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Complex Roots


Date: Tue, 1 Nov 1994 20:06:17 -0600
From: Phil C. Scott
Subject: Complex Roots

Dr. Math,

        We know it is possible to look at the graph of a polynomial
and tell a great deal about its real roots by looking at the
x-intercepts.  What can be discovered about a polynomial's
complex roots by looking at the graph?  There seem to be some
interesting "wiggles" at locations that appear to be related to
the "average" of the complex pairs.
        Do you have any insights regarding this topic?

        Best Regards,                  Pre-Calculus Students
                                       J.C.M. High School
                                       Jackson, TN

                                       c/o  Phillip Scott

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   


From: Dr. Ethan
Subject: Re: Complex Roots
Date: Tue, 1 Nov 1994 21:20:44 -0500 (EST)

To be honest, I and two of my Math Doctor buddies have never heard of 
any such relationship but we are checking it out. If we get divinely inspired
(i.e. find it in a book) we might have an answer soon. If not I am sure that 
one of the other math doctors will know.  Rest assured we are on it.
                Ethan

Subject: Re: Complex Roots
Date: Wed, 2 Nov 1994 11:44:11 -0500 (EST)

Just wanted to keep you posted.

        I graphed a lot of functions last night but didn't find any such wiggles
that you mentioned.  Could you send along a few functions for us to look at?
Also what exactly do you mean by the "average"? If you just mean normal 
old average, than what method are you using for graphing the complex 
number on the real plane?  I am still looking through a few books, and the
problem has been sent on to some more people who may be more help.
                Ethan - Doctor on Call


Date: Wed, 2 Nov 1994 18:33:57 -0600
From: Phil C. Scott
Subject: Wiggles

Dr. Math,

        We are using TI-82 calculators to graph the 
polynomials.  Sometimes the "wiggle" is very easily
seen.  Sometimes the "wiggle" is hardly noticable.
We have included some examples below.

        One example is:
        x^4 - 6x^3 - 22x^2 + 56x - 64

        This polynomial has a pair of complex roots
( 1 + i ) and ( 1 - i ).  The average of these two
roots  (R1 + R2)/2 is the real number 1.  If you look 
at the graph of the polynomial, there is "hill" very
near x = 1.  We used the window:
                Xmin  -10      Xmax  10
                Xmax -600      Ymax 100              
    
        Another example is:
        3x^4 - 4x^3 + 3x - 4

        This polynomial has a pair of complex roots
( 1 + i*sqrt(5) )/2 and ( 1 - i*sqrt(5) )/2.  The avg.
of these two roots is the real number (1/2).  If you 
look at the graph, there is a very slight "wiggle" near
x = (1/2).  We used the window:
                Xmin  -3    Xmax  3
                Ymin  -5    Ymax  5

        We suspect that there is some sort of clue about
the complex roots of a polynomial that are visible on the
real graph.  This is what our original message was asking.

        We would appreciate any further responses.

        Thank you for your replies so far.

                                J.C.M. Pre-Calculus Students

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   


Date: Thu, 3 Nov 1994 18:30:44 -0600
From: Phil C. Scott
Subject: Dents&wiggles

Dr. Math,

        The two previous polynomials we discussed were:
(1)  x^4 - 6x^3 - 22x^2 + 56x -64
(2)  3x^4 - 4x^3 + 3x - 4

        In both cases the "wiggle" was near the average
of the complex roots.  We used the word "near" in our last
message.  In polynomial (1) the "wiggle" was a hill which
according to our "maximum" program on the TI-82 was at the 
x-coordinate, x = .97051 (the avg. of complex roots = 1)
In polynomial (2) the wiggle was not a hill...more like a
dent.  Anyway, your value of x = (1/3) we agree with.  It 
fits what we mean by "near". (complex root avg. = .5 )

        We continue using the word "near" because we think
the effect of the complex roots on the real graph is hard
to formulate exactly.

        We have made the following study to look at the 
effect of moving the real root on out the number line.
In each of the following polynomials the complex roots are
( 1 + i ) and ( 1 - i ) .... an avg. of 1.  Notice what
happens to the x-coordinate of the "wiggle" as the real root
is moved to the right.  In each of these cases the "wiggle"
was a hill, and we used the TI-82 to find the x-coordinate
of the maximum.

