Why isn't $i * i = 1$?

Question

When we studied complex numbers they told us that $i * i = -1$ because $i = \sqrt -1$ and $i * i = i^2$, so the square removes the root.

However we can say as well that $i * i = \sqrt {-1} * \sqrt {-1} = \sqrt {-1 * -1} = \sqrt 1 = 1$.

In any case both are valid math, right?

Why can't I follow the second reasoming?!

Answer 1

You've learned the wrong definition of $i$. The true definition of $i$ is that $i\cdot i=-1$, and that's it. No $\sqrt{\phantom{-1}}$ in sight.

In fact, in my opinion, $\sqrt{\phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $\sqrt{ab}=\sqrt a\sqrt b$.

Answer 2

To be precise,

• you are right to state $i\cdot i=i^2=-1$, which is essentially a definition of $i$;

• $i=\sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-\sqrt{-1}$);

• $(\sqrt{-1})^2=-1$ also works, by definition of the square root;

• $\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $\sqrt a\sqrt b=\sqrt{ab}$;

• finally, $\sqrt1=1$ remains true.