GSTF Journal of Mathematics, Statistics and Operations Research (JMSOR) Vol.2 No.2, October 2014
The Number of Complex Roots of a Univariate
Polynomial Within a Rectangle
C.C. Cheng, G.H.J. Lanel
More explicitly, if P is a non-corner point then it is critical if and
only if there exists points A, B ∈ ∂ N on opposites of P such that
AB contains no axial point except P. If P is a corner point of R
then it is critical if not both AP and BP consists of axial points.
Define the pseudo argument change attributed to a critical point
P relative to ∂ N by
(
∆parg( f (A), f (B)), if ∆parg( f (A), f (B)) ≤ π/2
∆∂ N parg(P) =
−∆parg( f (B), f (A)), otherwise.
Keywords - polynomial; real root isolation; complex root isolation.
Abstract - Let f (z) be a nonzero complex univariate polynomial
and let R be a rectangle in the complex plane. The number of complex roots of f (z) inside R is given by the winding number of f (z)
on R if f (z) has no roots on the boundary of R. In this paper the
result is extended to all rectangles R even when f (z) has roots on
the boundary of R under the condition that f (z) is square-free. It
can also be used to formulate an algorithm that isolates the complex
roots of any polynomial.
Hence if one of f (AP) and f (PB) is and the other is not on
an axis then |∆∂ N parg(P)| = π/4. If both f (AP) and f (PB)
are not on an axis then |∆∂ N parg(P)| = π/2 or 0. Therefore
Throughout f (z) will be a polynomial with complex coef- ∆∂ N parg(P) = 0, ±π/4 or ±π/2. It is not hard to see that it is
ficients and R a rectangle in the complex plane. The real part independent of the choice of A and B.
and imaginary part of f are polynomials f1 , f2 ∈R[x, y] such that Since the pseudo argument changes attributed to critical points
f (z) = f (x + iy) = f1 (x, y) + i f2 (x, y). We shall identify a com- count the “number" of times f crosses an axis, the result below
plex point P with its coordinates (x, y) in the plane. A complex follows directly from the Argument Principle [4].
point P is axial if f1 (P) or f2 (P) is zero but not both. GeometriProposition 1. Suppose f (z) ∈C[z], N is a standard polygon and
cally, this means that f (P) lies on an axis but is not the origin.
suppose Pi , i = 1, . . . , s, are all the critical points of f on ∂ N. If
Let O be the origin and W1 ,W2 two nonzero points in the comthere is no root of f on ∂ N then the number of roots of f in the
plex plane. Then the pseudo argument change from W1 to W2 is
−
→
interior of N is given by
π
4 n where n is the least number of times the ray OP approaches
or leaves a half axis when the nonzero point P is revolved about
1 s
∑ ∆∂ N parg(Pi ).
O counterclockwise from the quadrant or the half axis contain2π i=1
ing W1 to that containing W2 . We shall denote this value by
∆parg(W1 ,W2 ). For example, if W1 ,W2 are in the same quad- Proof. Assume P1 , . . . , Ps are oriented counterclockwise on ∂ N.
rant (or on the same half-axis) then ∆parg(W1 ,W2 ) = 0. If W1 is We will choose points T1 , . . . , Ts on ∂ N such that Ti is between
on the positive x-axis and W2 in the second quadrant not on any Pi and Pi+1 , subscripts modulo s. If f (Pi ) and f (Pi+1 ) are on
5π
axis then ∆parg(W1 ,W2 ) = 3π
“neighboring" half-axes, i.e. |∆∂ N arg( f (Pi ), f (Pi+1 ))| = π/2
4 and ∆parg(W2 ,W1 ) = 4 .
