Why this book says that $ 2^{1/2} = ±\sqrt{2} $?

Question

Shouldn't it be: $ 2^{1/2} = \sqrt{2} $ ?

I know the problem is that there they are working with complex numbers, but I still don't understand.

The book is in the link, page 113, when they move from the claim 1 to 2. http://www.math.ucsb.edu/~wei/teach/122/Brown-Churchill-Complex%20Variables%20and%20Application%208th%20edition.pdf

In the first example they put, I do not know how they come to this:

$-iLog(1±\sqrt{2})$

Answer 1

In complex analysis, $\operatorname{Log} z = \log |z|+i\arg\theta$ is usually multivalued, so $x^{b} = e^{b\operatorname{Log} a}$ is, in general, multivalued as well.

In this case,

$$A = 2^{1/2} = e^{1/2 \operatorname{Log}2} = e^{1/2 (\log2 + 2 \pi i n)} = (-1)^n \sqrt{2}$$

i.e.

$$A = \pm \sqrt{2}$$