The fundamental theorem of algebra asserts that any polynomial p(z) with complex coefficients may be written...
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The fundamental theorem of algebra asserts that any polynomial p(z) with complex coefficients may be written in the form p(z) = a(z-21) (z - 22). (z-Zn), where a is a nonzero complex number, n is the degree of p, and z, are the (not necessarily distinct) roots of p(z). Let us assume that n ≥ 2, and all the z; are distinct. We are going to show n i=1 (you do not need to prove this). 4. Prove that For illustration, consider the case n = 2 first, so we may write p(z) = az² + bz+c. The condition that the roots should be distinct is equivalent to the dicriminant 6² - 4ac being nonzero. Hence conclude that 1 p'(zi) 1. Write down the roots 21, 22 of p(z) (in terms of a, b, c). Verify directly that (₁) + (2₂) = 0. Now we return to the general case, where the degree of p is n ≥ 2. 2. Show that the derivative of p evaluated at 2, is given by p' (zj) = a(zj — 2₁) (zj — 2₂) (zj — Zj-1) (zj — Zj+1) ··· (zj — Zn) (Here we are assuming that p(z) = a(z-2₁) (2-22) (z-Zn) as stated at the top of the page.) The quantity is a times the product of (z; - zi) for all i different from j; notice that the term (zj zj) has been omitted. 3. Let , be a closed curve enclosing z, (where j is an integer from 1 to n) and no other z₁. Prove that Jag = 0. p(z)` In the last few problem sets we have dissected various regions of integration so that they contain only one singularity. The outcome has been that the integral over a large region is equal to the sum of smaller integrals, each of which may be computed (e.g. by the Cauchy Integral Formula). We are still working our way towards a theorem that does not require this dissection. In any case, we will let the curve I be the circle centred at 0 with radius R large (larger than any zi]). Then it turns out that we can similarly dissect I into closed curves, each of which contains one z; in the interior, the integral of which was computed in the previous part. So we get -dz = n 1 -dz. √p(2) d= = Σ/₁₁ (2) d dz 2=1 lim R→∞ Jr P(2) 2πi p'(zj) p(z) j 1 1 p'(zi) dz = 0. = 0. Suppose that p(z) has degree 1, so that there is one root, 2₁. Then in fact, p) #0 (i.e the result does not hold). So the assumption that n ≥ 2 really was necessary. The fundamental theorem of algebra asserts that any polynomial p(z) with complex coefficients may be written in the form p(z) = a(z-21) (z - 22). (z-Zn), where a is a nonzero complex number, n is the degree of p, and z, are the (not necessarily distinct) roots of p(z). Let us assume that n ≥ 2, and all the z; are distinct. We are going to show n i=1 (you do not need to prove this). 4. Prove that For illustration, consider the case n = 2 first, so we may write p(z) = az² + bz+c. The condition that the roots should be distinct is equivalent to the dicriminant 6² - 4ac being nonzero. Hence conclude that 1 p'(zi) 1. Write down the roots 21, 22 of p(z) (in terms of a, b, c). Verify directly that (₁) + (2₂) = 0. Now we return to the general case, where the degree of p is n ≥ 2. 2. Show that the derivative of p evaluated at 2, is given by p' (zj) = a(zj — 2₁) (zj — 2₂) (zj — Zj-1) (zj — Zj+1) ··· (zj — Zn) (Here we are assuming that p(z) = a(z-2₁) (2-22) (z-Zn) as stated at the top of the page.) The quantity is a times the product of (z; - zi) for all i different from j; notice that the term (zj zj) has been omitted. 3. Let , be a closed curve enclosing z, (where j is an integer from 1 to n) and no other z₁. Prove that Jag = 0. p(z)` In the last few problem sets we have dissected various regions of integration so that they contain only one singularity. The outcome has been that the integral over a large region is equal to the sum of smaller integrals, each of which may be computed (e.g. by the Cauchy Integral Formula). We are still working our way towards a theorem that does not require this dissection. In any case, we will let the curve I be the circle centred at 0 with radius R large (larger than any zi]). Then it turns out that we can similarly dissect I into closed curves, each of which contains one z; in the interior, the integral of which was computed in the previous part. So we get -dz = n 1 -dz. √p(2) d= = Σ/₁₁ (2) d dz 2=1 lim R→∞ Jr P(2) 2πi p'(zj) p(z) j 1 1 p'(zi) dz = 0. = 0. Suppose that p(z) has degree 1, so that there is one root, 2₁. Then in fact, p) #0 (i.e the result does not hold). So the assumption that n ≥ 2 really was necessary.
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Related Book For
Fundamentals of Momentum, Heat and Mass Transfer
ISBN: 978-1118947463
6th edition
Authors: James Welty, Gregory L. Rorrer, David G. Foster
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