Alg1131W8T4 - Roots of unity

Saturday, 21 April 2018

11:24 AM

Machine generated alternative text: The remainder theorem (and the closely related factor theorem) gives us an immediate connection between the linear factors of a polynomial and its roots. Namely, if a e C is a root of the polynomial p(z) (i.e. p(a) = 0 ) then p(z) has (z — a) as a factor. Let's answer a few questions to test our understanding of these ideas. — 7 z + 10 by z — 2, we get a remainder of i) If p(z) — ii) If p(z) iii) If p(z) 2 —6z+5 then o 2 2 O , which shows that (z-1) O _ (1 + + i then p(i) O which shows that - 2z+ 1 then O which shows that (z- IS O O is a tactor of is a tactor of not a factor ot iv) If we divide p(z) v) If we divide p(z) vi) If we divide p(z) 2 2 2 — 7 z + 10 by z — 3, we get a remainder of we get a remainder of -2 18

 

Machine generated alternative text: One of the simplest quintic polynomials we can actually solve is This can be rewritten as meaning we can solve this by just finding the fifth roots of -1 i9 5 5 To do this we write z in polar form as z = re where r and 9 are real numbers and r > 0 and so, by De Moivre's Theorem, 5 i59 5 Now we can write our equation in polar form as 5 i59 = leiT Two numbers in polar form are equal when their moduli are equal and their arguments differ by an integer multiple of 2n 1/5 = 1 and 59 + 2kT where k is an integer 1 O (a shorthand for this is k e Z From this we find that 9 — (T + 2kT) and hence the 5th roots of 5 1 — (T+2kff)i —1 are Now; if we make our choices carefully, different values of k will give different solutions. Give a set of five values for k which give distinct fifth roots of —1. Reminder: a set in Maple is entered with curly braces, like {1 , 2 , 3}

 

Machine generated alternative text: 1m 12 11 10 For the complex number w we can compute directly that Using the diagram (click here for a larger view), or by direct computation; find the Cartesian form of the following: 10 11 12 Note: don't forget to use Maple notation, eg sqrt (3) for the square root of 3 and and I (capital i) for the imaginary unit Hint: Don't know how to use the diagram? Think about how each of these roots are symmetrically placed around the unit circle.

 

Ifw - he solutions to -- 1 are. 〔 〕 10 〔 〕 12 as in the previous questions then he solutions to are - 〔 〕 10 〔 〕 12 he solutionsto -1 十 i 〔 〕 12 Re he solutionsto 4 〔 〕 10 〔 〕 12

 

Here is a challenging problem. Consider the polynomial 5 4 2 — 5 z + 25 z — 36 z + 18_ Give the setof complex linear factors of p. To help you out, you are told that 1 + i is a root, and that three of the roots are integers. The set of factors is: {z-3, z-1, z-l-l, z-I+I, z+3] Note: Your set should be of a form like {z-1 , z- (1+2*1) _ Don't forget to use I (capital i) to represent the complex unit

 

 

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