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Visualizing Complex Roots
Jason M. Bryer
The College of Saint [email protected]
home.nycap.rr.com/jbryer
When one first encounters the quadratic polynomial, and shortly
thereafter the quadratic formula,
one encounters the square root of a negative number. Someone may
have told you that it is animaginary number, and that it represents
the imaginary (complex) roots of the polynomial. You
may have posed the question, if we have a means of representing
the real roots of a polynomialgeometrically, is there a way to
represent the complex roots geometrically? Lets see...
Real roots or complex roots
Consider a quadratic polynomial cbxaxxf++=
2
)( , a, b, c
. When a polynomial of thisform is plotted in the Cartesian
coordinate system, it is clear whether the polynomial has two,one
or no real roots. Geometrically, the roots of a quadratic are the
distance from the origin to
where the parabola crosses the x-axis.
Figure 1. Two real roots Figure 2. One (double) root Figure 3.
No real roots
When there are no real roots, we know thaty=f(x) has two complex
roots. But as shown above,
the complex roots are not visible using this geometric
interpretation.
Locating the complex rootsRecall that by the quadratic formula
we can find the roots of f(x) given by:
aacb
ab
24
2
2
However, if the discriminant is less than 0 (i.e. b2 4ac <
0), then there are two complex roots.These roots may be represented
as;
iIRa
baci
a
b=
2
4
2
2
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Where R and I are the real and imaginary parts, respectively.
The vertex has coordinates (-b/2a,
f(-b/2a)) or (R,aI2). We can then locate the line y = 2aI
2parallel to the x-axis by marking off
twice the distance of aI2
from (R,0) to (R,aI2) up to (R,2aI
2). However,
22)()( aIIRfIRf =+= . Therefore dropping perpendiculars from the
intersections of y =
2aI2
and the parabola yields the points (R-I,0) and (R+I,0). The
length R is measured from (0,0)
to (R,0) and the length I is measured from (R,0) to (R+I,0) or
from (R-I,0) to (R,0) as shown infigure 4. And now we have a method
to geometrically represent the complex roots of apolynomial. This
is a particularly nice representation because it is similar to our
traditional
method, that is, the roots are the distance from a point to
where the quadratic crosses a horizontalline.
Figure 4 Figure 5
Finding roots through a line reflectionThe points (R-I,0) and
(R+I,0) can be found in another, similar way. Since (R,aI
2) is the vertex
of the parabola, we can draw the line y = 2aI2 which is parallel
to the x-axis and tangent to thevertex. Now reflect the parabola
through this line as shown in figure 5 above. This new
parabola can be expressed asa
bcbxaxxf
2)(
22
+= . The points of intersection with the x-
axis are then (R-I,0) and (R+I,0). Again, the complex roots
become visible.
The Modulus Surface
For a third method of representing the complex
rootsgeometrically, consider the complex polynomial
cbzazzf ++= 2)( where a, b, c and z = x + iy. But first,
lets look at the real polynomial 1)( 2 ++= xxxf . From the
graph
in figure 6 we know that there must be two complex roots.
Because we are only interested in the zeros of the polynomial,
wecan treat the zeros as low points on a surface called the
modulus
surface. If w is the complex number w = u + iv, then its
modulus
is defined as 22|| vuw += , which is nonnegative. Now let us
aI2
aI2
y=2aI2
R I
R+IR-I
R+IR-I
y=2aI2
Figure 6. f(x)=x2+x+1
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consider the quadratic polynomial as a complex polynomial 1)( 2
++= zzzf where z = x + iy,
then the modulus off(z) is |f(z)|. The modulus surface of the
polynomial w =f(z) is the surface
defined by the three real variables, x, y, and |w|.
The zeros of a complex-valued function defined on the complex
plane are the points of the
modulus surface that touch the complex plane, since the complex
number w is zero if and only ifits modulus is zero.
Figure 7. |z2
+ z + 1| Figure 8. |z2
+ z 1|
Figure 8 above shows the modulus surface of z2
+z 1. Note that the two points where thesurface touches the
complex plane are on the real axis, hence there are two real roots.
Figure 7 is
the modulus surface of z2
+z + 1. Here, the surface touches the complex plane in exactly
two
points as well. These points, 0,3,1 and 0,3,1 are exactly the
complex roots, 31
of the quadratic polynomial 1)( 2 ++= xxxf .
These are only three methods of many to find the complex roots
of a quadratic polynomial. Andit should be clear that the roots of
a polynomial, if not real, are indeed not imaginary roots, but
complex roots.
References:Farnsworth, David, Measuring Complex Roots. Math and
Computer Ed., V. 17 (1983), No. 3,
pp. 191-194.
Long, Cliff and Thomas Hern, Graphing the Complex Zeros of
Polynomials Using ModulusSurfaces. College Math J., V. 20 (1988),
No. 1, pp. 33-45.
Page, Warren, Complex Roots Made Visible. College Math J., V. 15
(1984), No. 3, pp. 248-
249.
