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)

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

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

-3 -2 -1 1 2 3

9

4

1

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

Real y axis

Real x axis

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 !

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)

Real y

x (real)

x (imaginary)

THE GRAPH OF y = x2 with REAL y VALUES.

Normal parabola

Phantom parabola at RIGHT ANGLES to the normal one.

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

AUTOGRAPH VERSION.y = x²

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.

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”.

x = 1 + i

x = 1 – i

AUTOGRAPH VERSION.

y = (x - 1)² + 1

y = (x + 4)(x + 2

y = (x – 2)2

y = (x – 6)2 + 1

x = 2x = 6 + i

x = 6 – i

x = -2x = -4

AUTOGRAPH VERSION.

3 parabolas

   

http://autograph-maths.com/activities/philiplloyd/phantom.html

  

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 = 16have 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. y

Real x

Unreal x

Phantom graph

Usual graph

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.

1 2 3Real x

Unreal x

Solutions of x4 = 81

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = 16

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = 1

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = 0.0001

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = 0

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = –0.0001

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = –1

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = –16

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x4 = –81

-3 -2 -1

,

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)

y

x

z

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

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.

1 2 3Real x

Unreal x

Solutions of x3 = 27

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = 8

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = 1

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = 0.001

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = 0

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = – 0.001

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = –1

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = –8

-3 -2 -1

1 2 3Real x

Unreal x

Solutions of x3 = –27

-3 -2 -1

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

= (x2 – 1)(x2 – 1)

= x4 – 2x2 + 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)

– 2 – 1 1 2

1

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

(2,9)(-2,9)

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

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

END of PART 1

Please check out the Web site:

www.phantomgraphs.weebly.com