        POLYNOMIAL                       X-COORD. OF MAX
     (x^2 - 2x + 2)(x - 3)         1.333334
     (x^2 - 2x + 2)(x - 4)         1.183504
     (x^2 - 2x + 2)(x - 5)         1.131482
     (x^2 - 2x + 2)(x - 6)         1.103195
     (x^2 - 2x + 2)(x - 7)         1.085147

     (x^2 - 2x + 2)(x - 10)        1.056082
     (x^2 - 2x + 2)(x - 100)       1.005051
     (x^2 - 2x + 2)(x - 1000)      1.000501

        It appears to us that "wiggle" of these graphs is
always influenced by the complex roots.  What we are trying
do is develop a graphing technique that will let us find
the complex roots from the real graph.

        Thanks for your remarks so far.  Any further help 
would be appreciated.

                                J.C.M. Pre-Calculus students  

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   



Date: Thu, 3 Nov 1994 22:10:11 -0500
From: Stephen B Maurer
Subject: Re: Dents&wiggles

Dear Mr Scott's Pre-Calculus Class,

  Thank you for sending your very interesting question to Dr Math.
It's not one I had considered in all my years of looking at polynomials. I
haven't seen the other answers sent by students here (I am a professor),
so I hope I am not repeating things too much.

  Let c = a+bi and c* = a-bi be two conjugate roots of a polynomial p(x),  
Then the polynomial factors

   p(x) = A (x-c)(x-c*)(x-r1)...(x-rk),
        
where r1, r2,...,rk are the other roots.  Define

   f(x) = (x-c)(x-c*) = x^2 - 2ax + (a^2+b^2),
        
           g(x) = A (x-r1)...(x-rk),
        
so that   p(x) = f(x)g(x).  Finally, let K = g(a).  Then for x near a on
the number line,

   p(x) ~ f(x)g(a) = K f(x).     (*)
        
In other words, for x near a, p(x) looks approximately like K f(x), which
is a quadratic.  Since the apex of this quadratic is at x=a, it has a
wiggle (hump) near x=a.  Thus  p(x) ~ K f(x) also should have a hump
near x=a.

The second example you gave, p(x) = 3x^4 - 4x^3 + 3x - 4, only had a
slight bend, not a hump.  Why is that?  Well, the analysis in (*)
above is crude.  p(x) is actually f(x) g(x), not f(x) g(a).  If g(x)
stays close to g(a), in the sense that g(x)/g(a) stays close to 1,
then (*) is quite accurate, and there will be a hump.  (Whether it
opens up or down depends on the sign of K.)  Otherwise the change in
g(x) may overwhelm the quadratic nature of f(x).  This is what happens
in your second example.

I can show (using some calculus, so I won't show it here) that
the farther away the other roots are from a, and the smaller the
imaginary part b of a+bi, the better an approximation (*) is, that
is, the more likely you will see a hump instead of a mere bend.

One more thought for you.  You noticed that the critical point of
p(x) and that of f(x) are not the same.  On which side of the latter
will the former be?  (You need to know this to estimate the complex roots
from what you see, which is the former.)  The answer is determined by
the graph of g(x) around x=a.  For instance, if g(a)>0 and g(x)
is decreasing around x=a, then the local minimum of p(x) will be to
the right of the local minimum of f(x). Do you see why?

I hope this is helpful.

Stephen B Maurer
Professor of Mathematics
Swarthmore College



From: Dr. Ethan
Date: Thu, 3 Nov 1994 22:11:35 -0500 (EST)

     I'm impressed folks - this is good stuff.  Here are a few thoughts and 
some comments on maxima and minima.  First, do you all  know derivatives? 
If so, that could help.  

     I have included the end of your series and
it has given me a few thoughts that are forming but maybe there is 
something here that will help you in your very interesting search. 

     First let us look at the graph of a parabola.  The average of the
roots of a parabola (without an xy term--ignore this if you don't 
understand it) is the x value of its vertex.  Try a few; I think that you 
will agree. So for instance just look at x^2 - 2x +2 which has imag 
roots, and their average is one, as you noted.  And that is the minimum, 
as I stated above--it's true for all parabolas.  

     So now that you have tested a few parabolas and you believe me, 
let's look  at this sequence here.  As the numbers (x - n) get bigger the 
hump you noticed  gets closer and closer to one.  One way to think 
about this is that near zero  (let's say within -3 to 3) the x-n is just some 
big scaling factor that really doesn't  change the shape very much; it just 
scales it higher. I know that at x = 0 scaling will be -n, and at x = 3 it 
will be three minus n, but for n pretty large (>1000) this really isn't a 
very big scale difference.  