Let N be a standard polygon, i.e. a polygon whose edges are then (Pi , Pi+1 ) are of type I, and we use the Intermediate Value
either horizontal or vertical. If an edge E of N contains infinitely Theorem to choose Ti such that |∆∂ N arg( f (Pi ), f (Ti ))| = π/4
many axial points then f1 or f2 must be identically zero on E, i.e. and |∆∂ N arg( f (Ti ), f (Pi+1 ))| = π/4. Otherwise f (Pi ) and
f (E) lies on an axis. Let P be an axial point on the boundary ∂ N f (Pi+1 ) are on the same half-axis, i.e. ∆∂ N arg( f (Pi ), f (Pi+1 )) =
of the polygon N. Then there exists points A, B on the adjacent 0. Since there are no critical points between Pi and Pi+1 ,
edge(s) to P such that A, P, B are oriented counterclockwise on f∂ N (Pi Pi+1 ) = { f (P)|P is between Pi and Pi+1 on ∂ N} is ei∂ N and each of AP and BP either contains no axial points except ther entirely on an axis or none of its points except f (Pi ), f (Pi+1 )
P or contains infinite many axial points (thus must consist of are on the axis. In the first case, we say that (Pi , Pi+1 ) is of
axial points only). If AP and BP do not both contain infinitely type II, and in the second case, of type III. In both cases, we
many axial points then P is said to be critical and A, B isolate P. choose Ti to be any point on ∂ N between Pi and Pi+1 and so
I. INTRODUCTION TO PSEUDO ARGUMENT CHANGE
DOI: 10.5176/2251-3388_2.2.50
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GSTF Journal of Mathematics, Statistics and Operations Research (JMSOR) Vol.2 No.2, October 2014
∆∂ N arg( f (Pi ), f (Ti )) + ∆∂ N arg( f (Ti ), f (Pi+1 )) = 0. It is not dif- be used as those that isolate the roots and the critical points. Fificult to see that if (Pj−1 , Pj ) and (Pk , Pk+1 ) are of type I or II and nally ∆∂ R parg(Zi ) and ∆∂ R parg(Pi ) can be computed based on
(Pi , Pi+1 ) are of type III for i = j . . . k − 1 then
the signs of the real and imaginary parts of f (Ai ), f (Bi ) where
{Ai , Bi } isolates Zi or Pi .
∑ki= j ∆∂ N arg( f (Ti−1 ), f (Pi )) + ∆∂ N arg( f (Pi ), f (Ti ))
2. The hypothesis that f be square-free is needed for the
= ∑ki= j ∆∂ N parg(Pi ).
Theorem. For instance f (z) = z2 is not square-free with
real part f1 (x, y) = x2 − y2 and imaginary part f2 (x, y) = 2xy.
where j ≤ k and subscripts modulo s.
Let R be the rectangle [−1, 1] × [−1, 0]. Then (0, 0) is the
Therefore, using the principle of argument, the number of only zero on ∂ R and the critical points of f on ∂ R are
(−1, 0), (−1, −1), (0, −1), (1, −1), (1, 0). It is not difficult to
roots of f in the interior of N is given by
check that ∆R parg(0, 0) = 0, ∆R parg(−1, 0) = ∆R parg(1, 0) =
1
s
π/4 and ∆R parg(−1, −1) = ∆R parg(0, −1) = ∆R parg(1, −1) =
2π ∑i=1 ∆∂ N arg( f (Ti−1 ), f (Pi )) + ∆∂ N arg( f (Pi ), f (Ti ))
π/2 so the result of Main Theorem does not hold. In fact it can
1
= 2π
∑si=1 ∆∂ N parg(Pi ).
be shown that f (z) = zn has 4n critical points on the boundary
Remark. Suppose f (z) ∈C[z] and E is a straight line in the of a small rectangle containing (0, 0) each attributing pseudo arcomplex plane. Using the rotation of axes it is easy to see that if gument change of π/2. So if the Main Theorem is true for the
f (z) vanishes on infinitely many points of E then f (z) vanishes same rectangle R in this case then the pseudo argument change
at all points of the line. Using this observation, one can remove attributed by (0, 0) would have to be nπ.
the hypothesis from Proposition 1 that N must be standard.
Let R and T be standard rectangles such that the interior of T
contains
a root Z of f on ∂ R. We say that T is small relative to
Suppose Z is a root of f on ∂ R. If Z is on an edge E and
R
if
∂
R
∩
∂ T = {A, B} is contained in the edge(s) of R and A, B
not a corner point of the rectangle R then there exists points A, B
isolate
Z.
The inside boundary of T (relative to R) is the set of
of E on opposite sides of Z such that either there are no axial
all
points
of
∂ T which lie inside R or on ∂ R.
points on AB or f (AB) lies on an axis. If Z is a corner point of
In
order
to prove the Main Theorem we need the following
R then there exists points A, B such that each of AZ and BZ satLemma
whose
proof will be postponed until the end of the paper.
isfy the following: either all points except Z are axial or none of
them are. For convenience we shall say that A, B isolate Z and
Lemma 2. Suppose R is a standard rectangle in the complex
assume that A, Z, B are oriented counterclockwise on ∂ R. Deplane, f (z) ∈C[z] is square-free and Z = x0 + iy0 is a root of f
fine the pseudo argument change attributed to Z relative to ∂ R
on ∂ R. Then there exists a rectangle T such that the following
by ∆∂ R parg(Z) = −∆parg( f (B), f (A)). It is not hard to see that
holds:
this is independent of the choice of A, B.
