We can use here De Moivre's theorem, which states that if z=r(costheta+isintheta)z=r(cosθ+isinθ), then z^n=r^n(cosntheta+isinntheta)zn=rn(cosnθ+isinnθ).
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)(1+i)13.
For that let us first write 1+i1+i in polar form. As 1+i=1+1i1+i=1+1i, r=sqrt(1^2+1^2)=sqrt2r=√12+12=√2 and theta=tan^(-1)(1/1)=pi/4θ=tan−1(11)=π4
Hence 1+i=sqrt2(cos(pi/4)+isin(pi/4))1+i=√2(cos(π4)+isin(π4))
and therefore root(3)(1+i)=(sqrt2)^(1/3)(cos((2npi+(pi/4))/3)+isin((2npi+(pi/4))/3))3√1+i=(√2)13(cos(2nπ+(π4)3)+isin(2nπ+(π4)3))
Now, choosing n=0,1,2}n=0,1,2}, we get three roots as
root(6)2(cos(pi/12)+isin(pi/12))6√2(cos(π12)+isin(π12)),
root(6)2(cos((3pi)/4)+isin((3pi)/4))6√2(cos(3π4)+isin(3π4))
and root(6)2(cos((17pi)/12)+isin((17pi)/12))6√2(cos(17π12)+isin(17π12))
