How do you calculate root(n)(i), ninRR?

1 Answer
Jul 20, 2017

root(n)i=cos(pi/(2n))+isin(pi/(2n))

Explanation:

The polar form of z=i is given by

z=0+i1=cos(pi/2)+isin(pi/2)=e^(ipi/2)

Now according to De Moivre's theorem

if z=r(costheta+isinteta)

z^n=r^n(cosntheta+isinntheta)

or root(n)z=z^(1/n)=r^(1/n)(cos(theta/n)+isin(theta/n))

Hence root(n)i=cos(pi/(2n))+isin(pi/(2n))