Complex Numbers, Understanding Square Roots

Question

Since $i^2 = -1$, then doesn't $i = \pm \sqrt[2]{i}$? How does $i$ only equal the plus part?

Answer 1

We cannot say $i = \pm \sqrt[2]{-1}$ without first defining $\sqrt[2]{-1}$. It's true that if you have a number $a$ such that $a^2 = -1$, then $b^2 = -1$ for $b=-a$. Whether we choose to give $a$ the name $i$ (and then $b=-i$) or give $b$ the name $i$ (then $a = -i$) is arbitrary and the mathematics that ensue are equivalent.

Answer 2

It's actually $\pm i=\sqrt {-1}$, not $i=\pm \sqrt {-1}$.

For any complex number $z$ and any positive integer $n$, $z^{1/n}$ has $n$ distinct values in the set of complex numbers. e.g. $1^{1/3}$ has three distinct values : $1,\omega,\omega^2.$ Similarly $(-1)^{1/2}$ must also have two distinct values, $i$ and $-i.$ For an imaginary number, it is meaningless to say it is positive or negative, so you cannot say " $i=\sqrt {-1}$, taking the positive square root" (as we normally do for the square root of a positive real number).

By a pure convention one may write $i = \sqrt{-1}$, but this convention is not logically tenable from various considerations, and mathematically there is no need for such a convention. The symbol $i$ standing for the unique complex number $(0,1)$ and understood by $i^2=-1$ is good enough for any mathematical purpose.