Show that the roots of the equation z^5 - (z-i)^5 =0 are (1/2)(cot (k *pi/5) + i) where k = 1,2,3,4
thanks
z^5 = (z-i)^5 right?
(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 = (1/2) ( [ cos (kπ/5) - i sin (kπ/5) ] / sin (kπ/5) ) + i
z = (1/2) ( cot (kπ/5) - i + 2i)
z = (1/2) ( cot (kπ/5) + i ), k =1,2,3,4 (k=0 rejected because cot (0) is undefined) (shown)
hi,
thanks for making the effort to type everything out :) really appreciate it