We can use here De Moivre's theorem, which states that if #z=r(costheta+isintheta)#, then #z^n=r^n(cosntheta+isinntheta)#.
It may be worth mentioning that the theorem is valid for fractions as well. As we are going to find cube roots, what we seek is #(1+i)^(1/3)#.
For that let us first write #1+i# in polar form. As #1+i=1+1i#, #r=sqrt(1^2+1^2)=sqrt2# and #theta=tan^(-1)(1/1)=pi/4#
Hence #1+i=sqrt2(cos(pi/4)+isin(pi/4))#
and therefore #root(3)(1+i)=(sqrt2)^(1/3)(cos((2npi+(pi/4))/3)+isin((2npi+(pi/4))/3))#
Now, choosing #n=0,1,2}#, we get three roots as
#root(6)2(cos(pi/12)+isin(pi/12))#,
#root(6)2(cos((3pi)/4)+isin((3pi)/4))#
and #root(6)2(cos((17pi)/12)+isin((17pi)/12))#
