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Complex numbers and polynomials - University College Cork

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3 Linear <strong>and</strong> quadratic <strong>polynomials</strong>Here, we focus on linear <strong>and</strong> quadratic <strong>polynomials</strong> whose coefficients are real <strong>numbers</strong>,<strong>and</strong> restrict their domain of definition to be the set of real <strong>numbers</strong>. Wecan then consider their graphs <strong>and</strong> thence examine their properties in a geometricmanner. These are functions of the formwhere a, b, c are real, <strong>and</strong> a ≠ 0.3.1 Graphs of linesax + b, ax 2 + bx + c, x ∈ R,While the Greeks worked out the geometry of the straight line, they had no awayof representing them in an algebraic manner. It fell to Descartes to describe a wayof doing this: one introduces the cartesian plane R 2 of points with two coordinates,a procedure with which you are familiar. Graphs of functions are then identified assubsets of this plane whose intersection with every vertical line consists of at mosta single point. As a consequence, a circle is not the graph of a function, nor is theparabola {(x, y) ∈ R 2 : y 2 = 4x}.If c + mx is a linear polynomial, its graph in R 2 is the set of points{(x, y); y = mx + c, x ∈ R} = {(x, mx + c) : x ∈ R},which is abbreviated to y = mx + c; m is its slope, <strong>and</strong> c its y-intercept.Given two distinct points (x 1 , y 1 ), (x 2 , y 2 ) in its graph, so thaty 1 = mx 1 + c, y 2 = mx 2 + c,in which case x 2 ≠ x 1 , we can solve these equations for m, c <strong>and</strong>, doing so, find thatm = y 2 − y 1, c = y 1x 2 − y 2 x 1.x 2 − x 1 x 2 − x 1Substituting these into the equation, <strong>and</strong> rearranging the resulting expression, weobtain the familiar formula for the equation of a line through two points which arenot on the same vertical line:Equivalently in this case,y − y 1 = y 2 − y 1x 2 − x 1(x − x 1 ).(x 2 − x 1 )(y − y 1 ) = (y 2 − y 1 )(x − x 1 ).But this formula works even if x 2 = x 1 , provided that y 2 ≠ y 1 ; or if y 2 = y 1 <strong>and</strong>x 2 ≠ x 1 . But we can’t have y 2 − y 2 = x 2 − x 1 = 0 unless we wish to regard the entireplane as a line!To cover all possibilities, then, the equation of a line in R 2 is an expression of theformax + by + c = 0,where a 2 + b 2 > 0. This is the graph of a linear polynomial iff ab ≠ 0. It is thegraph of a constant polynomial if a = 0, b ≠ 0.5

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