MEI Conference 2014
Phantom graphs
Catherine Berry
catherine.berry@mei.org.uk
Roots of quadratics: the pictorial approach
y
4
y
8
4
−2
y
The quadratic formula:
6
2
−4
Roots of quadratics: the algebraic approach
x
2
4
2
2
4
−4
y = x² - 4
The equation
x² - 4 = 0
has two real roots
−4
−2
2
4
y = x²
The equation
x² = 0
has one real root
−4
−2
b  b 2  4ac
2a
x
x
−2
x
2
4
y = x² + 4
The equation
x² + 4 = 0
has no real roots
Complex roots of equations: the algebraic approach
b 2  4ac  0
Two real roots
b  4ac  0
One real root
b  4ac  0
No real roots
2
2
Complex roots of equations: the pictorial approach
The fundamental theorem of algebra (one form):
Every polynomial equation of degree n with
complex coefficients has n roots in the
complex numbers.
???
In practice, students use the factor theorem
and polynomial division to solve polynomial
equations with complex roots.
The roots of the equation x² = 4 are 2 and -2.
Equivalently, the points (2, 4) and (-2, 4) lie on the graph
y = x². They are the intersections of y = x² and y = 4.
The roots of the equation x² = -4 are 2j and -2j.
Do the points (2j, -4) and (-2j, -4) lie on the graph y = x²?
Where are the intersections of the graphs y = x² and y = -4?
1
In the general quadratic case, consider the equation x² = k.
We can illustrate the roots of this equation by looking at
where the graph y = x² meets the line y = k.
Suppose x is allowed to be a complex number ….
Instead of the x-axis, we will need a plane – the complex
plane.
So now, consider the equation x² = k, where x can be
complex (but k is real).
We can illustrate the roots of this equation by looking at
where the graph z = (x + jy)² meets the plane z = k.
We need to be able to plot z = (x + jy)², where z is real.
So we will use x and y for the real and imaginary axes
on the horizontal plane, and use z for the vertical axis
(which is real).
z  ( x  jy ) 2
x = t, y = 0, z = t²
and x = 0, y = t, z = -t²
z  x 2  2 jxy  y 2
z is real, so the imaginary part must be zero….
2 xy  0
So either y = 0:
z  x2
or x = 0:
z   y2
y  ( x  1) 2  2
 x2  2x  3
x = t, y = 0, z = x² - 2x + 3
and x = 1, y = t, z = 2 – t²
z  ( x  jy ) 2  2( x  jy )  3
z  x 2  2 jxy  y 2  2 x  2 jy  3
z  x 2  y 2  2 x  3  2 jy ( x  1)
z is real, so the imaginary part must be zero….
So either y = 0:
or x = 1:
z  x2  2x  3
z  1 y2  2  3  2  y2
2
y  x3
y  ( x  1) 2 ( x  1) 2  x 4  2 x 2  1
z  ( x  jy )
3
z  ( x  jy ) 4  2( x  jy ) 2  1
z  x  3jx y  3xy  jy
3
2
2
3
z  x 4  4 jx3 y  6 x 2 y 2  4 jxy 3  y 4  2 x 2  4 jxy  2 y 2  1
z  x 4  6 x 2 y 2  y 4  2 x 2  2 y 2  1  4 jxy ( x 2  y 2  1)
z  x3  3xy 2  jy (3x 2  y 2 )
z is real, so the imaginary part must be zero….
z is real, so the imaginary part must be zero….
zx
z  8 x3
z  8 x3
3
So either y = 0:
or y  x 3 :
or y   x 3:
So either y = 0:
z  x4  2x2  1
or x = 0:
z  y4  2 y2 1
or x 
y2 1
or x  
z  4 y 4  4 y 2
:
y2 1
:
z  4 y 4  4 y 2
e  2
z  e x  jy  e x e jy
z  e x (cos y  jsin y )
e z  2e j
z  e x cos y  je x sin y
e z  eln 2 e j
z is real, so the imaginary part must be zero….
What about non-polynomial equations?
z
ln 2  j
sin y  0  y  n
e e
z  ln 2  j
z
For even n, cos n = 1:
For odd n, cos n = -1:
In fact there are an infinite number of solutions:
z  ln 2  j(2n  1)
e jz  e  jz
2
2
z  cos( x  jy )
e j  cos   jsin 
z  cos x cos jy  sin x sin jy
e  j  cos   jsin 
z  cos x cosh y  jsin x sinh y
cos  
e 2 jz  4e jz  1  0
e j  e  j
2
Imaginary part must be zero….
e  2 3
jz

jz  ln 2  3


z   jln 2  3
or

z  ex
z  e x
So every equation of the form ex = k has an infinite
number of solutions.
What about non-polynomial equations?
cos z  2
e j  cos   jsin 
sinh y  0
sin x  0
e j  e  j
2
e   e
cos j 
 cosh 
2
cos  
sin  
e j  e  j
2j
sin j 
e   e
 jsinh 
2j
 y  0  z  cos x
 x  n  z  cosh y for odd n
and z   cosh y for even n
3
So cos z = k has an infinite number of roots for any
real k… where did we lose the others?
cos z  2
e e
2
jz
 jz
You can find out more about phantom graphs at
2
http://phantomgraphs.weebly.com/
e 2 jz  4e jz  1  0
e jz  e 2 n jeln(2
e jz  2  3

jz  ln 2  3


z   jln 2  3
3)
 e2 n jln(2 
3)
jz  2n j  ln(2  3)

z  2n  jln(2  3)
4