2. PHANTOM GRAPHS PART 1. Philip Lloyd Epsom Girls Grammar School Web site: www.phantomgraphs.weebly.comwww.phantomgraphs.weebly.com

3. We often say that “Solutions of a quadratic are where the graph crosses the x axis”. y = x 2 – 4x + 3 y = x 2 – 2x + 1 y = x 2 – 2x + 2 1 3 1 1 Crosses twice Crosses once Does not cross at all so x 2 – 4x + 3 = 0 so x 2 – 2x + 1 = 0 so x 2 – 2x + 2 = 0 has 2 (real) solutions has 1 (real) solution has no (real) solutions (or 2 equal solutions)

4. But then we say that this last equation has “complex” or “imaginary” solutions. We can find where y actually CAN equal zero by “completing the square” : x 2 – 2x + 2 = 0 x 2 – 2x = – 2 x 2 – 2x + 1 = – 2 + 1 (x – 1) 2 = – 1 x – 1 = ± i x = 1 + i and x = 1 – i

5. Clearly, the graph does not actually cross the x axis at these points. It does not cross the x axis at all! So, physically, where are these “imaginary” solutions?

6. It is probably better to consider a simpler case using just y = x 2 For positive y values : we get the usual points… ( ± 1, 1) ( ± 2, 4) ( ± 3, 9) etc But we can also find negative, real y values even though the graph does not seem to exist under the x axis: If y = – 1 then x 2 = – 1 and x = ± i If y = – 4 then x 2 = – 4 and x = ± 2 i If y = – 9 then x 2 = – 9 and x = ± 3 i -3 -2 -1 1 2 3 941941

7. The big breakthrough is to change from an x AXIS ….… Real y axis Real x axis

8. Real y axis Real x axis Unreal x axis Complex x plane or “Argand plane” (Instead of just an x axis) The big breakthrough is to change from an x AXIS ….… to an x PLANE !

9. This produces another parabola underneath the usual y = x 2 but at right angles to it! (A sort of phantom parabola “hanging” from the usual y = x 2 graph)

10. Real y x (real) x (imaginary) THE GRAPH OF y = x 2 with REAL y VALUES. Normal parabola Phantom parabola at RIGHT ANGLES to the normal one.

11. There are still only two variables, x and y but, even though x values may be complex, y values are always REAL.

12. AUTOGRAPH VERSION. y = x² y = x²

13. Going back to y = x 2 – 2x + 2 ……. This can be written as: y = (x – 1) 2 + 1 Normally we say the MINIMUM VALUE of y is 1 But the REAL Minimum value of y is not 1! We just showed y can be 0 ! (ie when x = 1 + i and x = 1 – i ) (1,1)

14. In fact y can equal any real value! Suppose y = – 3 So ( x – 1) 2 + 1 = – 3 ( x – 1) 2 = – 4 x = 1 + 2 i and x = 1 – 2 i Similarly if y = – 8 ( x – 1) 2 + 1 = – 8 ( x – 1) 2 = –9 x = 1 + 3 i and x = 1 – 3 i

15. In fact there is NO MINIMUM REAL y VALUE because all complex x values of the form x = 1 ± K i will actually produce more REAL VALUES of y to – ∞. These values are all in the same PLANE at right angles to the basic graph. No other complex x values will produce real y values. The result is another parabola “hanging” from the vertex of the normal graph.

16. So, instead of saying … “Solutions of quadratics are where the graph crosses the x AXIS ” we should now say … “Solutions of quadratics are where the graph crosses the x PLANE”.

17. x = 1 + i x = 1 – i

18. AUTOGRAPH VERSION. y = (x - 1)² + 1 y = (x - 1)² + 1

19. y = (x + 4)(x + 2 y = (x – 2) 2 y = (x – 6) 2 + 1

20. x = 2 x = 6 + i x = 6 – i x = - 2 x = - 4

22. AUTOGRAPH VERSION. 3 parabolas 3 parabolas http://autograph-maths.com/activities/philiplloyd/phantom.html

23. Now consider y = x 4 For positive y values we get the usual points ( ± 1, 1), ( ± 2, 16), ( ± 3, 81) but equations involving x 4 such as : x 4 = 1 or x 4 = 16 have 4 solutions not just 2. (This is called the Fundamental Theorem of Algebra.)

