sqrt(i)=(1+i)/sqrt(2)

[Happypig375]

The package I want to suggest the idea to: AngouriMath

sqrt(i)=(1+i)/sqrt(2) should simplify to True https://www.wolframalpha.com/input/?i=sqrt%28i%29%3D%281%2Bi%29%2Fsqrt%282%29 https://www.youtube.com/watch?v=Z49hXoN4KWg

[WhiteBlackGoose]

I smell https://github.com/asc-community/AngouriMath/issues/204 here

[Happypig375]

Still applies for i^(1/2)=(1+i)/2^(1/2). https://www.wolframalpha.com/input/?i=i%5E%281%2F2%29%3D%281%2Bi%29%2F2%5E%281%2F2%29

[WhiteBlackGoose]

That's not what I meant. There are two numbers a and b such that ... = a^2 = b^2. But if we say that sqrt(...) = a and sqrt(...) = b, we would imply a = b, while a = -b. This is a problem of how we should think about it