Still Confused

Steven G

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Dec 30, 2014
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Can you simplify sqrt(4/-1) for me. I get 2i and -2i depending on how I do this?
 
Simple enough: [MATH]\sqrt{-4} = \pm2i[/MATH]. Both answers are correct -- or neither is, since the square root, in the context of complex numbers, is not a single-valued function.

Or, if you choose to define [MATH]\sqrt{-1} = i[/MATH] only, as many algebra books do when they introduce imaginary numbers, then you forfeit the rule that [MATH]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/MATH] for all real numbers a, b. Then the answer has to be [MATH]\sqrt{-4} = 2i[/MATH], and not [MATH]\frac{\sqrt{4}}{\sqrt{-1}} = \frac{2}{i} = -2i[/MATH].
 
Simple enough: [MATH]\sqrt{-4} = \pm2i[/MATH]. Both answers are correct -- or neither is, since the square root, in the context of complex numbers, is not a single-valued function.

Or, if you choose to define [MATH]\sqrt{-1} = i[/MATH] only, as many algebra books do when they introduce imaginary numbers, then you forfeit the rule that [MATH]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/MATH] for all real numbers a, b. Then the answer has to be [MATH]\sqrt{-4} = 2i[/MATH], and not [MATH]\frac{\sqrt{4}}{\sqrt{-1}} = \frac{2}{i} = -2i[/MATH].
Prof, are you saying that in the context of complex numbers that \(\displaystyle \sqrt{25}\) =\(\displaystyle \pm5\) as well?
 
Prof, are you saying that in the context of complex numbers that \(\displaystyle \sqrt{25}\) =\(\displaystyle \pm5\) as well?
Of course I do not speak for Prof. Peterson, but \(\displaystyle \sqrt{25}=5\) is one number. Now twenty has two square roots they are \(\displaystyle \pm\sqrt{25}=\pm5\) For this prof in his complex variables classes this is what we use: "we add(enlargement of the reals) one number \(\displaystyle \mathit{i}\) that a solution for the equation \(\displaystyle x^2+1=0\). Notice that means \(\displaystyle -\mathit{i}\) is also a solution.
 
There are several different contexts one could talk about, and I can't speak for all of them. I'd want to see your specific context. I'm primarily concerned with the difficulties algebra students have when complex numbers are introduced, and they do things like your example. My perspective here is not that of a complex variables class.

But as I see it, if z is a complex variable, [MATH]\sqrt{z}[/MATH] is (at root, so to speak) a multi-valued function, even if it happens that [MATH]z = 25[/MATH].

It's true that we define a principal value for the square root even in complex numbers, which is a single-valued function, and that is what the symbol represents; so I'm certainly overstating my case. Yes, [MATH]\sqrt{25} = 5[/MATH] only. What's important to me is that students not forget that in using that, a somewhat arbitrary choice has been made, one result of which is that some of the rules familiar from real numbers are lost. And that's what your example shows.
 
If I'm not mistaken your original problem only has one simplification: 2i.

I think that a good way to look at this is to remember that we don't define [math]i = \sqrt{-1}[/math]. We define [math]i^2 = -1[/math].

Then [math]\sqrt{ \dfrac{4}{-1} } = \sqrt{ 4 \cdot -1} = \sqrt{4i^2} = \sqrt{4} \cdot \sqrt{i^2} = 2i[/math] using the usual "+" only definition of the square root, no matter where you put the -1 inside the radical.

-Dan
 
Dan,
Hold on a minute. If you do exactly what you said but you keep the -1 in the denominator and still use the definition i2=-1 then you get 2/i. And 2/i = -2i
 
No, there's never an i in the denominator unless you use the "quotient rule" ([MATH]\sqrt{a/b} = \sqrt{a}/\sqrt{b}[/MATH]), which doesn't apply when non-real numbers are involved (that is, when a or b are negative). He didn't use that rule, which is why his work is correct.
 
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