My maths teacher claims that $(\sqrt{-4})^2$ is not defined if we consider only those numbers in the real number plane.
Even if we exclude imaginary numbers, IMO the statement can be written as
$$-4^{(2 \times \frac{1}{2})}$$
Then the powers obviously cancel and we are left with $-4$. Even using WolframAlpha results in $-4$.
According to Wikipedia, $(b^n)^m = b^{nm}$ for all $b \neq 0$. This is what makes me not believe my teacher's claim, even if we are not considering complex numbers.
My teacher's logic is that we need to first evaluate $\sqrt{-4}$ before we can proceed with squaring the result. He also claims that the $(b^n)^m = b^{nm}$ identity is true only for $b > 0$, contradicting what Wikipedia states.
So, my two questions, summarized:
- Is $(\sqrt{-4})^2$ defined if we do not consider complex numbers?
- Is $(b^n)^m = b^{nm}$ for only $b > 0$ or $\forall$ $b \neq 0$?
$\sqrt{-4}$ is not defined in real numbers, so neither is $\sqrt{-4}^2$
That Wikipedia page says $(b^n)^m=b^{nm}$ for integer exponents $m$ and $n$.
When $b<0$, it may not hold for fractional exponents such as $\frac12$ (as in your case).