roots of a complex number

[Sonal7]

I am trying to solve Z^8=1. I can see on argand diagram that the roots are what they, as the roots add to 0. But I cant figure out how you would derive them with the equation. r [cos (2kpi)/8+i sin(2kpi)/8]
as when k is 1 you do get (1/4) pi but this doesnt seem correct when you use K=2, then theta is pi/2- this is incorrect it seems. See the attachment for the ans.

 

[Harry_the_cat]

Theta = pi/2 leads to the solution z = i.

 

[pka]

I am trying to solve Z^8=1. I can see on argand diagram that the roots are what they, as the roots add to 0. But I cant figure out how you would derive them with the equation. r [cos (2kpi)/8+i sin(2kpi)/8] as when k is 1 you do get (1/4) pi but this doesnt seem correct when you use K=2, then theta is pi/2- this is incorrect it seems. See the attachment for the ans.

Hint: \(\displaystyle \frac{2k\pi}{8}=\frac{k\pi}{4}\) now let \(\displaystyle k=0,1,\cdots,7\).

 

[Sonal7]

I got it now, thank you very much. So silly of me not to think of i as the solution when theta is pi/2.