How do you find three cube roots of $1+i$ ?

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Answer 1 · 2017-03-06T09:24:28

$root(6)2(cos(pi/12)+isin(pi/12))$ , $root(6)2(cos((3pi)/4)+isin((3pi)/4))$ and $root(6)2(cos((17pi)/12)+isin((17pi)/12))$

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 $root(6)2(cos((17pi)/12)+isin((17pi)/12))$

How do you find three cube roots of 1+i1+i?