     I recommend that you take two calcs and set both x ranges to be 
xmin=0 xmax=3 and on the first on ymin=1 ymax=3.5 and graph 
x^2 - 2x +2 and on the second ymin=1000 ymax=3500(scale 1000) 
and graph
                                        (x^2 -2x +2)(x+1000)
I used + instead of minus so that they would both be pointing up - 
it doesn't really affect the idea.

     This is not a full answer but I think that this is a step in the right
direction.  I think these correlations have something to do with the 
fact that anyplace on a curve can be approximated by a parabola. 
(See Taylor polynomials--ignore if you don't understand.) So in 
particular we would expect the wiggles to be close to the average of the 
roots of the parabola that approximates it, which in some particular 
case and at some particular places happens to be the one gotten by 
(x - r1)(x-r2), where r1 and r2 are the imaginary roots in quesion.

     Hope that this is somewhat helpful.  If not, please write back and 
give us another try. I will keep thinking, as I know will many other 
people.

        Ethan Doctor on Call (but not really knowing what he is doing)



Date: Sun, 6 Nov 1994 12:05:17 -0500
From: "John Conway"
 steve@mathforum.org (Stephen Weimar))

(Well, you guys really hit pay dirt in the responses you've received!  
You may not realize just how amazing these are until you hear that 
Prof. Maurer is a professor at Swarthmore who is a leader in the field 
of Discrete Math and John Conway, who wrote this letter, is one of the 
most famous mathematicians living. I hope you'll contact both of them 
personally to let them know what you've done with their answers and 
what you think of their communication to you. -- steve)

A very interesting question!  At first I misread it as asking what
we can tell about the non-real roots by looking at the x-intercepts
of its graph, so let me answer that one first - essentially nothing!
(this is because if a,b,c,... are the x intercepts of  y = f(x), then
all we can tell is that  f(x) = (x-a)(x-b)(x-c)... times some g(x)
which has only non-real roots, and of course, you can make the roots
of g(x) be any collection of pairs of conjugate non-real complex numbers.

   Now back to the original question.  I think we must add the 
information that the degree of the polynomial is some known number, 
if we are to read off information very easily.

    Also, let's think what we mean by "looking at the graph".  If we 
are allowed to make very precise measuresments on the graph, then 
we can work out (in theory) just what function f(x) is, and so all its 
roots are actually determined by the EXACT shape of the graph.  
If you know the degree is n, you only need to measure n+1 points to 
be able to work out, at least in theory, all the coefficients of f(x).

   So are question is really a rather loose one - just take a casual
glance at the picture, for example, see where the wobbles are, and
use what you see to produce an "engineer's guess" at the complex
roots.

   Let's start with n = 2, and the equation  y = (x-a)^2 + b.

I'll suppose b > 0 to stop the roots from being real.

   Of course the roots are  a +- root(-b), so we look for geometrical
interpretations of a  and  b.  Easy - a  and b are the x and y
coordinates of the minimum.  So

   First easy theorem.  If  y = x^2 + ...  has non-real roots, these
are  a +- root(-b),  where  (a,b) is its minimum.

   Corollary:  If  y = kx^2 + ...  has non-real roots, these are

   a +- root(kb), where (a,b) is the minimum.

So we need also an engineer's way of evaluating k.

   Here it is - k is the amount y would increase if you went 1
unit left or right of the minimum.

   This gives us the engineer's rule for quadratics:

   The real part is the x-coordinate of the minimum.  The
geometric part is the geometric mean of two numbers you can see
in this little picture:

            |          |
            |          |
            \          /
       ______\________/______
              \      /
               \____/



       _______________________   x-axis

The two numbers I mean are the height of the minimum, and
the "extra height" above that, that's cut off by the horizontal
line that meets the curve 1 unit left and right of the minimum.

    I never knew this before, although I did know that something
vaguely like it was true.

  Now what about more general polynomials f(x).  Well, really
the quadratic case is all there is, in a sense, if you're only
taking a rough glance.  Let me explain why.

    Take your curve  y = f(x), and fix your attention on a
given minimum.  Let  y = q(x)  be the best quadratic approximation
near that minimum.  Let's suppose for simplicity that the minimum
is at x = 0.

