How to evaluate absolute square $|i|^{2}$

Question

By definition, the complex conjugate of a complex number $z$ is $\bar z$, and is usually found by changing the sign of the complex part.

Example: $|x+iy|^2 = (x+iy)(x-iy)=x^2+y^2$

So what about $|i|^2$? Is it $-1$ or $1$?

Answer 1

As noted in the comments, $|i|=1$ and so $|i|^2 = 1$.

Recall that $|x+iy|^2 = x^2 + y^2$. $i$ fits this form for $x=0,y=1$.

If you prefer to use $|z|^2 = z \overline{z}$, then $\overline{i}=-i$, so

$$|i|^2 = i(-i) = -i^2 = -(-1)=1$$