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.


Anonymous said...

That business about the vertex was just the extension I needed for tomorrow. Thank you!


Pat's Blog said...

good luck... I usually get about a dozen blank stares, and two or three start nodding and smiling... Fishing is about the ones you get in the boat... Hope it works for you