Basic exponent problem from facebook

Question

Given :

$$\sqrt{\sqrt{(-4)^2}}$$

a) $2$

b) $-2$

c) $\pm2$

d) $2i$

e) $-2i$

f) $\pm 2i$

My opinion, the answer is $2$. Because the result of the root can't be negative. For example if we have $\sqrt x = -2$, the answer is there's no solution right?

But, there's something that bothers me if i express them like this :

$$\left((-4)^{\frac 2 2}\right)^{\frac 1 2}$$

It gives me $2i$ instead. To disagree with this answer, i'm thinking i should do $(-4)^2$ first. So the answer will be $2$.

Actually i'm too embarrassed to ask this. But, i hope i can get new knowledge from you guys that changes my mind.

Answer 1

The answer is $2$ because, when $x\in[0,\infty)$, $\sqrt x$ is the only non-negative real number such that $\sqrt x^2=x$. So\begin{align}\sqrt{\sqrt{(-4)^2}}&=\sqrt{\sqrt{16}}\\&=\sqrt2\\&=2.\end{align}

When you think in terms of complex numbers, don't use the equality $(a^b)^c=a^{bc}$. It is tempting and, yes, it works for positive real numbers. But it doesn't even make sense in general. What is $a^{1/2}$? It is a square root of $a$, yes, but which one?

Answer 2

When $x$ is negative, $x^{\frac{1}{2}}$ is ambiguous, and the rules of exponents, such as $(x^2)^{\frac{1}{2}} = x^{2 \cdot \frac{1}{2}} = x$, do not necessarily apply. The correct answer is $2$.

Note: This is similar to how, for example, the rules of surds (radicals) do not apply to negative numbers. Otherwise, you could write $1 = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot\sqrt{-1} = i \cdot i = -1$, which is wrong. (In fact, even Euler was puzzled by this at the time when the theory of complex numbers was relatively new).