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I am a little bit bummed that I have this question as I'm sure it has been asked before (I couldn't find the answer) but...

If $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only true for positive reals $a$ and $b$. Then what allows us to say the following? $$\sqrt{-a} = \sqrt{-1\cdot a} = \sqrt{-1}\sqrt{a} = i\sqrt{a}$$

I don't know what allows the second equal sign. Is this just convention?

Blue
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Chris Christopherson
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  • I think you are most likely to meet $\sqrt{-a}$ when solving quadratic equations. In that case you always have $\pm\sqrt{-a}$ anyway, which means it doesn’t matter which square root you choose, because the other one will be included too. – Milten Oct 19 '21 at 06:19

3 Answers3

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The square root function is usually understood by convention to return the "principal" square root of a complex number. For real numbers, this just means that the square root of a positive real is a positive real and the square root of a negative real is imaginary with positive imaginary part. Under this convention, it happens to be valid that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if $a$ and $b$ are real with at least one of them nonnegative. But the equality fails if both are negative. For example $$ 1=\sqrt{(-1)(-1)}\neq\sqrt{-1}\sqrt{-1}=i\cdot i=-1. $$

subrosar
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  • Ahh ok so it really is convention. Based off what you are saying, for example, the following holds yes? $$\sqrt{-12} = \sqrt{-3\cdot 4} = \sqrt{-3}\sqrt{4} = i\sqrt{3}\cdot 2 = 2i \sqrt{3}?$$ You are saying that this third "=" IS valid? Further question, does this idea extent to $$\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}?$$ – Chris Christopherson Oct 19 '21 at 00:51
  • @ChrisChristopherson Yes, yes, and yes. – subrosar Oct 19 '21 at 01:56
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Be aware of the setting $i=\sqrt{-1}.$ The law $\sqrt{a}\sqrt{b}=\sqrt{ab}$ does only apply to real numbers. It is wrong for complex numbers because $$ -1= i^2=i\cdot i =\sqrt{-1} \cdot \sqrt{-1}\neq \sqrt{(-1)^2}=1 $$ The best way to see it is in my opinion to operate with the equation $i^2=-1$ only; not roots. They are an abbreviation for real numbers.

Another possibility is to use Euler's formula $-1=e^{i \pi}.$ Thus $$ -1=e^{i\pi}=\left(e^{i\pi/2}\right)^2=i^2\text{ and }\sqrt{(-1)^2}=\sqrt{\left(e^{i\pi}\right)^2}=\sqrt{e^{2i\pi}}=\sqrt{1}=1 $$ This uses the power $1/2$ instead of roots and you avoid formulas which simply do not hold for complex numbers.

