Indeed, usually you would want to avoid a notation of sqrt(-1) or (-1)^(1/2). You would use e^(1/2 log(-1)) instead because mathematicians have already decided on a “natural” way to define the logarithm of complex numbers. The problem here lies with choosing a branch of the logarithm as e^z = x has infinitely many complex solutions z. Mathematicians have already decided on a default branch of the logarithm you would usually use. This matters because depending on the branch you choose sqrt(-1) either gives i or -i. A square-root is usually defined to only give the positive solution (if it had multiple values it wouldn’t fit the definition of a function anymore) but on the complex plane there isn’t really a “positive” direction. You would have to choose that first to make sure sqrt is defined as a function and you do that via the logarithm branch.
So, just writing sqrt(-1) leaves ambiguity as you could either define it to give i or -i but writing e^(1/2 log(-1)) then everyone would just assume you use the default logarithm branch and the solution is i.
Indeed, usually you would want to avoid a notation of
sqrt(-1)or(-1)^(1/2). You would usee^(1/2 log(-1))instead because mathematicians have already decided on a “natural” way to define the logarithm of complex numbers. The problem here lies with choosing a branch of the logarithm ase^z = xhas infinitely many complex solutionsz. Mathematicians have already decided on a default branch of the logarithm you would usually use. This matters because depending on the branch you choosesqrt(-1)either givesior-i. A square-root is usually defined to only give the positive solution (if it had multiple values it wouldn’t fit the definition of a function anymore) but on the complex plane there isn’t really a “positive” direction. You would have to choose that first to make suresqrtis defined as a function and you do that via the logarithm branch.So, just writing
sqrt(-1)leaves ambiguity as you could either define it to giveior-ibut writinge^(1/2 log(-1))then everyone would just assume you use the default logarithm branch and the solution isi.