square root of a complex number?

[apple2357]

This is probably one for Dr. Peterson but imagine others might be interested....
In this bit ( from 14 years ago!) http://mathforum.org/library/drmath/view/66677.html

you write about the square root defined to only refer to the positive square root so that we have a function and avoid ambiguity. This makes total sense. You then talk about that this is not possible when thinking about the square root of a complex number and i am not sure why?

For example, the square root of 8-6i is either + or - (3-i). Why can't we choose to define this as +(3-i) and keep the square root as a function into the complex world too?

 

[JeffM]

This, I believe, is the result of a confusing nomenclature based on the historical development of mathematics. The word "root" was being used by mathematicians in its current meaning long before people were thinking about negative or complex numbers.

Obviously, plus 4 has two real square roots, plus 2 and minus 2. But of course we define a function as having a unique resultant.
Thus, if we want to talk about a square root function, we must choose either the positive or negative root. Because we frequently are interested only in the positive root (for example in the Euclidean distance formula), the mathematical community calls the non-negative root the principal square root and defines the principal square root the resultant of the square root function when talking about real numbers.

Things get a little weird with complex numbers. The definition of i is frequently [MATH]MISSTATED[/MATH] as

[MATH]i = \sqrt{-\ 1}[/MATH].

The concepts of positive and negative do not apply to complex numbers. Therefore we cannot determine the meaning of that definition because there are 2 square roots of minus 1, and neither is non-negative.

Mathematicians could have decided to define a principal square root of minus 1, but they have not done so. I myself do not know whether the reason is that the attempt to do so causes difficulties somewhere, that there is no practical need to use one rather than the other, or both.

http://mathworld.wolfram.com/PrincipalSquareRoot.html

Or as wikipedia says at https://en.wikipedia.org/wiki/Imaginary_unit

The imaginary number i is defined solely by the property that its square is −1:

[MATH]i^{2}=-1[/MATH]
With i defined this way, it follows directly from algebra that i and −i are both square roots of −1.

 

[apple2357]

Ok, so we simply cannot talk about positive/ negative complex numbers?

 

[pka]

This is probably one for Dr. Peterson but imagine others might be interested....

I realize that I am not Prof Peterson. But I did teach complex variables for forty years . So...

you write about the square root defined to only refer to the positive square root so that we have a function and avoid ambiguity. This makes total sense. You then talk about that this is not possible when thinking about the square root of a complex number and i am not sure why?
For example, the square root of 8-6i is either + or - (3-i). Why can't we choose to define this as +(3-i) and keep the square root as a function into the complex world too?

Because the complex field is not ordered we do not talk about positive or negative roots.
However iit is true that every non zero complex number has two square roots.
Let's take the example you suggest: \(\displaystyle 8-6\bf{i}\) That is not one of those "nice" numbers we love to use as examples.
So we will assign its argument as \(\displaystyle \arg(8-6\bf{i})=\theta=-\arctan\left(\frac{3}{4}\right)\)
The modulus or absolute value is \(\displaystyle |8-{6\bf{i}}|=10\) So the polar form is \(\displaystyle 10\exp(\theta{\bf{i}})\)
The ease of this notation is clear here, the two square roots are \(\displaystyle \sqrt{10}\exp\left(\frac{\theta i}{2}\right)~\&~\sqrt{10}\exp\left(\frac{(\theta+\pi) i}{2}\right)\)

 

[Dr.Peterson]

This is probably one for Dr. Peterson but imagine others might be interested....
In this bit ( from 14 years ago!) http://mathforum.org/library/drmath/view/66677.html

When you want to interact specifically with the people from Ask Dr. Math, you can go to the new site https://www.themathdoctors.org/ask/ , to which some of us have moved. But this is, indeed, a question of general interest. And pka is definitely more qualified than I on complex variables. (That particular question was not asked at that level.)

you write about the square root defined to only refer to the positive square root so that we have a function and avoid ambiguity. This makes total sense. You then talk about that this is not possible when thinking about the square root of a complex number and i am not sure why?

For example, the square root of 8-6i is either + or - (3-i). Why can't we choose to define this as +(3-i) and keep the square root as a function into the complex world too?

It is possible to define something called the principal square root; see http://mathworld.wolfram.com/nthRoot.html.

What's not possible is to define a principal square root that behaves as simply as what we are familiar with from real numbers. As has been said, you can't just talk about "positive" complex numbers; instead, you have to arbitrarily choose some range of arguments (angles). This will be a function; it's just not as nice as one would wish. The key word in my comment is "consistently".

To check your claim that the square root of 8 - 6i is ±(3 - i), we can square the latter: (3 - i)^2 = 9 - 6i - 1 = 8 - 6i. So you are correct there. (And pka's answer can be simplified to that form, with a little work.)

Note that there is nothing "positive" about 3 - i; you could just as well say that i - 3 deserves that name as opposed to -(i - 3). In fact, by the definition mentioned in Wolfram, the principal square root of 8 - 6i is -3 + i. (Can you see why?)

Ultimately, it is an arbitrary choice what you call the principal root, and, as I said, any choice you make doesn't make everything work happily ever after, in particular the fact that [MATH]\sqrt{ab} = \sqrt{a}\sqrt{b}[/MATH], which is a primary reason for the choice of the positive real root.

