What is $\sqrt{(-1)^2}$

Question

This question is primarily terminology based. In that $\sqrt{}$ denotes the principal square root. Here are two reasoning

$\sqrt{(-1)^2}=1$ since $\sqrt{(-1)^2}=\sqrt{1}$ which we know has a principal square root of $1$.
Or $\sqrt{(-1)^2}=-1$ since $\sqrt{(e^{\pi i})^2}=\sqrt{e^{2\pi i}}=e^{\pi i}=-1$

Which reasoning is correct and why? Also if possible can you leave a source.

$\sqrt{a^2}=|a|$ in general, for $a$ a real number. Try to avoid using complex numbers with $\sqrt{z}$, unless you want to abandon a lot of the usual features of $\sqrt{z}$, or you want to use a multi-valued function. — Thomas Andrews, Jun 20 '15 at 05:29
The issue with the second derivation is that it is not generally true that $\sqrt{e^x}=e^{x/2}$; this holds for real $x$ but for complex $x$ it can fail if you interpret $\sqrt{}$ as the principal square root. — Eric Stucky, Jun 20 '15 at 05:36

score 3 · Answer 1 · answered Jun 20 '15 at 05:35

3

The first one is correct.

In the second one, $\sqrt{e^{2\pi i}}=e^{\pi i}$ is wrong. Should be $\sqrt{e^{2\pi i}}=|e^{\pi i}|=1$ (This stands of course under that assumption of knowing $e^{\pi i}$ is real).

answered Jun 20 '15 at 05:35

Indominus

1,166

score 1 · Answer 2 · edited Jun 20 '15 at 06:08

1

$\sqrt{(-1)^2} = \sqrt{1} = 1$

edited Jun 20 '15 at 06:08

Jyrki Lahtonen

133,153

answered Jun 20 '15 at 05:29

DeepSea

77,651

The last equality is incorrect. – guest Jun 20 '15 at 05:59
@guest: The OP gave a correct answer. A misguided editor added the $\pm$ in there. – Jyrki Lahtonen Jun 20 '15 at 06:09
1

@HarishChandraRajpoot. Why did you edit this and replaced a correct answer by an incorrect one? – Jyrki Lahtonen Jun 20 '15 at 06:10
Actually, there was no information about the question. It was simply $\sqrt{(-1)^2}=\sqrt{1}=?$ So. mathematically, I corrected it as $\pm 1$ but for this particular question, answer is absolutely $1$ as you gave. It isn't my fault but still I was misled by the statement only. – Harish Chandra Rajpoot Jun 20 '15 at 06:16
@JyrkiLahtonen thanks for the clarification. – guest Jun 20 '15 at 08:03

score 0 · Answer 3 · answered Jun 20 '15 at 05:41

0

$(e^{2\pi i})=(\cos(2\pi)+i\sin(2\pi))=1+0i=1$

$\sqrt{1}=1$

answered Jun 20 '15 at 05:41

tzamboiv

454

score 0 · Answer 4 · answered Jun 20 '15 at 06:08

0

according to the absolute definition
$$\sqrt{x^2}=|x|=|-1|=1$$

answered Jun 20 '15 at 06:08

What is $\sqrt{(-1)^2}$

4 Answers4