to understand for students. Unfortunately many of the students fail in learning mathematics because teachers do not pay attention to the basic concepts and use formulas without explaining where they came from. In order to overcome these difficulties, teachers should use different methods. Nowadays technology is able to provide us the help that we need: computer aided instruction is one of the most efficients methods in teaching mathematics, very effective when it comes to visualize abstract concepts. In this article we try to discuss some methods of visualizing complex roots, with examples created by us, using Matlab. When teaching complex roots, teachers do not deal with identifying complex roots graphically and usually use the quadratic formula to solve a quadratic equation. The question is: where are the complex roots located and how can they be visualised? 2. Preliminaries The roots of the general quadratic equation 푦 = 푎푥2 + 푏푥 + 푐 (푎, 푏, 푐 ∈ ℝ) are known to occur in the following sets: real and distinct (the discriminant is greater than zero); real and coincident (the discriminant is equal to zero); complex conjugate pair(the discriminant is less than zero. In this case, a simple x–y plot of the quadratic equation does not reveal the location of the complex conjugate roots.) 3. The auxiliary polynomial method , Example: We have a quadratic function with complex roots 푓(푥) = 푥2 − 2푥 + 5 or 푓(푥) = (푥 − 1)2 + 4. The minimum point (vertex) of the graph is at 푉(1,4), see Figure 1, so it doesn’t cross the x-axis. Because it lies entirely above the x-axis, we know it has no real roots. The discriminant is: Received April 2014. 38 Ágnes Orsolya Páll-Szabó ∆= 푏2 − 4푎푐 = 4 − 20 = −16 and the roots are: −푏 ± √∆ 푥 = = 1 ± 2푖 1,2 2푎 8 6 V 4 Y 2 0 -2 -2 0 2 4 6 8 X Figure 1. 풇(풙) = 풙ퟐ − ퟐ풙 + 5 and vertex at 푽(ퟏ, ퟒ) We reflect the quadratic horizontally over the vertex, in Figure 2, we do this by transforming 풇(푥) = (푥 − 1)2 + 4 into 풇(푥) = −(푥 − 1)2 + 4 , and find the x-intercepts of this new graph 푅(−1,0) and 푃(3,0). 8 6 V 4 Y 2 R P 0 -2 -2 0 2 4 6 8 X Figure 2. f(x) = (x − 1)2 + 4 (blue) and f(x) = −(x − 1)2 + 4 (green) If we rotate the 푅푃 segment about his midpoint, then the 푅 and 푃 points will describe a circle, with centre 퐴(1,0) and radius of 2, see Figure 3, 4. In order to graphically see the complex roots of the original function, we need to rotate the reflected image 90 degrees to place the quadratic into the Argand plane. Acta Didactica Napocensia, ISSN 2065-1430 Where are the quadratic’s complex roots ? 39 8 8 6 6 V 4 V 4 Y Y 2 B 2 R A P A 0 0 -2 -2 C -2 0 2 4 6 8 -2 0 2 4 6 8 X X Figure 3. Figure 4. The points 퐵(1,2푖) and 퐶(1, −2푖) on the diagram are a representation of the complex roots. The complex roots of the initial equation are therefore given by 푥 = 1 ± 2푖. [2], [13]. 4. General case (auxiliary polynomial method) Let 푦 = 푎푥2 + 푏푥 + 푐 represent any parabola.First we complete the square. 푏 푏2 푏2 푦 = 푎 (푥2 + 푥 + ) + 푐 − 푎 4푎2 4푎 푏 2 푏2 푦 = 푎 (푥 + ) + 푐 − 2푎 4푎 푏 2 푏2 Its vertical reflection over its vertex is = −푎 (푥 + ) + 푐 − , which expands to 푦 = 2푎 4푎 2 2 푏 −푎푥 − 푏푥 + 푐 − .Its complex roots will be at 4푎 2 √ 2 푏 푏 푏 − 4(−푎)(푐 − 2푎) 푥 = ± −2푎 −2푎 which is 푏 √4푎푐−푏2 푥 = ± . −2푎 2푎 Since (– 푏2 + 4푎푐) is negative, these complex roots of the flipped quadratic can be written as: 푏 푖√푏2 − 4푎푐 푥 = − ± , (ퟏ) 2푎 2푎 and real roots of the quadratic we started with 푏 √푏2 − 4푎푐 푥 = − ± . (ퟐ) 2푎 2푎 Volume 8 Number 1, 2015 40 