How do you find fifth roots of 11?

1 Answer
Jun 26, 2018

=> z = e^(2/5 k pi i ) z=e25kπi

k ={0,1,2,3,4} k={0,1,2,3,4}

Explanation:

We are hence find zz if z = root5 1 z=51

=> z^5 = 1z5=1

Use how e^(2kpi i ) =1 , AA k in ZZ

Hence z^5 = e^(2kpi i )

=> z = e^(2/5 k pi i )

For k ={0,1,2,3,4} , As any other k will give related solutions

Where re^(i theta ) = rcostheta+ irsintheta