One of the simplest quintic polynomials we can actually solve is 25+1=0 This can be rewritten...

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One of the simplest quintic polynomials we can actually solve is 25+1=0 This can be rewritten as z5 === meaning we can solve this by just finding the fifth roots of To do this we write z in polar form as z = re 20 where r and are real numbers and r > 0 and so, by De Moivre's Theorem, 25 === = (reio) 5 =r5 ei50 5150 lei = Now we can write our equation in polar form as Two numbers in polar form are equal when their moduli are equal and their arguments Sor=11/51 and 50=+2k where k is shorthand for this is k€ Z). From this we find that = (기 (+2k) and hence the 5th roots of -1 are (π+2km)i 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}. (a One of the simplest quintic polynomials we can actually solve is 25+1=0 This can be rewritten as z5 === meaning we can solve this by just finding the fifth roots of To do this we write z in polar form as z = re 20 where r and are real numbers and r > 0 and so, by De Moivre's Theorem, 25 === = (reio) 5 =r5 ei50 5150 lei = Now we can write our equation in polar form as Two numbers in polar form are equal when their moduli are equal and their arguments Sor=11/51 and 50=+2k where k is shorthand for this is k€ Z). From this we find that = (기 (+2k) and hence the 5th roots of -1 are (π+2km)i 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}. (a

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