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
1
3
y = x2 – 2x + 1
1
Crosses twice
Crosses once
so x2 – 4x + 3 = 0
so x2 – 2x + 1 = 0
has 2 (real) solutions has 1 (real) solution
(or 2 equal solutions)
y = x2 – 2x + 2
1
Does not cross at all
so x2 – 2x + 2 = 0
has no (real) 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
x2 – 2x
x2 – 2x + 1
(x – 1)2
x–1
=0
= –2
= –2 + 1
=–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
9
4
For positive y values :
we get the usual points…
(±1, 1) (±2, 4) (±3, 9) etc
1
-3
-2
-1
1
2
3
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
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.
Real y
Normal parabola
x (real)
x (imaginary)
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
(1,1)
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 )
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 – 6)2 + 1
y = (x + 4)(x + 2
y = (x – 2)2
x = -4
x = -2
x=2
x=6+i
x=6–i
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 = 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.
Usual graph
y
Phantom 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.
Unreal x
Solutions of x4 =
-3
-2
-1
1
2
3
81
Real x
Unreal x
Solutions of x4 =
-3
-2
-1
1
2
3
16
Real x
Unreal x
Solutions of x4 =
-3
-2
-1
1
2
3
1
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
-3
-2
-1
Solutions of x4 = –0.0001
1
2
3
Real x
Unreal x
-3
-2
-1
Solutions of x4 = –1
1
2
3
Real x
Unreal x
-3
-2
-1
Solutions of x4 = –16
1
2
3
Real x
Unreal x
-3
-2
,
-1
Solutions of x4 = –81
1
2
3
Real x
y = x4
ordinary form of the equation
z = (x + iy)4 form of the equation for “Autograph”
z = x4 + 4x3yi + 6x2y2i2 + 4xy3i3 + y4i4
z
z = x4 + 4x3yi – 6x2y2 – 4xy3i + y4
Re(z) = x4 – 6x2y2 + y4
y
Im(z)= 4yx(x2 – y2)
If Im(z) = 0 then y = 0 or x = 0 or y = ±x
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)
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.
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 =
-3
-2
-1
1
2
3
0
Real x
Unreal x
Solutions of x3 =
-3
-2
-1
1
2
3
– 0.001
Real x
Unreal x
Solutions of x3 =
-3
-2
-1
1
2
3
–1
Real x
Unreal x
Solutions of x3 =
-3
-2
-1
1
2
3
–8
Real x
Unreal x
Solutions of x3 =
-3
-2
-1
1
2
3
–27
Real x
Now consider the graph y = (x + 1)2(x – 1)2
= (x2 – 1)(x2 – 1)
= x4 – 2x2 + 1
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)
(-2,9)
(2,9)
y = (x + 1)2(x – 1)2
=
x4
–
2x2
–2
+1
–1
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
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