1. PHANTOM GRAPHS PART 1. Philip Lloyd Epsom Girls Grammar School
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2. We often say that “Solutions of a quadratic are where the graph crosses the x axis”.
y = x2 – 4x y = x2 – 2x y = x2 – 2x + 2
Crosses twice Crosses once Does not cross at all
so x2 – 4x + 3 = so x2 – 2x + 1 = so x2 – 2x + 2 = 0
has 2 (real) solutions has 1 (real) solution has no (real) solutions
(or 2 equal solutions)

3. x2 – 2x = – 2 x2 – 2x + 1 = – 2 + 1 (x – 1)2 = – 1 x – 1 = ± i
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” :
x2 – 2x + 2 = 0
x2 – 2x = – 2
x2 – 2x = –
(x – 1)2 = – 1
x – = ± i
x = 1 + i and x = 1 – i

4. So, physically, where are these “imaginary” solutions?
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?

5. It is probably better to consider a simpler case using just y = x2
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 x2 = – 1 and x = ±i
If y = – 4 then x2 = – 4 and x = ±2i
If y = – 9 then x2 = – 9 and x = ±3i 9 4 1

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

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

8. This produces another parabola underneath the usual y = x2 but at right angles to it!
(A sort of phantom parabola “hanging” from the usual y = x2 graph)

9. THE GRAPH OF y = x2 with REAL y VALUES.
Normal parabola
x (real)
x (imaginary)
Phantom parabola at RIGHT ANGLES to the normal one.

10. AUTOGRAPH VERSION.

11. Normally we say the MINIMUM VALUE of y is 1
Going back to y = x2 – 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)

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

13. 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.

14. “Solutions of quadratics are where the graph crosses the x AXIS”
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”.

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

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

17. AUTOGRAPH VERSION. 3 parabolas
      

18. but equations involving x4 such as : x4 = 1 or x4 = 16
Now consider y = x4
For positive y values we
get the usual points
(±1, 1), (±2, 16), (±3, 81)
but equations involving x4 such as :
x4 = 1 or x4 = 16
have 4 solutions not just 2.
(This is called the Fundamental Theorem of Algebra.)

19. If y = 1, x4 = 1
so using De Moivre’s Theorem:
r4cis 4θ = 1cis (360n)
r = 1 and θ = 360n
θ = 0, 90, 180, 270
x1 = 1 cis = 1
x2 = 1 cis 90 = i
x3 = 1 cis 180 = – 1
x4 = 1 cis 270 = – i

20. If y = 16, x4 = 16
so using De Moivre’s Theorem:
r4cis 4θ = 16cis (360n)
r = 2 and 4θ = 360n
θ = 0, 90, 180, 270
x1 = 2 cis 0 = 2
x2 = 2 cis 90 = 2i
x3 = 2 cis 180 = – 2
x4 = 2 cis 270 = – 2i

21. This means y = x4 has another phantom part at right angles to the usual graph.
Phantom graph
Usual graph
Unreal x
Real x

22. BUT WAIT, THERE’S MORE !!! The y values can also be negative.
If y = –1, x4 = –1
Using De Moivre’s Theorem:
r4cis 4θ = 1cis ( n)
r = 1 and 4θ = n
θ = n
x1 = 1 cis 45
x2 = 1 cis 135
x3 = 1 cis 225
x4 = 1 cis 315

23. Similarly, if y = –16, x4 = –16
Using De Moivre’s Theorem:
r4cis 4θ = 16cis ( n)
r = 2 and 4θ = n
θ = n
x1 = 2 cis 45
x2 = 2 cis 135
x3 = 2 cis 225
x4 = 2 cis 315

24. This means that the graph of y = x4 has TWO MORE PHANTOMS similar to the top two curves but rotated through 45 degrees.

25. AUTOGRAPH VERSION. y = x^4

26. 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.

