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As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.

FIRST

\begin{align*} \sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\ &= \sqrt{16}\\ &=4 \end{align*}

SECOND

\begin{align*} \sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\ &= 4i \times i\\ &=4i^2\\ &=-4 \end{align*}

Incidentally if the above is input in complex mode of Casio scientific calculator, the result comes out as $-4$.

Which of the above solutions is correct?

Vincenzo Tibullo
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Masroor
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  • Both, since $(-4)^2=4^2=16=(\sqrt{-16}*\sqrt{-1})^2$ – cirpis Jul 23 '14 at 17:04
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    Speaking generally, $\sqrt{-16}$ and $\sqrt{-1}$ are not single numbers; they are $\pm 4i$ and $\pm i$ respectively. As such you can't multiply them like you can multiply ordinary complex numbers. Instead you have to select all 4 combinations and check what the possible values are in each of the combinations. In particular, $(4i)(i) = (-4i)(-i) = -4$ and $(-4i)(i) = (4i)(-i) = 4$. – Ian Jul 23 '14 at 17:07
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    $\sqrt[2]{16}=\pm4,\sqrt[2]{-16}=\pm4i$ – barak manos Jul 23 '14 at 17:24
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    This horse has already been beaten to death multiple times on the site, hasnt it? – Did Jul 23 '14 at 18:41
  • @Did I was just typing the same thing (down to the beaten to death comment.) having trouble finding an appropriate duplicate though... so many similar questions. – rschwieb Jul 23 '14 at 18:45

5 Answers5

5

$$\sqrt a\cdot\sqrt b=\sqrt{ab}$$ only work if $a,b\ge0$

lab bhattacharjee
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  • unless he works in the complex domain, where $\sqrt{-a}=\sqrt{a}i$ makes complete sense – cirpis Jul 23 '14 at 17:05
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    @cirpis, Even in complex domain, for $a,b>0 \sqrt{-a}\sqrt{- b}=i\sqrt a\cdot i\sqrt b=-ab$ right? and not $\sqrt{-a}\sqrt{- b}=\sqrt{(-a)(-b)}=\sqrt{ab}$ – lab bhattacharjee Jul 23 '14 at 17:07
  • @cirpis, Also, when we write $a\ge0,$ we implicitly assume $a$ to be real, right as we can not say $P+iQ\ge$ or $\le0$ for real $P,Q$ – lab bhattacharjee Jul 23 '14 at 17:10
  • well, $a$ is real. $\sqrt{a}$ isnt and we arent making a comparison of $\sqrt{a}$, so everything is fine. also, both $-\sqrt{ab}$ and $\sqrt{ab}$ are valid solutions. even when handling square roots of positive numbers, we encounter two solutions, its just a convention to take the positive square root. – cirpis Jul 23 '14 at 17:13
  • @cirpis, http://en.wikipedia.org/wiki/Principal_value#Principal_values_of_standard_functions – lab bhattacharjee Jul 24 '14 at 03:26
2

In the real numbers it's possible to define a square root function that is injective: for $x\ge0$, $\sqrt{x}$ is the unique nonnegative real number $y$ such that $y^2=x$.

In the complex numbers this is not really possible: an injective square root function can be defined, for instance declaring that $\sqrt{0}=0$ and for a nonzero number $$ x=r(\cos\alpha+i\sin\alpha)\ne0, $$ with $0\le\alpha<2\pi$ and $r>0$, defining $$ \sqrt{x}=\sqrt{r}(\cos(\alpha/2)+i\sin(\alpha/2)), $$ where $\sqrt{r}$ is the unique square root of a positive number defined above. However, this function has no good algebraic property, apart from $$ (\sqrt{x})^2=x. $$ For instance, with this definition, $\sqrt{-1}=i$, because $-1=1(\cos\pi+i\sin\pi)$, but $$ -1=\sqrt{-1\mathstrut}\sqrt{-1\mathstrut}\ne\sqrt{(-1)(-1)}=1 $$ which instead would be a decent property to have.

Indeed, if instead of the interval $[0,2\pi)$ we chose $[-\pi,\pi)$ for the argument, we would have to write $-1=1(\cos(-\pi)+i\sin(-\pi))$ and so the square root would suddenly become $-i$. Why choosing one interval for the argument and not another one? In any case the product rule wouldn't hold.

Thus $\sqrt{-16}\cdot\sqrt{-1}$ doesn't really make sense except perhaps to denote two values. But this opens other problems: how many values would $\sqrt{-4}+\sqrt{-4}$ have? Of course three: $-4i$, $0$ and $4i$. On the other hand, $2\sqrt{-4}$ would have only two, namely $-4i$ and $4i$.

You see that there's no way out. Well, there is, and is called “going to a Riemann surface”. Not at all elementary.

egreg
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1

$$\sqrt[n]{a b} = \sqrt[n]{a} \, \sqrt[n]{b} \quad (*)$$

if $a$ and $b$ are negative, then $(*)$ works only for $n$-th roots with odd $n$, alas $n = 2$ is even.

mvw
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1

The answer to each sqrt after removing i is +-, so for a simple explanation you have either (+)(+) or (-)(+) or (-)*(-), depending on the domain or context of the particular problem.

user121955
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0

The 'square root function' (the function we represent by $\sqrt{\cdot}$) is actually a branch cut of the multiple valued 'square root relation'. The typical square root function returns a complex value with an argument in $[0, \pi)$.

So $\sqrt{-16}\cdot\sqrt{-1}=4i\cdot i=-4$ because $i$ has an argument of $\pi/2$. The other value that we would reasonably see as being "$\sqrt{-16}$" is $-4i$, but this complex number has an argument of $3\pi/2$.

This pattern also persists in the real numbers. We have that $\sqrt{4}=2$. But we could also reasonably say that $\sqrt{4}=-2$. But $2$ has an argument of $0$ while $-2$ has an argument of $\pi$.