24. If y = 1, x 4 = 1 so using De Moivre’s Theorem: r 4 cis 4θ = 1cis (360n) r = 1 and 4θ = 360n θ = 0, 90, 180, 270 x 1 = 1 cis 0 = 1 x 2 = 1 cis 90 = i x 3 = 1 cis 180 = – 1 x 4 = 1 cis 270 = – i

25. If y = 16, x 4 = 16 so using De Moivre’s Theorem: r 4 cis 4θ = 16cis (360n) r = 2 and 4θ = 360n θ = 0, 90, 180, 270 x 1 = 2 cis 0 = 2 x 2 = 2 cis 90 = 2 i x 3 = 2 cis 180 = – 2 x 4 = 2 cis 270 = – 2 i

26. This means y = x 4 has another phantom part at right angles to the usual graph. y Real x Unreal x Phantom graph Usual graph

27. BUT WAIT, THERE’S MORE !!! The y values can also be negative. If y = –1, x 4 = –1 Using De Moivre’s Theorem: r 4 cis 4θ = 1cis (180 + 360n) r = 1 and 4θ = 180 + 360n θ = 45 + 90n x 1 = 1 cis 45 x 2 = 1 cis 135 x 3 = 1 cis 225 x 4 = 1 cis 315

28. Similarly, if y = –16, x 4 = –16 Using De Moivre’s Theorem: r 4 cis 4θ = 16cis (180 +360n) r = 2 and 4θ = 180 + 360n θ = 45 + 90n x 1 = 2 cis 45 x 2 = 2 cis 135 x 3 = 2 cis 225 x 4 = 2 cis 315

29. This means that the graph of y = x 4 has TWO MORE PHANTOMS similar to the top two curves but rotated through 45 degrees.

31. AUTOGRAPH VERSION. y = x^4 y = x^4

32. Consider a series of horizontal planes cutting this graph at places such as : y = 81 So we are solving the equation: x 4 = 81 The result is a series of very familiar Argand Diagrams which we have never before associated with cross sections of a graph.

33. 1 2 3 Real x Unreal x Solutions of x 4 = 81 -3 -2 -1

34. 1 2 3 Real x Unreal x Solutions of x 4 = 16 -3 -2 -1

35. 1 2 3 Real x Unreal x Solutions of x 4 = 1 -3 -2 -1

36. 1 2 3 Real x Unreal x Solutions of x 4 = 0.0001 -3 -2 -1

37. 1 2 3 Real x Unreal x Solutions of x 4 = 0 -3 -2 -1

38. 1 2 3 Real x Unreal x Solutions of x 4 = –0.0001 -3 -2 -1

39. 1 2 3 Real x Unreal x Solutions of x 4 = –1 -3 -2 -1

40. 1 2 3 Real x Unreal x Solutions of x 4 = –16 -3 -2 -1

41. 1 2 3 Real x Unreal x Solutions of x 4 = –81 -3 -2 -1,

42. y = x 4 ordinary form of the equation z = (x + iy) 4 form of the equation for “Autograph” z = x 4 + 4x 3 yi + 6x 2 y 2 i 2 + 4xy 3 i 3 + y 4 i 4 z = x 4 + 4x 3 yi – 6x 2 y 2 – 4xy 3 i + y 4 Re(z) = x 4 – 6x 2 y 2 + y 4 Im(z)= 4yx(x 2 – y 2 ) If Im(z) = 0 then y = 0 or x = 0 or y = ±x Subs y = 0 and Re(z) = x 4 ( basic curve ) Subs x = 0 and Re(z) = y 4 ( top phantom ) Subs y = ±x and Re(z) = – 4 x 4 ( bottom 2 phantoms) y x z

44. Next we have y = x 3 Equations with x 3 have 3 solutions. If y = 1 then x 3 = 1 so r 3 cis 3θ = 1cis (360n) r = 1 θ = 120n = 0, 120, 240 x 1 = 1 cis 0 x 2 = 1 cis 120 x 3 = 1 cis 240

