Give #Z# complex, what is #Z^(1/n)# ?

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Answer 1 · 2016-12-01T12:25:51

See below.

Any complex number #Z# can be represented as

#Z = r e^(i (phi + 2k pi))#.

For instance, if #Z = x + i y# then

#r = sqrt(x^2+y^2)# and #phi= arctan(y/x)#.

Here we are using de Moivre's identity which reads

#e^(i phi) = cos(phi)+i sin(phi)#

so

#Z^(1/n)=r^(1/n) e^(i (phi + 2k pi)/n)# or equivalently

#Z^(1/n)=r^(1/n)(cos((phi + 2k pi)/n)+isin((phi + 2k pi)/n))#

for #k = 0,1,cdots,n-1#