 

[JeffM]

Ok, so we simply cannot talk about positive/ negative complex numbers?

Correct.

There are a number of ways to explain this. The most intuitive for me is to recognize that we do not have a complex number line. Instead we have a complex number plane (the Argand plane). So 4 > 2 on the real number line, and 3 < 7 on the real number line,
Consider then 4 + 3i and 2 + 7i. Does the fact that 4 > 2 or the fact that 3 < 7 control.

When I think about a number line that is horizontal and say a < b, I mean that a is to the left of b. When I think about a number line that is vertical and say a < b, I mean that a is below b. But a point on a plane is defined relative to a horizontal and a vertical axis, and a may be to the left of b with respect to the horizontal axis, but a may be above b with respect to the vertical axis. A concept that was useful in the simplicity of one dimension is not necessarily useful in the greater complexity of two dimensions.

 

[apple2357]

Note that there is nothing "positive" about 3 - i; you could just as well say that i - 3 deserves that name as opposed to -(i - 3). In fact, by the definition mentioned in Wolfram, the principal square root of 8 - 6i is -3 + i. (Can you see why?)

I am thinking..

 

[Dr.Peterson]

Here's the sentence I was referring to:

For complex numbers z, the root of interest (generally taken as the root having smallest positive complex argument) is known as the principal root.​

 

[pka]

For complex numbers z, the root of interest (generally taken as the root having smallest positive complex argument) is known as the principal root.

This is a tricky question as to what is a principal root. As an undergraduate eons ago. I learned from Marie Weiss's text about roots of unity.
I have used the basic idea in all my classes sinse . I have debated myself as to using a general example or a particular. I came down on a particular due to the nature of this site. So suppose \(\displaystyle z=-3+4\bf i\). Let's say \(\displaystyle \theta=\arctan\left(\frac{4}{-3}\right)\)
Moreover, suppose that the task is to find the six sixth roots of \(\displaystyle z\). The polar form of \(\displaystyle z\) is \(\displaystyle 5\exp(\theta{\bf i})\).
Now define \(\displaystyle \tau =\sqrt[6]{5} \exp\left(\frac{\bf i \theta}{6}\right)\) Now it should be clear that \(\displaystyle \bf\tau^6=z\) Thus we have the principle value.
To go further: define \(\displaystyle \rho=\exp\left(\frac{2\pi\bf i }{6}\right)\) so now the set \(\displaystyle \{\tau\cdot \rho^k,~k=0,1,\cdots,5\}\) is the set of the six sixth roots of \(\displaystyle z=3+4\bf i\)
BUT most importunately we can define \(\displaystyle \tau\cdot \rho^0=\tau\) as the principal sixth root.

 

[apple2357]

Thank you all for your contributions. it has given me plenty to think about.

So i am left thinking despite the possibility to defining a principal square root in the complex world we choose not to when doing so.

So square root is a function when working in the real world, i.e. sqrt(4) = 2 ( single output) but when we move into complex numbers
sqrt (8-6i) = +-(3-i) ( two outputs), even though we could retain the notion of a function by picking the output as -3+i ( the principal square root which has the smallest positive argument).

Is that a correct summary?

 

[JeffM]

Thank you all for your contributions. it has given me plenty to think about.

So i am left thinking despite the possibility to defining a principal square root in the complex world we choose not to when doing so.

So square root is a function when working in the real world, i.e. sqrt(4) = 2 ( single output) but when we move into complex numbers
sqrt (8-6i) = +-(3-i) ( two outputs), even though we could retain the notion of a function by picking the output as -3+i ( the principal square root which has the smallest positive argument).

Is that a correct summary?

No.

All positive real numbers have two real square roots. Zero has one real square root. Negative real numbers have no real square roots.

All complex numbers have two complex square roots, except for zero, which has one square root.

But functions, by definition, have unique resultants. So we must chose which root is to be defined as the resultant of the square root function.

EDIT: I want to thank Dr. P and pka for having clarified my own understanding of the difference between the square root function and the principal square root.

 

[Dr.Peterson]

Thank you all for your contributions. it has given me plenty to think about.

So i am left thinking despite the possibility to defining a principal square root in the complex world we choose not to when doing so.

So square root is a function when working in the real world, i.e. sqrt(4) = 2 ( single output) but when we move into complex numbers
sqrt (8-6i) = +-(3-i) ( two outputs), even though we could retain the notion of a function by picking the output as -3+i ( the principal square root which has the smallest positive argument).

Is that a correct summary?

I would say that (a) there is always a square root relation (which is not a function); (b) in the reals, we additionally define a square root function by taking the non-negative root, and use that consistently; (c) in the complex numbers, we can define a principal root, but that is not typically used as "the" square root, because it is less useful. I imagine pka can say more about the reality of that within the field of complex analysis.

I left you another trail to follow by mentioning that [MATH]\sqrt{ab} = \sqrt{a}\sqrt{b}[/MATH] does not hold for all complex numbers under any particular definition of a principal value, implying that you might look for a counterexample under the definition I mentioned. You might find this interesting to experiment with.