Do you believe in √(-1)?

[muon2]

When I was young, I found Isaac Asimov's rebuttal to the unreality of imaginary numbers one of the most compelling. The story has stayed with me throughout my academic career, but I am afraid I might not get it all verbatim. So, I found this rendition of it from his later autobiography, In Memory Yet Green (1979):

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Mathematics deals in abstract quantities that include thing like zero, negative numbers, variables, functions, groups, vector spaces, algebras, and a host of other concepts. One of them happens to be complex numbers which includes the square roots of -1, and for historical reasons they were dubbed imaginary numbers. They are no more imaginary than any of the other abstract objects in mathematics.

[muon2]

No.  Imaginary numbers are, by definition, imaginary, i.e. products of the imagination.  Therefore, they are not real; if they were, they would be real numbers.  Since the square root of -1 is clearly not a real number but an imaginary number, then by its very definition it cannot exist.

The name has little to do with its meaning and more to do with its history. That would be like saying a strange quark has less reality than an electron because of its fanciful name.

But I would claim the imaginary numbers to be just as functional as any real number. If you believe that a line segment can be rotated in a plane then you have worked with imaginary numbers, just not in their algebraic form. For instance, consider the line segment in the 2-dimensional plane from (0,0) to (3,4). A 90 degree rotation around the origin takes the other end to (-4,3). But this is just multiplying the 2-dimensional point by the imaginary unit i. If I wrote it with algebra it would read (3+4i)i = (-4+3i), and I would have used imaginary numbers.

[muon2]

No. I believe in quaternions.

Hamilton's greatest achievement. Too bad we quit teaching and using them in favor of vector and matrix formalism. Matrices always work as a representation, but the beauty of simple elements like quaternions directly shows symmetry.

[muon2]

No.  Imaginary numbers are, by definition, imaginary, i.e. products of the imagination.  Therefore, they are not real; if they were, they would be real numbers.  Since the square root of -1 is clearly not a real number but an imaginary number, then by its very definition it cannot exist.

The name has little to do with its meaning and more to do with its history. That would be like saying a strange quark has less reality than an electron because of its fanciful name.

Imaginary numbers were therefore called lateral numbers after their discovery.
They are nowadays called "imaginary" because they are too complex for too many people to understand... 🤓

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯

Why would the result seem so unusual? Is it because there is more than one value in the answer?

The natural log is the inverse of the exponential, and is often used to find the result of a real as well as a complex exponential. Since complex exponentials are equivalent to certain trigonometric functions, the complex logarithm behaves in part like an inverse trig function. Inverse trig functions with real arguments have multiple values for the answer, offset by 2pi. Similarly offsets by 2pi appear in the complex logarithm, so it may have multiple values. Hence complex exponentiation may also have multiple values as in this case.

[muon2]

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯

Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.

But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

[muon2]

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯

Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.

But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i ⁱ.

As am I. My point is that i ^z is real for a number of values of z, and is defined based on its exponential to a specific real value of z = 2. Why is is surprising that there are many other values of z, including complex values such as i, that also give real results?

If the problem is one of visualization, then I have to start by asking how you visualize i ². If it is only as an algebraic formula, I can see where that is hard to extend. If instead you visualize points in a plane that can shift and rotate then it is easier to see how some can end up on the real axis.

[muon2]

It still blows my mind, though, is the exponentiation of √(-1) with itself. 🤯

Why would the result seem so unusual? Is it because there is more than one value in the answer?

No, because the result is a real number. The proof of that is very easy to understand, but it's hard to visualize why the result comes into being.

But it is defined such that exponentiation results in a real number - square it and you get -1, fourth power gives +1. Why would other powers, including i, also being real be a surprise?

I was talking about i ⁱ.

As am I. My point is that i ^z is real for a number of values of z, and is defined based on its exponential to a specific real value of z = 2. Why is is surprising that there are many other values of z, including complex values such as i, that also give real results?

If the problem is one of visualization, then I have to start by asking how you visualize i ². If it is only as an algebraic formula, I can see where that is hard to extend. If instead you visualize points in a plane that can shift and rotate then it is easier to see how some can end up on the real axis.

It's just too hard to understand for me what "to the power of i" even means.
Is there even a real number x that, exponentiated with itself, delivers the same result as iⁱ = x^x?

Philosophically it should be just as hard to understand what it means "to the power of π (pi)". Exponentiation by integers is natural in that 5³ means 5 x 5 x 5. Exponentiation by fractions can be inferred from the product rule: x^ax^b = x^a+b so 5^1/2 means the square root because 5^1/2 5^1/2 = 5¹ = 5. But if a or b is an irrational number, then it isn't at all clear what it means in an exponent.

Exponentiation to an irrational power can be defined one of two ways. It can be found as the limit of powers to an infinite sequence of rational numbers that converge to the real number. That's fine if you grasp infinite convergent sequences, but philosophers have had problems with that going back to Zeno's paradoxes. The alternative is to recognize that exponentiation is the inverse of a logarithm, so you can find 5^π = e^{π ln 5} by using a table of natural logarithms and inverses. But from your statement, it seems this doesn't cause you any philosophical problems.

So if exponentiation and logarithms are inverse and you accept Euler's formula from knowledge of the infinite series of the trig functions, then there should be no philosophical problem with treating 5ⁱ as one would 5^π. e ^{i ln 5} = cos(ln 5) + i sin(ln 5). Here the table of natural logs is augmented by tables of sine and cosine, and the knowledge that a complex exponential can produce a complex value.

To answer your last question, the principal value of iⁱ is approximately 0.2079. With a little calculus it can be shown that the lowest real value for x>0 is at (1/e)^(1/e) = 0.6922 and x^x is not generally defined with real number values for x<0. So there is no real value x that solves x^x = iⁱ.