Thursday, 4 November 2010

Complex and Other Conjugates

We had covered the Fundamental Theorem of Algebra in pre-calc and after a few days of trying to make sure they understood how to decide how many real and non-real solutions there were to a real valued polynomials and that they could be factored over linear and quadratic factors , I was trying to get them to see how easy it was to work back and forth from the quadratic polynomial and the complex zeros. Finally I told them, "It's almost like the Pythagorian Theorem. a2+ b2 = c " [They will later learn that this is really just the product of two equal pythagorean results for each of the conjugates, but not yet]

I showed them that the complex solutions always came in conjugate pairs, a + bi and a - bi... and that if they knew the zeros, they could create a quadratic which had those solutions. It seems that students are no longer exposed to what I call the sum and product property of quadratic zeros... that if the zeros of a quadratic are r and s, a quadratic with those zeros is given by x2 - (r+s)x + (rs)... (If you knew the zeros were at x=3 and x= -1, then a quadratic with those zeros would be x2 - (3-1)x + (3*-1) or x2 - (2)x -3).

They had to get it somewhere if they were ever taught to factor, but it never seems to find purchase in their mathematical memory. I pointed out that for complex conjugates, r+s was real since the complex +bi and - bi eliminated each other, and so r+s was just twice the real coefficient of either conjugate. I connected this to the graph by showing that this "a" in the a+bi was the axis of symmetry of the graph. For example a quadratic with complex zeros at 3+2i and 3-2i will have a linear term of -(3+3)x [assuming we use the easiest case with A=1]. This also means that the of the quadratic will have its vertex at a point with an x value of 3.

It is also easy to see the constant term of the quadratic. If you multiply (a+bi)(a-bi) the distribution gives a2 +abi - abi - b2i2... As with the sum, the two imaginary terms will cancel out, and since i2=-1, we see that the product of the two complex conjugates will always be a2 + b2. So for the example 3+2i and 3-2i, the product of the conjugates is just 9+4=13. Putting this in as the constant term, we see that the quadratic x2 - 6x + 13 has zeros at 3+2i and 3-2i.

This gives another way to solve a quadratic equation with complex roots. If we have a function x2 - B x + C that we know has complex solutions, we know that the real coefficient, a, of the conjugate pair will be -B/2 (shades of the quadratic formula)... and we know that a2 + b2 = C, so we can figure out b pretty simply. If we had a quadratic such as x2 - 8 x + 25 we would quickly see that the real coefficient is -(-8)/2 = 4... and know that 42 + b2 = 25, so b must be 3, and the two solutions are at 4 +/- 3i. I finally admit to the students that this is just a variant of the quadratic formula, and sooner or later someone will ask if it is not possible to do the same thing with a quadratic with real zeros... I let them suggest a quadratic, and only press for the use of an even B in the linear coefficient for ease of explanation. For example with x2 - 6 x + 5 (ok, I picked an easy one).. we would see that the "a" part was three again... but now we have 32 + b^2 = 5.

If we remember that b was the imaginary coefficient, it makes sense that the b^2 would be negative if the zeros were real... Kids translate that into, "just subtract instead of add, and take the square root of that for the "b" value. So 9 - 4 = 5, and the square root of 4 is two. Our real solutions are 3 +/- 2; or 5 and 1.
At this point I show them that with real rational coefficients, not only do non-real zeros come in conjugate pairs, but so do the real irrational zeros. If we had x2 - 8 x + 10; the solutions of 4 +/- sqrt(16-10) would be the conjugate irrational solutions.

If we graph y=x2 - 6x + 13, we notice that not only is the vertex at x=3, but its y coordinate is at 4... the square of 2, which was the b part of our complex conjugates... It seems that functions with complex solutions a +/- bi, the vertex is at (a, b2). This also works for real valued zeros if we with the a +/- b approach above. The vertex of x2 -6x + 5 would be at (3, - 22).

I'm not sure I would try to drag a really weak class throuh all this, but it often helps bright kids see how interconnected the zeros, factors, and graphs of all quadratics are, not just the complex numbers.
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