0

I have just started learning complex numbers and am now confused about one property.

We know that $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$ is only valid if both $a$ and $b$ are not negative simultaneously. My teacher told me that the relationship breaks if both the numbers are negative to solve the following contradiction.

$\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)(-1)}=1$

Now, that makes me wonder if the relationship also holds if $a$ or $b$ are complex numbers with non-zero imaginary parts. Any help would be greatly appreciated.

Asher2211
  • 3,406
  • 10
  • 31
Habib
  • 145
  • $\sqrt{-a}\cdot \sqrt{-b}=\sqrt{a}\cdot{i}\cdot\sqrt{b}\cdot{i}=\sqrt{ab}\cdot(-1)=-\sqrt{ab}$, where $a,b$ are positive real numbers. – E. Huang Jun 21 '21 at 03:34

1 Answers1

2

Taking "to the $0.5$" power makes much less sense for complex numbers than for real numbers. For real numbers, you can get away with defining square roots by choosing the positive square root. But in the complex numbers, how do you decide if $-3+4i$ or $3-4i$ is the square root of $-7 - 24i$? You can't do this in a continuous way, and so asking about such relations is difficult to make sense of, since you need to specify how you're square rooting.

  • No, the square root $\sqrt z$ is a bi-valued function (other than for $z=0$). – Ted Shifrin Jun 21 '21 at 04:21
  • @TedShifrin I'm not sure but what I've been told in the past is that the convention is to use $z^{\frac12}$ to represent the multivalued function and $\sqrt{z}$ to represent a particular branch of the branched power function. However I could very well be wrong. – Stephen Donovan Jun 21 '21 at 06:00
  • @StephenDonovan That may have been your professor's convention, but I’ve never heard of it. And if you choose any branch of $\sqrt z$ the multiplicative property will fail to hold in general. – Ted Shifrin Jun 21 '21 at 06:05