Download Complex Numbers and Vectors (MathWorks for Teachers) by Evans, Les. PDF
By Evans, Les.
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Extra info for Complex Numbers and Vectors (MathWorks for Teachers)
Example text
1: Graph of a quadratic with two points of intersection with the x-axis Important aspects of this graph are the points at which it crosses the xaxis—its x-intercepts. We know that the x-axis is the line on the cartesian plane where y = 0. Thus, when we want to find the possible x-intercepts we are seeking values for x that allow y to be equal to zero. That is y = a]x - hg2 + k = 0 . Before we attempt to solve this equation, we should consider how values of k might affect our results. There are three possible sets of values for k: k > 0, k = 0 and k < 0.
11 - 1 t= Cardano published a similar result in Ars Magna. However, he did not have the benefit of modern notation, which makes his achievement more remarkable. While he did not fully understand complex numbers, he presented the first calculation using complex numbers. To quote him directly: Dismissing mental tortures, and multiplying 5 + - 15 by 5 - - 15 , we obtain 25 - ]- 15g. Therefore the product is 40 … and thus far does arithmetical subtlety go, of which this, the extreme, is, as I have said, so subtle that it is useless.
If we let z = 0, then: 0 3 + 4 # 02 + 7 # 0 + 4 = ]0 + ag]0 + bg]0 + cg 4 = abc This suggests that a, b and c are all factors of 4. We also know that if P] z g = 0 then ]z + ag]z + bg]z + cg = 0 . From the Null Factor Law, we know that z =- a, - b or - c , so P]- ag = P]- bg = P]- cg = 0 . As a consequence, if we can find a value d such thatP]d g = 0 , then z - d must be a factor of P] z g. Putting all the above together, we have a method of searching for factors of a polynomial. For P] z g = z 3 + 4z2 + 7z + 4 , we could use the most obvious factors of 4 to find the first root of P] z g.