Ágnes Orsolya Páll-Szabó They are the same except for the 푖 in the complex roots.So we can think of (ퟐ) as √푏2−4푎푐 푏 √푏2−4푎푐 representing a circle of radius , centred at (− , 0). Therefore multiplying 2푎 2푎 2푎 √푏2−4푎푐 by 푖 has the effect of rotating the point ( , 0 ) 90 degrees anti-clockwise around the 2푎 푏 point (− , 0). [11] 2푎 5. The tangent line method 푏 Let (푥) = 푎푥2 + 푏푥 + 푐 , then its derivative is 푓′(푥) = 2푎푥 + 푏. Let 훼 = − , 훽 = 2푎 √4푎푐−푏2 and we consider two points from the graph 2푎 푆2(훼 + 훽, 2훾), 푆1(훼 − 훽, 2훾). 푏 √4푎푐−푏2 √4푎푐−푏2 푓′(훼 + 훽) = 2푎(훼 + 훽) + 푏 = 2푎 (− + ) + 푏 = 2푎 = 2푎훽 ,this mean that the 2푎 2푎 2푎 slope of the tangent line to the parabola at 푆2 is 2푎훽. The tangent line in 푆2 to the parabola which contains the point (훼, 0) is: 푦2 = 2푎훽푥 − 2푎훼훽, and in 푆1 : 푦1 = −2푎훽푥 + 2푎훼훽, see Figure 6. 푥 +푥 푥 −푥 푟 = 푆2 푆1 = 훼 , 푟 = 푆2 푆1 = 훽 , 푥 = 푟 ± 푖푟 are the complex roots of the quadratic. For further 1 2 2 2 1 2 information see [2]. Example: Let 푓(푥) = 푥2 − 2푥 + 5 , see Figure 5., 푓′(푥) = 2x − 2 , 훼 = 1, 훽 = 2 , 푓′(훼 + 훽) = 4, 푦2 = 4푥 − 4, 푦1 = −4푥 + 4 , 푆2(3,8), 푆1 (−1,8). The roots are : 푥 +푥 푥 −푥 푥 = 푟 ± 푖푟 = 1 ± 2푖 , where 푟 = 푆2 푆1 = 훼 = 1 , 푟 = 푆2 푆1 = 훽 = 2 1 2 1 2 2 2 Figure 5. 풇(풙) = 풙ퟐ − ퟐ풙 + ퟓ Figure 6. 6. Lill’s circle With this method we can find both real and imaginary solutions, and will work for equations of any degree.We consider a general quadratic equation: 푎푥2 + 푏푥 + 푐 = 0. Acta Didactica Napocensia, ISSN 2065-1430 Where are the quadratic’s complex roots ? 41 Figure 7. 풚 = 풙ퟐ + ퟑ풙 + ퟐ If 푎 ≠ 1, we divide the equation by leading coefficient. From the starting point (0, −1) we lay out counter-clockwise perpendicular segments with lengths equal to the 푎, 푏 and 푐 coefficients.If we have negative coefficient we go upwards on the vertical path, and downwards if the coefficient is positive. On the horizontal path we go left if the coefficient is positive, and right in the case of negative one. When the quadratic path is finished we draw a line from the starting point to the finishing point, and we draw a circle that has this diameter. The solutions of the equation are the points where this circle intersects the x-axis. Quadratic equations with complex solutions: Figure 8. 풚 = 풙ퟐ + ퟑ풙 + ퟔ First of all, we construct the 푎-line, 푏-line and 푐-line like in the previous case. Then we translate the 푎- line in order to add to the 푐-line. We draw the blue circle using the 푐 + 푎 lines as diameter. The center of the yellow circle will be the point where the 푏-line and 푐-line intersect, and the radius will be given by the intersection of the 푏-line and the blue circle. Where the yellow circle intersects 푥 = −푏/2 line, there we find the two complex roots. [5] Volume 8 Number 1, 2015 42 Ágnes Orsolya Páll-Szabó 7.Surface plot Figure 9. Surface A (red) and Re Y=0 (green) Figure 10. Surface B (blue) and Im Y=0 (green) Figure 11. 푓(푥) = 푥2 − 2푥 + 5 , complex roots at 1+2i,1-2i “The reader is now asked to leave the comfort zone of two-dimensional plots and extend his/her thinking into the third dimension.” said N.Bardell in [3]. We consider 푦 = 푎푥2 + 푏푥 + 푐 and 푋 = 퐺 + 푖퐻 a complex value. If we substitute 푋 in the equation the result is 푦 = 푎(퐺 + 푖퐻)2 + 푏(퐺 + 푖퐻) + 푐 푦 = 푎퐺2 + 2푎퐺퐻푖 − 푎퐻2 + 푏퐺 + 푏퐻푖 + 푐 푦 = ( 푎퐺2 − 푎퐻2 + 푏퐺 + 푐) + 푖(2푎퐺퐻 + 푏퐻) 푦 = 퐴 + 푖퐵 퐴 = 푎퐺2 − 푎퐻2 + 푏퐺 + 푐 , B=2푎퐺퐻 + 푏퐻 where 푅푒 푋 = 퐺, 퐼푚 푋 = 퐻, 푅푒 푦 = 퐴, 퐼푚 푦 = 퐵 Then we plot the surfaces.
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