Marius S.L.
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  • Yes this is why I am being cautious and asking the question. Although, the reason I am asking the question is to be able to explain it to an Algebra 2 class. Unfortunately, Euler's famous equation is beyond the curriculum. – Chris Christopherson Oct 19 '21 at 00:55
  • Also I too prefer the definition $$i^2 = -1$$ but does this not imply that $$i=\pm \sqrt{-1}?$$ – Chris Christopherson Oct 19 '21 at 00:56
  • The square root is ambiguous for complex numbers and suggest formulas which simply do not hold. $i^2=-1$ means $i^2=(e^{i\pi/2})^2=e^{i\pi }=-1$ so $\sqrt{i^2}=e^{i\pi/2}=i$ or $\sqrt{i^2}=e^{i(\pi/2+2k\pi)}$ to be exact. But neither of these values yields $i\stackrel{?}{=}-\sqrt{-1}$ because $i\neq -i.$ – Marius S.L. Oct 19 '21 at 01:11
  • Woah good to know so $y^2=x$ does not imply $y=\pm\sqrt{x}$ for complex $y$ and $x$? – Chris Christopherson Oct 19 '21 at 01:19
  • This is the difficulty with $\pm \sqrt{.}$ over complex numbers. It leads to confusion. It is ok for real numbers, but it is not for complex numbers. $z=\sqrt{y+ix} $ normally refers to the main branch of the square root as $\sqrt{4}$ refers to $+2.$ $i=\pm\sqrt{-1}$ would assign two values to one number, a specific algebraic number, namely one of the solutions to $x^2+1=0.$ How can it be two numbers? It is really best to work with Euler's formula as standard calculus. Better to think of $\sqrt{z}$ as $z^{1/2}.$ – Marius S.L. Oct 19 '21 at 01:33
  • Hmmm I'm still a little lost. Are you saying that $z^2 = y+ix$ for real $x$ and $y$ is a well defined function in the complex world? – Chris Christopherson Oct 19 '21 at 01:42
  • Let's see: set $z=:r\cdot e^{i\varphi }.$ Then $z^2=r^2e^{2i\varphi }.$ The same goes for the RHS so $x+iy=s\cdot e^{i\omega }.$ Our solution is now $r^2=s$ with real numbers $r,s$ and $e^{2i\varphi}=e^{i \omega }.$ Thus $e^{i[2\varphi -\omega ]}=1$ which is true for all $2\varphi = \omega +2k\pi$ for all $k\in \mathbb{Z}.$ This is why we have many possible solutions. In the complex number plane we only see the points $z=\pm \sqrt{s}\cdot e^{\omega /2}$ since all terms $e^{2k\pi}$ are rotations on the unit circle. ... – Marius S.L. Oct 19 '21 at 02:03
  • .. But it is the reason why complex calculus is very different from real calculus. If you would write $x$ instead of $i$ and complex numbers as linear polynomials, and only use $x^2=-1,$ then you would not get confused by $\sqrt{x}.$ – Marius S.L. Oct 19 '21 at 02:03
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$\DeclareMathOperator{\Log}{Log} $ Let $a \in \mathbb R$ and define a square root via $\sqrt{z} := \exp(\tfrac 12\Log z)$ with $\Log z$ having a branch cut along the negative imaginary axis, i.e. $\Log(z) := \log|z| + i \theta$ with $\log$ the real logarithm and $\theta \in [-\pi/4, 3\pi/4)$. Then your formula above says: $$\exp(\tfrac 12\Log(-a)) = \exp(\tfrac 12\Log(-1\cdot a))= \exp(\tfrac 12\Log(-1))\exp(\tfrac{1}{2}\Log a)=i\exp(\tfrac{1}{2}\Log a)$$ Thus the two things to check here are that $\Log(-1\cdot a) = \Log(-1) + \Log a$ and that $\exp(\tfrac 12\Log(-1)) = i$. For the latter, note $\Log(-1) = \log(1) + i\pi= i\pi$ so that $\exp(\tfrac 12\Log(-1))=e^{i \pi/2}=i$. For the former, note that if $x,y \in \mathbb C$ with $-\pi/4 \le \arg(x)+\arg(y)<3 \pi/4$ then $\arg(x)+\arg(y) = \arg(xy)$ (note this does not hold if we cross a branch... Can you see why? ) so we have: $$\Log(x)+\Log(y)=\log|x|+\log|y|+i(\arg(x) + \arg(y)) \stackrel{!}{=}\log(|xy|)+i\arg(xy) := \Log(xy)$$ where equality $!$ follows from the multiplicativity of the real logarithm, the fact $|x||y| = |xy|$, and the fact $\arg(x) + \arg(y) = \arg(xy)$.

Applying this result with $x = -1$ and $y = a$ we see $\arg(x) = \pi$, $\arg(y) = 0$, and $\arg(xy) = \arg(-a) = \pi$ so the above lemma holds and we are done. Note the choice of the branch along the negative imaginary axis was arbitrary; the argument above holds without any problem for any branch cut from $0$ to $\infty$ lying strictly in the open lower half-plane. If you want to choose a branch intersecting the closed upper half-plane (e.g. the standard branch) you'll have to keep track of the extra $2 \pi i$ factor added or removed every time you cross the branch, though the argument will go through in essentially the same way.

Note: As @Subrosar says, we always have $\Log(x) + \Log(y) = \Log(xy)$ when one of $x$ or $y$ is nonnegative and the choice of branch cut is a straight line from $0$ to $\infty$, and this happens simply because one of $\arg(x)$ or $\arg(y)$ is zero; assuming WLOG $\arg(x) = 0$, we have $\arg(x) = 0 \le \arg(xy) = \arg(y)$, so $xy$ does not cross the branch cut since it is simply a scaled copy of $y$. Note this fails if the branch cut is not a straight line, as it could be that $y$ and $xy$ lie on separate sides of the branch cut.

Brevan Ellefsen
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