   Then the curve  y = a + bx + cx^2 + dx^3 + ...

will be quite well-approximated by

                   y = a + bx + cx^2    (period)

near zero.   So at the complex roots of the latter equation,
f(x) will be quite near zero, so we can expect nearby roots
of  f(x) itself.

    So my exact "engineer's rule" for the quadratic will work
roughly for any polynomial.

    However, you should be prepared to work at the appropriate
scale.  Let me discuss this.  If I take a horizontal line a bit
above the minimum  (a,b)  that cuts the curve roughly where
x = a +- c,  then the imaginary parts of the corresponding pair
of roots will be the geometric mean of the two heights I mentioned,
MEASURED IN UNITS OF c.

   How can you tell roughly what is the right scale?  Well, at
least you can tell if it's going wrong.  If the horizontal line
you take hits the curve at  a - c1  and   a + c2, where  c1 and c2
are very different, you know the quadratic part is NOT a sufficiently
good approximation out to this distance, and you'd better beware.

    On the other hand, if the curve looks very like a parabola for
quite a long time around the minimum, you can guess the corresponding
pair of roots with fair confidence.

   In summary:  if you are allowed to make precise measurements on
the curve, then all the roots are in fact completely determined.
If near some minimum the curve looks like a parabola out to a
distance d (say), and my quadratic "engineer's rule" gives roots
whose imaginary parts are less than d, then it's probably going to
be pretty close to the truth.

    What information can you get from other features, say from
points of inflection?  Well, these first arise when  f(x)  is a
cubic polynomial, so you'd answer this question by working out
in detail what happens in the cubic case, and then if  f(x)
doesn't look like a quadratic for long enough, but does look
like a cubic, you could confidently use your "cubic engineer's
rule".

             John Conway


Date: Tue, 8 Nov 1994 22:39:43 -0500
From: Stephen Weimar
Subject: Re: Conway Thanked

         Both my classes sent short thank you notes to
 Professor Conway.
         The students were thrilled to receive the replies
 from the professors.

Great!  This makes it all the easier to give more.
We look forward to future conversations.

         Thanks to all those involved with the Dr. Math
 project.  The students will remember this for a long
 time.

Our pleasure. -- steve

          Best Regards               Phil Scott


Date: Sun, 6 Nov 94 15:42:11 EST
From: "John Conway"
Subject: Re:  Complex Roots

Yes, I learned something, too!  I already vaguely "knew" something
about how the complex roots were related to the shape of the graph,
but had never thought about it in detail.

    But to explain something to somebody else, you MUST really 
understand it yourself, and in full detail.  This is why I LIKE
all of elementary mathematics, even though I really AM a
professional mathematician.  

    In this case, I sat at this machine typing out an answer (I'm
quite proud of the fact that I just came up with the answer as I
was writing it - I only had to step away breifly to the blackboard
to draw the general quadratic - otherwise it came straight out).

   It might help to follow my thoughts, which were:

  i) CAN you do this? Is the desired information really there?

    This led to my remarks that from the x-intercepts you CAN'T
    determine the non-real roots, but from the exact curve you CAN.
 
       (That, to me, was the trivial bit.)

 ii) A minimum just above the x-axis obviously (to me) corresponds
    to a conjugate-complex pair of roots with small imaginary part
   (because if it were just below, they'd be real, and coincide if
just one, and I know how pairs of real roots become pairs of complex
ones "through" a coalescence).

     The standard case of this is the quadratic one, so we'd better

 iii) work out the exact rule in the quadratic case!  (which I'd
never done before).   This wasn't too hard.

    Then

  iv) (which I really was aware of during ii) and iii)), the general
case really IS the quadratic case, because in any given small
region, your polynomial will be quite well approximated by a 
quadratic.

    Finally

  v)  (which I'd never thought about before) can we guess roughly 
how far this quadratic approximation will work?

    If  f(x) = q(x) + r(x),  where  q  is the quadratic approximation
to f,  then  (supposing the real part of the roots is 0), we can expect
the size of  r(iy)  to be about the same as that of  r(y)  [because
the dominant term of  r(y)  will probably be the first one, eg  5y^3,
and this doesn't change in size if we replace  y  by iy].

   This led to my rule of thumb that if it looks like a parabola out
to a distance of about d, then our engineer's guess will be OK so
long at it gives imaginary parts less than about  d  in size.

     Then as a sort of PS I added

   vi)  some general remarks about how one would develop better
approximate guesses corresponding to higher-degree approximations.

   [All immediate to a professional.]