27. Real x
Unreal x
Solutions of x4 = 81

28. Real x
Unreal x
Solutions of x4 = 16

29. Real x
Unreal x
Solutions of x4 = 1

30. Real x
Unreal x
Solutions of x4 =

31. Real x
Unreal x
Solutions of x4 = 0

32. Real x
Unreal x
Solutions of x4 = –0.0001

33. Real x
Unreal x
Solutions of x4 = –1

34. Real x
Unreal x
Solutions of x4 = –16

35. Real x
Unreal x
Solutions of x4 = –81 ,

36. y = x4 ordinary form of the equation
z = (x + iy)4 form of the equation for “Autograph”
z = x4 + 4x3yi + 6x2y2i2 + 4xy3i3 + y4i4
z = x4 + 4x3yi – 6x2y2 – 4xy3i + y4
Re(z) = x4 – 6x2y2 + y4
Im(z)= 4yx(x2 – y2)
If Im(z) = 0 then y = 0 or x = 0 or y = ±x
Subs y = 0 and Re(z) = x4 ( basic curve )
Subs x = 0 and Re(z) = y4 ( top phantom )
Subs y = ±x and Re(z) = – 4 x4 ( bottom 2 phantoms) z x y

37. Next we have y = x3
Equations with x3 have 3 solutions.
If y = 1 then x3 = 1
so r3cis 3θ = 1cis (360n)
r = 1 θ = 120n
= 0, 120, 240
x1 = 1 cis 0
x2 = 1 cis 120
x3 = 1 cis 240

38. Similarly, if y = 8 then x3 = 8
so r3cis 3θ = 8cis (360n)
r = 2 θ = 120n
= 0, 120, 240
x1 = 2 cis 0
x2 = 2 cis 120
x3 = 2 cis 240

39. Also y can be negative.
If y = –1, x3 = –1
r3cis 3θ = 1cis ( n)
r = 1 and 3θ = n
θ = n
x1 = 1 cis 60
x2 = 1 cis 180
x3 = 1 cis 300

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

41. AUTOGRAPH VERSION. y = x³

42. So we are solving the equation: x3 = 27
Again consider a series of horizontal Argand planes cutting this graph at places such as :
y = 27
So we are solving the equation:
x3 = 27
The result is a series of very familiar Argand Diagrams which we have never before associated with cross sections of a graph.

43. Real x
Unreal x
Solutions of x3 = 27

44. Real x
Unreal x
Solutions of x3 = 8

45. Real x
Unreal x
Solutions of x3 = 1

46. Real x
Unreal x
Solutions of x3 = 0.001

47. Real x
Unreal x
Solutions of x3 = 0

48. Real x
Unreal x
Solutions of x3 = – 0.001

49. Real x
Unreal x
Solutions of x3 = –1

50. Real x
Unreal x
Solutions of x3 = –8

51. Real x
Unreal x
Solutions of x3 = –27

52. Now consider the graph y = (x + 1)2(x – 1)2 = (x2 – 1)(x2 – 1)
– –
Any horizontal line (or plane) should cross this graph at 4 places because any equation of the form x4 – 2x2 + 1 = “a constant” has 4 solutions.
(Fundamental Theorem of Algebra)

53. We can find “nice” points as follows: If x = ±2 then y = 9
(-2,9)
(2,9)
y = (x + 1)2(x – 1)2
= x4 – 2x2 + 1
We can find “nice” points as follows:
If x = ±2 then y = 9
so solving x4 – 2x2 + 1 = 9
we get : x4 – 2x2 – 8 = 0
so (x + 2)(x – 2) (x2 + 2) = 0
giving x = ±2 and ±√2 i
– –

54. x4 – 2x2 – 63 = 0 (x + 3)(x – 3) (x2 + 7) = 0 y = (x + 1)2(x – 1)2
Similarly if x = 3 then y = 64
so solving x4 – 2x2 + 1 = 64
x4 – 2x2 – 63 = 0
(x + 3)(x – 3) (x2 + 7) = 0
giving x = ±3 and ±√7 i

55. 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 ± ni
This means that a phantom curve, at right angles to the basic curve, stretches upwards from the maximum point.

56. However if y = – 1 then x4 – 2x2 + 1 = – 1 (x2 – 1)2 = – 1 x2 – 1 = ±i
To solve this we change it to polar form.
so x2 = √2cis( n) or √2cis( n)
Using De Moivre’s theorem again:
x 2 = r2cis 2θ = √2 cis(45 or 405 or 315 or 675)
x = 2¼ cis(22½ or 202½ or 157½ or 337½)
If y = – 1, x = –1.1 ± 0.46i , 1.1 ± 0.46i

57. If y = – 1, x = –1.1 ± 0.46i , 1.1 ± 0.46i
If y = – 2, x = –1.2 ± 0.6i , 1.2 ± 0.6i
If y = – 4, x = –1.3 ± 0.78i , 1.3 ± 0.78i
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.

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

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END of PART 1
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