1. T is arbitrarily small (in length/width) and is small relative to
R such that Z is the only root of f contained in the interior of T .
II. MAIN THEOREM
Theorem. Suppose f (z) ∈C[z] is square-free and suppose R
is a rectangle in the complex plane. If Zi , i = 1, . . . , m, are the
roots of f on ∂ R and Pi , i = 1, . . . , n, the critical points of f on
∂ R then the number of roots of f in the interior of R is given by
!
m
n
1
∑ ∆∂ R parg(Zi ) + ∑ ∆∂ R parg(Pi ) .
2π i=1
i=1
Remarks.
1. The above result can be used to symbolically computing the
number of roots of a given polynomial within a rectangle. One
first computes the largest square-free divisor f of the given polynomial. Then its real part f1 and the imaginary part f2 are computed. After computing gcd( f1 , f2 ) and the cofactors of f1 , f2 ,
an existing real root isolation procedure can be used on them to
find the isolating intervals of the roots and critical points on the
boundary of the rectangle [3]. Next, bisection is used to make
these intervals disjoint. Then the endpoints of the intervals can
2. Either both f1 , f2 are strictly monotonic or one of them is in
the upper-lower cone and the other in the left-right cone formed
by the lines y − y0 = ±(x − x0 ).
3. The pseudo argument change attributed by each critical point
on ∂ T is π/2 .
III. PROOF OF MAIN THEOREM
Proof. By (1) of Lemma 2, there exists disjoint rectangles
Ti small relative to R and each Zi is the only root of f contained
in the interior of Ti such that (2)-(3) hold. Let N = R − ∪m
i=1 Ti
and let Qi, j , j = 1, . . . , `i , be the critical points on the inside
boundary of Ti . Each of these Qi, j is also a critical point on ∂ N.
Using (1) and (2) of Lemma 2, one sees that there are at most
three critical points on the inside boundary of each Ti . By (3)
of Lemma 2, ∑ j ∆∂ N parg(Qi, j ) is between 0 and −3π/2. Hence
∆∂ R parg(Zi ) = ∑ j ∆∂ N parg(Qi, j ). Then the number of roots of f
in the interior of R is the same as those in the interior of N which,
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GSTF Journal of Mathematics, Statistics and Operations Research (JMSOR) Vol.2 No.2, October 2014
by Proposition 1, is
1
2π
m
IV. DISCUSSION AND CONCLUSION
`i
!
n
∑ ∑ ∆∂ N parg(Qi j ) + ∑ ∆∂ N parg(Pi )
i=1 j=1
1
=
2π
m
i=1
n
!
∑ ∆∂ R parg(Zi ) + ∑ ∆∂ R parg(Pi )
i=1
i=1
since ∆∂ R parg(Pi ) = ∆∂ N parg(Pi ).
Proof of Lemma 2. We first prove that if Z = x0 + iy0 is
a root of f then (x0 , y0 ) is a non-singular point for both f1
and f2 . Suppose ∂ f1 /∂ x(x0 , y0 ) = ∂ f1 /∂ y(x0 , y0 ) = 0. Then,
by Cauchy-Riemann, ∂ f2 /∂ y(x0 , y0 ) = ∂ f1 /∂ x(x0 , y0 ) = 0 and
∂ f2 /∂ x(x0 , y0 ) = −∂ f1 /∂ y(x0 , y0 ) = 0. Hence f 0 (z0 ) = 0, contradicting the fact that f is square-free. The same for f2 can be
proved similarly.
Without loss of generality we may translate the axes so that
(x0 , y0 ) = (0, 0), i.e., Z is the origin. Therefore only two cases
need to be considered: case (i) both partials of f1 , f2 are nonzero
at (0, 0); case (ii) one of f1 , f2 has its x-partial vanish at (0, 0)
and y-partial nonzero at (0, 0), the other has its x-partial nonzero
at (0, 0) and y-partial vanish at (0, 0).