    What's hard about learning mathematics (for a student), or
teaching it (for a teacher) often is this "feel of the problem"
stuff.  What's this problem REALLY about? - sort-of-thing.

    It seems to be easy to learn how to manipulate formulae, but
very very very hard to develop this "feeling" skill, and so it
must obviously be very hard for teachers to learn how to teach
this "feely" activity.  Some ideas I use in this connection might
be valuable.

    [Of course, I usually teach VERY bright undergraduates and
graduate students - not exactly a typical audience - but I'm
also quite good at teaching more typical "students", right down
to very young children.  I honestly believe that the real teaching
problems are almost always the same, so I'll pass on my ideas
anyway.]

    When you teach something involving a formula (or, if we're 
talking about very young children, some concept that's a bit
more abstract than they're used to), always

   1)  Find a picture that relates to the formula, and teach 
     both at once.  If there are several really different
     "pictures", teach a few of them (unless this might lead
to "overload".   

                      PLEASE do that!

   2)  Have lots of examples involving these pictures, and 
     get the students to understand how the picture changes
     as the parameters in the formula  [or maybe, the formula
     itself]  changes.

         
   3)  Find what are the most important parameters!  I mean
     the ones the engineer wants to know first, so that he can
     tell whether the cost is going to be millions of dollars,
     or only thousands.  Don't worry too much about the things
     that will only affect the "conceptual price" by a few cents.

       [To use an example that came up in another one of these
     problems, every adult knows which are the most important
     digits in a price of  $496:25.  By the way, in that 
     "rounding" problem, one should of course use the children's
     own developing "feel" for real prices  - this might very
     well help them to stop rounding (say) this price to the
     nearest $10  and getting  $90.]

   4)  Use any "feel" for things that your students have already
     developed, for example this feeling for real prices, or
     what you've taught them in similar contexts earlier.  [This
     will also help them to firm up on those earlier things.]

   5)  Occasionally ask the students for information about teaching
     methods!  (I've done this with 3-year-olds to very good effect.)

       Of course, you don't just ask them how you should teach
     such-and-such.  And only for quite articulate students should
you even ask which way they thought was best, when you've done
 something two ways, or done two similar things in different ways.
But any good teacher will find all sorts of ways to "ask" the
students which way was best.  Children love to see their suggestions
taken seriously, and affect the entire class.  

     Marx was a great philosopher, even though not all his ideas
have stood the test of time.  One I think that really has is when
he said something like

   "Honesty, and Sincerity are two of the most important things
in life."

   "So, if you can fake those, you've got it made!"

I often follow Groucho's advice in teaching.  I often teach when
I'm tired, or teach subjects with which I'm thoroughly bored.  So
I just fake liveliness, or fake a total fascination with the subject
I'm bored with.

    Often, when a student suggests some way I should do things, I
"fake" a way of following this suggestion.  The easiest one is
when Sally suggests something I was going to do anyway.  I say -
"great"  "We'll do it Sally's way!   What a really wonderful
idea!"  and then every now and then call this "Sally's method".

    I have no moral qualms whatever about that one.  The one
that might get me burnt in Hell, or at least earn the disapproval
of other teachers, now I'm letting the cat out of the bag, is when
Jim suggests something, and I really don't do it, but find a way
to pretend I do.  In defence, let me say that I only do this sort
of thing when Jim's suggestion is really a good one, given Jim's
knowledge, but as a practical matter wouldn't quite work.  The
fakery is usually to modify Jim's idea a bit, to the nearest
idea that does work, and quietly ignore the necessary modifications.

    In the hope that it will help me to gain absolution for this sin,
let me say that I'm usually not quite so enthusiastic about Jim's
ideas as I was for Sally's.  I might say

    "That's a very good idea, Jim.  Let's see how it works."
but after that, it's just like my treatment of Sally's.

   Of course the thing that really pleases me no end is when
Manuel produces a teaching idea (or any other idea) that I've
never even heard of before.  But strangely enough, Manuel's 
idea doesn't get too much more of a star billing than
Sally's.  The reason is partly that it really was very clever
indeed for Sally to come up with even the standard idea,. so
she fully deserves her praise.  Another reason is that (probably)
Manuel is a student who's getting plenty of praise already.  

    Often more than half of the students  produce ideas that
really do affect the way the course goes, and most of the rest
get that impression even though it might not quite be true.
Suddenly, I'm feeling terribly guilty about this little fakery
- maybe I should stop doing it?

