Complex RootsDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/