A lvl H2 Maths: Complex Numbers

Question from http://www.sgforums.com/forums/2297/topics/361489

Show that the roots of the equation z⁵ - (z-i)⁵ = 0 are ½(cot (kπ/5) + i) where k = 1, 2, 3, 4

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Answer:

z⁵ = (z-i)⁵

(z / (z-i))⁵ = 1

(z / (z-i)) = e^(2kiπ/5), k = 0,1,2,3,4

1 + i/(z-i) = e^(2kiπ/5)
i/(z-i) = e^(2kiπ/5) - 1
= e^(kiπ/5) (e^(kiπ/5) - e^(-kiπ/5))
= e^(kiπ/5) (2 i sin kπ/5)

Divide by i throughout
1/(z-i) = e^(kiπ/5) (2 sin kπ/5)
z-i = e^(-kiπ/5) [1/(2 sin kπ/5)]
z = ½ ( [ cos (kπ/5) - i sin (kπ/5) ] / sin (kπ/5) ) + i
z = ½ ( cot (kπ/5) - i + 2i)
z = ½ ( cot (kπ/5) + i ), k =1, 2, 3, 4 (k=0 is rejected because cot (0) is undefined) (shown)

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