      John Conway



Date: Wed, 9 Nov 1994 12:17:06 -0500
From: Stephen Weimar
Subject: re: fakery

I think an essential part of teaching is acting, and there are many 
forms of it which are just right for the classroom, but probably the 
most important is recreating the experience of the learner. If I didn't 
do this, I would not find the language that works for my audience 
half the time and I would also misinterpret more than I already do 
what the student meant.

For those situations where we are covering well-trodden ground, 
acting is called for and it's a great art that not only is good for the 
students but renews the possibility of new insights for the actor 
as well.

One of my main interests as a teacher is to model the process of 
thinking and this means externalizing what is most often hidden 
from sight.  Running excitedly with Jim's idea but taking it to a place 
different than he anticipated is not fakery but representing the process 
we've been through.

By being enthusiastic about Jim's idea you convey the notion that it's 
good to trust one's instincts, than a good starting place is not necessarily
what one uses in the final analysis.  By thinking out loud and showing 
what to do with it to make it work, you show how to shape good ideas 
from a reasonable beginning.

If I were you I would keep up the acting and let them see the process 
of deciding why something needs to be modified.

They would not be very bright students if they did not know you 
were acting and did not appreciate the performance.

--steve


Date: Tue, 8 Nov 1994 15:44:06 -0600
From: Phil C. Scott
Subject: Thank You

Steve,
        Could you forward the following notes to:
Professor Stephen B. Maurer

Dear Professor Maurer,

        Thank you for your response to our question
about polynominals that have complex roots.  We are still
working and hope to find out more information concerning
this problem.  We will keep you updated on any further
conclusions.  

                               Sincerely,
                               Amy Britt
                               Cheryl Cole
                               LaShunda Smith
 
P.S. Thanks again on behalf of Mr. Scott's Pre-Calculus
class. 

Professor Maurer,

    We would like to add our thanks to the thank you from
second period for the help you have given us with our
question.  We know how valuable your time is.  We are
not, however, finished working with our problem.  We
will send you our further results as we  find them.
We Hope!!!

                         SINCERELY,
                                   Tavares Newsom
                                   Chris Boyd
                                   (4th period)

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   


Date: Fri, 11 Nov 1994 15:32:55 -0600
From: Phil C. Scott
Subject: More Wiggles

Dr. Math

   We are still working on looking at real graphs and deciding 
something about the complex roots from the wiggles on
that graph. Two pairs of complex polynomials produce a 'W'
shaped graph with relationships shown below.

Effect of two pairs of the same complex numbers on the graph:

    polynomial             complex roots   values of extrema
--------------------------------------------------------------
(x^2+4x+5)(x^2-4x+5)        -2+-i   2+-i   (+-sqrt(3), 16)
(x^2+6x+10)(x^2-6x+10)      -3+-i   3+-i   (+-sqrt(8), 36)
(x^2+8x+17)(x^2-8x+17)      -4+-i   4+-i   (+-sqrt(15), 64)
(x^2+10x+26)(x^2-10x+26)    -5+-i   5+-i   (+-sqrt(24), 100)
(x^2+12x+37)(x^2-12x+26)    -6+-i   6+-i   (+-sqrt(35), 144)
(x^2+20x+101)(x^2-20x+101) -10+-i  10+-i   (+-sqrt(99), 400)
(x^2+30x+226)(x^2-30x+226) -15+-i  15+-i   (+-sqrt(224), 900)
                                           (+-a, +-i)      (+-h, k)

  Complex Roots are: (a+-i),(-a+-i) where a = (sqrt(h^2 + 1))

This works as long as the complex pairs are the same distance from
zero. We are still working on what happens to the graphs of 4th
degree polynomials when the pairs of complex roots are different
distances from zero.

Other members of our class have looked at other special cases.  We
suspect there will be no simple method for always finding complex
roots from a real graph, but is has been interesting looking at this
question.

Thanks to all the people at Dr. Math for taking time to respond.
We learned some helpful things from all your messages.  We wanted
to let you know something about what we had discovered since our
last message.

Sincerely,                  J.J. Guy, Brian Benton, Andy Moorehead
                                                    J.C.M.  Pre-Calculus Class

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   



Date: Mon, 28 Nov 1994 15:01:17 -0600
From: Phil C. Scott
Subject: Complex Roots

Dr. Math,

        MORE ON WIGGLES.

