The root of -1 is i. But what is the root of i? Ie what x is such that x^2 = i.

Do you need a new variable, say j? Does this recurse infinitely?

No new numbers are needed, the complex numbers are algebraically closed (any polynomial of degree n with complex coefficients has n complex roots, counting multiplicities). Just use demoivre, the square roots of i are (1+i)/sqrt(2) and it's conjugate.

Just a minor correction, the two roots of i are (1+i)/sqrt(2) and -(1+i)/sqrt(2). Note that the conjugate of (1+i)/sqrt(2) is (1-i)/sqrt(2) and it is a root of -i, not i.

By the way, if you want to verify that (1+i)/sqrt(2) is the square root of i, just square it:

(1+i)(1+i)/2 = (1 + 2i + i²)/2 = (1 + 2i - 1)/2 = 2i/2 = i

Just a minor correction, the two roots of i are (1+i)/sqrt(2) and -(1+i)/sqrt(2). Note that the conjugate of (1+i)/sqrt(2) is (1-i)/sqrt(2) and it is a root of -i, not i.

Yes, thank you for the correction. I had a moment of stupidity and was planning to unnecessarily talk about the 4th roots of -1, followed by a larger moment of stupidity and lazy rewriting after abandoning this poor idea.