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

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Answer 1 · 2017-07-20T06:30:00

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

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))#