        Our textbook gives examples of graphs of polynomials
with roots of multiplicity 3.  For example, y=(x-2)^3 where
the slope of the graph levels out at (2,0).  We have noticed
a curious "wiggle" which we are calling "pseudomultiplity 3"
when the real root remains at x=2 and a pair of complex
roots 2+bi, 2-bi is included.

        It seems that the slope of the polynomials is the
product of the "i's" of the complex roots of "pseudo-
multiplicity 3" problems, as you can see in our table.
For example, in (2+i)(2-i), the product of the i's,
+i and -i is 1.  Here are some other problems we used in 
our research.

 Equation              Conjugate pair       Slope at (2,0)
_______________________________________________________
 (x-2)(x^2-4x+5)         (2+i)(2-i)               1
 (x-2)(x^2-4x+8)        (2+2i)(2-2i)              4
 (x-2)(x^2-4x+13)       (2+3i)(2-3i)              9
 (x-2)(x^2-4x+20)       (2+4i)(2-4i)             16
 (x-2)(x^2-4x+17/4)   (2+1/2i)(2-1/2i)          1/4
_______________________________________________________

Example problem:

          Graph x^3-6x^2+16x-16   
          Window Xmin=0   Ymin=-6 
                 Xmax=3   Ymax=1 
          Use the TI-82 tangent feature to find the slope. 
             the slope is very close to 1
          Since this is the product of the "i's"
             the "i's" are then +i and -i
          Because 2 is an obvious root, the complex 
             roots are (2+i) and (2-i).

    Do you see any problems in our reasoning?

    So we believe that if a graph of this type intersected
the x axis at (-4,0) and the slope was say 1/4 then the
three roots would be -4, -4 + .5i, -4 - .5i.

    We realize that this method works only for a special type
of graph, but we think it is interesting.

    We just wanted to thank all the nice people at Dr. Math
for their previous messages...and to let you know that we
had some more fun with the problem.  Thanks Again!

                                        Tabitha Long
                                        Houston Ballard
                                        Laketha Maclin
                                        Darius Brown
                                        Andy Lino
                                        Danny Joiner
                                        JCM H.S. Pre-Calculus

Phillip C. Scott              
Mathematics Teacher           
pcs2036@jackson.freenet.org   


Date: Wed, 30 Nov 1994 12:08:12 -0500
From: Stephen B Maurer
Subject: Re: Complex Roots

To JCM H.S. Pre-Calculus,

  Hmmm, you keep discovering interesting things.

  Questions about slopes of curves are usually best answered using
calculus, and using calculus I was able to prove that your conjecture 
is correct: the slope of
(x-a)(x-(a+bi))(x-(a-bi)) at x=a is always b^2.

  In fact, the relationship you discovered is more general than that.  
The real root doesn't have to be the same as the real part of the 
complex roots.  E.g.,

  f(x) = (x-17)(x-3-4i)(x-3+4i)          (1)

has slope 4^2 at x=3.

  Indeed, f doesn't have to be a cubic, or have to be "monic" 
(coefficient of highest power is 1), and there is still a relationship, 
though more complicated.

  However, as I understand it, you are trying to see if you can identify
complex roots just by looking at the real-valued graph.  Once you 
move the real root away from the complex roots (as in Eq (1) above), 
then the relationship I am talking about doesn't seem to help you 
identify the real part of the complex roots.

 Somehow you must get that information, and only then do you know 
where to look at the slope, and only then does this slope tell you the 
imaginary part of the complex roots.

  Keep up the good work.

Prof Maurer


Date: Sat, 23 Sept 2000 11:40
From: Francis Edward Su
Subject: Complex roots made visible

I thought I would send you a reference that gives a very cute answer 
for quadratics that is simpler than the description of Professor Conway.  
It is:

	 A. Norton and B. Lotto, "Complex Roots made Visible,"
	 College Math J., 1984, number 3, pp. 248-249.

Basically, just take the quadratic, reflect it about its vertex (min/max),
see where it intersects the x-axis, and rotate those points 90 degrees
along a circle centered halfway between them. The resulting points, when
viewed in the complex plane are the resulting roots.  A student of mine
found this reference.  A helpful picture he created, and a more detailed
description can be found at the "Mudd Math Fun Facts" site:

   Complex Roots Made Visible
   http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml   

I hope that's helpful!  The Math Forum site is a really great site.

Best regards,
Francis Su
    
Associated Topics:
High School Imaginary/Complex Numbers
High School Polynomials

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