45. Similarly, if y = 8 then x 3 = 8 so r 3 cis 3θ = 8cis (360n) r = 2 θ = 120n = 0, 120, 240 x 1 = 2 cis 0 x 2 = 2 cis 120 x 3 = 2 cis 240

46. Also y can be negative. If y = –1, x 3 = –1 r 3 cis 3θ = 1cis (180 +360n) r = 1 and 3θ = 180 + 360n θ = 60 + 120n x 1 = 1 cis 60 x 2 = 1 cis 180 x 3 = 1 cis 300

47. The result is THREE identical curves situated at 120 degrees to each other!

49. AUTOGRAPH VERSION. y = x³ y = x³

50. Again consider a series of horizontal Argand planes cutting this graph at places such as : y = 27 So we are solving the equation: x 3 = 27 The result is a series of very familiar Argand Diagrams which we have never before associated with cross sections of a graph.

51. 1 2 3 Real x Unreal x Solutions of x 3 = 27 -3 -2 -1

52. 1 2 3 Real x Unreal x Solutions of x 3 = 8 -3 -2 -1

53. 1 2 3 Real x Unreal x Solutions of x 3 = 1 -3 -2 -1

54. 1 2 3 Real x Unreal x Solutions of x 3 = 0.001 -3 -2 -1

55. 1 2 3 Real x Unreal x Solutions of x 3 = 0 -3 -2 -1

56. 1 2 3 Real x Unreal x Solutions of x 3 = – 0.001 -3 -2 -1

57. 1 2 3 Real x Unreal x Solutions of x 3 = –1 -3 -2 -1

58. 1 2 3 Real x Unreal x Solutions of x 3 = –8 -3 -2 -1

59. 1 2 3 Real x Unreal x Solutions of x 3 = –27 -3 -2 -1

61. Now consider the graph y = (x + 1) 2 (x – 1) 2 = (x 2 – 1)(x 2 – 1) = x 4 – 2x 2 + 1 Any horizontal line (or plane) should cross this graph at 4 places because any equation of the form x 4 – 2x 2 + 1 = “a constant ” has 4 solutions. (Fundamental Theorem of Algebra) – 2 – 1 1 2 1

62. y = (x + 1) 2 (x – 1) 2 = x 4 – 2x 2 + 1 We can find “nice” points as follows: If x = ± 2 then y = 9 so solving x 4 – 2x 2 + 1 = 9 we get : x 4 – 2x 2 – 8 = 0 so (x + 2)(x – 2) (x 2 + 2) = 0 giving x = ± 2 and ± √ 2 i – 2 – 1 1 2 (2,9)(-2,9)

63. y = (x + 1) 2 (x – 1) 2 Similarly if x = 3 then y = 64 so solving x 4 – 2x 2 + 1 = 64 x 4 – 2x 2 – 63 = 0 (x + 3)(x – 3) (x 2 + 7) = 0 giving x = ± 3 and ± √7 i

64. Using this idea to find other “nice” points: If y = 225, x = ± 4 and ± √14 i and if y = 576, x = ± 5 and ± √23 i The complex solutions are all 0 ± n i This means that a phantom curve, at right angles to the basic curve, stretches upwards from the maximum point.

65. However if y = – 1 then x 4 – 2x 2 + 1 = – 1 (x 2 – 1) 2 = – 1 x 2 – 1 = ±i x 2 = 1 ± i To solve this we change it to polar form. so x 2 = √2cis(45 0 + 360n) or √2cis(315 0 + 360n) Using De Moivre’s theorem again: x 2 = r 2 cis 2θ = √2 cis(45 or 405 or 315 or 675) x = 2 ¼ cis(22 ½ or 202½ or 157½ or 337½) If y = – 1, x = – 1.1 ± 0.46 i, 1.1 ± 0.46 i

66. If y = – 2, x = – 1.2 ± 0.6 i, 1.2 ± 0.6 i If y = – 4, x = – 1.3 ± 0.78 i, 1.3 ± 0.78 i Notice that the real parts of the x values vary. This means that the phantom curves hanging off from the two minimum points are not in a vertical plane as they were for the parabola.

69. AUTOGRAPH VERSION. y = (x - 1)²(x + 1)² y = (x - 1)²(x + 1)²

70. END of PART 1 Please check out the Web site: www.phantomgraphs.weebly.com