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

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

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 + 1 = – 2 + 1 (x – 1)2 = – 1 x – 1 = ± i x = 1 + i and x = 1 – i

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?

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 -3 -2 -1 1 2 3

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

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)

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)

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

AUTOGRAPH VERSION.

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)

In fact y can equal any real value! Suppose y = – 3 So (x – 1)2 + 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

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.

“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”.

AUTOGRAPH VERSION. y = (x - 1)² + 1

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

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

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.)

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

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

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

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 (180 + 360n) r = 1 and 4θ = 180 + 360n θ = 45 + 90n x1 = 1 cis 45 x2 = 1 cis 135 x3 = 1 cis 225 x4 = 1 cis 315

Similarly, if y = –16, x4 = –16 Using De Moivre’s Theorem: r4cis 4θ = 16cis (180 +360n) r = 2 and 4θ = 180 + 360n θ = 45 + 90n x1 = 2 cis 45 x2 = 2 cis 135 x3 = 2 cis 225 x4 = 2 cis 315

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

AUTOGRAPH VERSION. y = x^4

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.

Real x Unreal x Solutions of x4 = 81 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = 16 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = 1 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = 0.0001 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = 0 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = –0.0001 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = –1 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = –16 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x4 = –81 -3 -2 -1 1 2 3 ,

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

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

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

Also y can be negative. If y = –1, x3 = –1 r3cis 3θ = 1cis (180 +360n) r = 1 and 3θ = 180 + 360n θ = 60 + 120n x1 = 1 cis 60 x2 = 1 cis 180 x3 = 1 cis 300

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

AUTOGRAPH VERSION. y = x³

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.

Real x Unreal x Solutions of x3 = 27 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = 8 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = 1 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = 0.001 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = 0 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = – 0.001 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = –1 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = –8 -3 -2 -1 1 2 3

Real x Unreal x Solutions of x3 = –27 -3 -2 -1 1 2 3

Now consider the graph y = (x + 1)2(x – 1)2 = (x2 – 1)(x2 – 1) – 2 – 1 1 2 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)

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 – 2 – 1 1 2

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

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.

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(450 + 360n) or √2cis(3150 + 360n) 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

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.

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

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