The original question was: How do you do xi (x to the i power), and how on Earth was it developed? There isn’t really anything to base xi on from previous rules of exponents as it is a completely new idea.
Physicist: Euler, the dude who originally came up with using the imaginary unit, 
Leonhard "Lenny" Euler. A squinty genius.
Euler’s (surprising) equation provides a way to talk about complex exponents (“complex” = “involves
So, for example (using Euler’s equation and some log properties):
So why is this formalism used, instead of some other set of rules? Like everything else in mathematics, it’s a matter of convenience and self consistency. You’d hope that the usual, old rules would apply in a natural, convenient way. For example, you’d want 
![\begin{array}{ll}x^i x^{-i}\\=\left[e^{\ln{(x^i)}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} + i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} + i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 + \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7Dx%5Ei+x%5E%7B-i%7D%5C%5C%3D%5Cleft%5Be%5E%7B%5Cln%7B%28x%5Ei%29%7D%7D%5Cright%5D%5Cleft%5Be%5E%7B%5Cln%7B%28x%5E%7B-i%7D%29%7D%7D%5Cright%5D%5C%5C%3D%5Cleft%5Be%5E%7Bi%5Cln%7B%28x%29%7D%7D%5Cright%5D%5Cleft%5Be%5E%7B-i%5Cln%7B%28x%29%7D%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5Cleft%5B%5Ccos%7B%28-%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28-%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+-+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2+-+i%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D-i%5E2%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2-i%5E2%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2+%2B+%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D1%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}x^i x^{-i}\\=\left[e^{\ln{(x^i)}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} + i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} + i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 + \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}")
(Here I’ve used: Cos(-x) = Cos(x), Sin(-x) = -Sin(x), i2=-1, and the Pythagorean identity.)
Even more important, this technique is used because it recovers the usual rules for real exponents (exponents that don’t involve 
You’ll find (at least, those people who are so inclined will find) that Euler’s equation, and in particular the method for finding imaginary exponents above, is consistent with all the rules of exponentiation. Specifically, 
how does e^(i*pi) = -1 ??
Q: Why is e to the i pi equal to -1?
Here’s another way you could think about this, based on “analytic continuation.” For the real numbers, e^x = \sum_{k=0}^\infty x^k/k! . For complex numbers, simply define e^z = \sum_{k=0}^\infty z^k/k!. That gives everything you want!
thanks, another question
whats wrong with this proof:
e^ipi=-1
-e^ipi=1
(-1)*e^ipi=e^0
e^ipi*e^ipi=e^0
e^ipi+ipi=e^0
ipi+ipi=0 (this is false)
The exponential function is multi-valued for non-real solutions. e^(2iπ) = cos(2π) + i sin(2π) = cos(0) + i sin(0) = 1. With this method you could also “prove” 0 = 2iπ = 4iπ = 6iπ = … In reality you can only take the log of both sides of the equation after line 5 when “log” is a single-valued function, or if both arguments are in the same phase (sorry don’t know the technical term)…just as acos(cos(2π))=0 not 2π.
Nitpick: You mean “The logarithm is multi-valued.” The exponential function is still single-valued, but it is horribly not injective. The technical term is “branch,” but you have the correct idea — you need to choose the values of log with the same phase.
i understand how 2pi(radians) =360=0 (degrees)
so could the 2pi be in degrees?
Just cuz i was board, I wanted to find “i” like it was a variable and got weird results
e^ypi=-1 (find y to see if it is i)
(e^pi)(e^y)=-1
(23.1407)(e^y)=-1
(e^y)=-1/(23.1407)
y=ln(-1)+ln(1/23.1407)
y= ln(-1) + (-pi)
y= ln(e^ypi) -pi
y=ypi*ln(e)-pi
y=ypi-pi; pi=ypi-y; pi= y(pi-1)
y=pi/(pi-1)=1.4669ish ( what did i do wrong?)
ops my bad, i did the second line wrong so ignore my past comment
e^ypi=-1 (find y to see if it is i)
ypi=ln(-1)
y=ln((-1)^(1/pi)) [i don’t know how to simplify further than this]
From your second line:
πy=Ln(-1)=ln(1)+i(π+2πn)=πi(1+2n) (where n is an integer)
Then, y=i(1+2n).