In the first case, by the Implicit Function Theorem, there exists S1 = [−a, a]×R and S2 =R×[−b, b] for positive real a, b over
which f1 is a strictly monotonic. We may choose each Si such
that only one root of f , i.e. (0, 0), is in it. Let T1 = S1 ∩ S2 .
Then f1 is strictly monotonic in this arbitrarily small T1 . Similarly there exists arbitrarily small rectangle T2 over which f2
is strictly monotonic. Let T = T1 ∩ T2 . Then f1 , f2 are strictly
monotonic over T .
In the second case, we may assume that ∂∂fx1 (0, 0) =
∂ f1
0, ∂ y (0, 0) 6= 0, ∂∂fx2 (0, 0) 6= 0, ∂∂fy2 (0, 0) = 0. Since ∂∂fy1 (0, 0) 6= 0
and ∂∂fx2 (0, 0) 6= 0, by the Implicit Function Theorem, there exist functions y = φ1 (x) and x = φ2 (y) defined on (−h1 , h1 ) and
(−h2 , h2 ) that agree with f1 , f2 around (0, 0). Since ∂∂fx1 (0, 0) = 0
and ∂∂fy2 (0, 0) = 0, we have φ10 (0) = 0 and φ20 (0) = 0. Hence
there exists arbitrarily small positive δ1 , δ2 such that |φ1 (x)| < |x|
for −δ1 < x < δ1 and |φ2 (y)| < |y| for −δ2 < y < δ2 . Let S1 =
[−δ1 , δ1 ] × [−δ1 , δ1 ], S2 = [−δ2 , δ2 ] × [−δ2 , δ2 ] and T = S1 ∩ S2 .
Then T is arbitrarily small, small relative to R. So one of f1 , f2
is in the upper-lower cone and the other in the left-right cone
created by the lines y = ±x. This proves (1) and (2).
From (1) and (2), one sees that there are exactly four critical points on ∂ T . Using Proposition 1, one sees that the sum of
pseudo argument changes of the critical points on ∂ T must be
2π. Since each pseudo argument change of a critical point is between −π/2 and π/2, ∆∂ T parg(P) = π/2 for each critical point
P on ∂ T .
Isolation of polynomial roots is an important area in Computer Algebra [1]. Isolation of complex roots is discussed first
in [5] although the algorithm does not solve the problem completely, e.g. in case when all coefficients are integral. Using the
Main Theorem, one can formulate a symbolic algorithm for isolating complex roots of a complex polynomial as follows. Using a root bound theorem, find a rectangle in which all roots
of the given polynomial lie. Subdivide the rectangle into two
sub-rectangles and count the number of roots in the interiors
and boundaries of the sub-rectangles. Repeat these on all subrectangles created, keeping track of the number of roots in each
sub-rectangle and outputting it when it has exactly one root in
it. In any case, the hypothesis that the given polynomial must be
square-free cannot be dropped. [6] avoids the problem by stopping the algorithm when a fixed number of subdivisions found
roots on the boundary.
REFERENCES
[1] G. E. Collins, "Infallible calculation of polynomial zeros to specified precision," Mathematical Software, Academic Press. New York, pp. 35-68, 1977.
[2] G. E. Collins and W. Krandick, "An efficient algorithm for infallible polynomial complex root isolation," in Proceedings of International Symposium
on Symbolic and Algebraic Computation. Berkeley, California, PP. 189-194,
1992.
[3] J. R. Johnson, "Algorithms for polynomial real root isolation," Quantifier
Elimination and Cylindrical Algebraic Decomposition, Springer, PP. 269299, 1998.
[4] M. Marden, "Geometry of Polynomials," American Mathematical Society,
Providence, 1966.
[5] J. Pinkert, "An exact method for finding the roots of a complex polynomial,"
ACM Transactions on Mathematical Software, 2(4):351–363, 1976.
[6] H. Wilf, "A global bisection algorithm for computing the zeros of polynomials in the complex plane," J.A.C.M., 25(3):415-420, 1978.
AUTHORS' PROFILE
G.H.J. Lanel is a Senior Lecturer in Mathematics at University of Sri
Jayewardenepura, Sri Lanka. He obtained his first degree in Mathematics with
first class honours at Open University of Sri Lanka, thereafter he obtained his
first master degree in Industrial Mathematics with a merit from the University
of Sri Jayewardenepura and second master degree in Applied Mathematics from
Oakland University in Michigan, USA. He later obtained his PhD from the Oakland University. His research interests mainly lie in Computer Algebra as well as
in applications of Graph Theory.
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