Okay, first of all you have to remember that the only reason “ln” makes sense for real numbers is that the exponential function is one-to-one. If you have two real numbers a and b, e^a = e^b if and only if a = b.
In the complex numbers, you no longer have this nice property. Instead, e^z = e^w if and only if Re(z) = Re(w) and Im(z) – Im(w) = 2k\pi for some integer k. Since you can find many different z and w with e^z = e^w, you cannot(!) define a logarithm function! Therefore, it makes no sense to take ln of both sides of your second equation.
To solve e^[y\pi] = -1 for y, instead you need to let y be a complex number u + iv. Then write -1 = e^[y\pi] = e^[u\pi]e^[iv\pi]. Conclude that u = 0, so that -1 = e^[iv\pi]. What can v possibly be?
Neal,
Ok that makes sense, thanks for taking your time, my math knowledge is at a low level but I am still intellectually curious. I have one more weird thing that I am confused on
e^ipi=-1
e^ipi=i^2
e^(i*(pi/2))=i
e^(i*2*pi)=i^4
535.49165^i=1 ; I understand that 2k\pi for some integer k implies here so “i” is not 0, but it acts like it. My real question is that lets say integer k was a really high negative # (like -infinity), would we then assume that 0^i=1?
Your computation is correct, although it misses the detail that the square root (you raised both sides to the 1/2 power) is, like the logarithm, not well-defined, so you implicitly made a choice about which square root to use. But your choice was consistent, so there was no problem.
Anyway, yes, e^[i2\pi] = 1. You may therefore conclude that (e^[2\pi])^i = 1.
What’s really going on behind the scenes here is not that i is (or acts like) 0, it’s that 2\pi acts like 0. Remember from the post that e^[it] = cos(t) + isin(t). What is e^[i2\pi]? e^[i4\pi]? e^[i*-500\pi]?
This answers your question! As angles, 2k\pi = 0 regardless of what k is — it can be as high negative as we want. Geometrically, it doesn’t matter how many times you’ve walked around a circle, you’ll always end up back where you started. Thus, for any high negative number, (small)^i = 1.
[I may post later with more details … this is actually pretty interesting 🙂 ]
e^(-inf*pi*i)=indeterminate [we don’t know where in the circle it stops or ever stops], but only real (non-“i”) values are 1 and -1
therefore, lets just assume that:
e^(-inf*pi*i)= positive or negative 1
0^i= positive or negative 1
[this is so interesting that I am procrastinating my work and not even feeling guilty 🙂 ]
Okay, let’s back up. It’s not clear what “0^i” means. In fact, to define complex exponents, you have to use the exponential function and the logarithm: z^w = e^[wlog(z)]. What does “log(z)” mean? It means “pick one of the preimages of z under the exponential map.”
Fact: the exponential map has an “essential singularity” at infinity. This means that as you approach infinity from any direction, you can get close to any value you want.
What does this have to do with 0^i? If you try to write down what the symbols mean, 0^i = e^[ilog(0)].
Problem: log(0) doesn’t make sense because the exponential function is always nonzero. Nothing maps to zero. So instead of log(0), you have to take log(some sequence that tends to zero). Depending on your choice of sequence and choice of branch of log, when you write down the limit, you can get 0^i = anything you want.
For example, if you take the sequence it, t\to 0, since log(it) = t + ik\pi/2 (where k is some integer), (it)^i = e^[ilog(it)] = e^[i(t + ik\pi/2)] = e^[it]e^[-k\pi/2] limits to e^[-k\pi/2], not zero, as t goes to 0. So here we see that 0^i = e^[-k\pi/2].
“0^i = e^[-k\pi/2]”
so 0^i can equal 1??? that is weird
Yep! That’s why it doesn’t make any sense to talk about “0^i” 🙂 It’s like “0/0” — it can be anything you want!
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A complex number is ill-defined. To say that i^2=-1 implies that sqrt(-1) is a real magnitude (not modulus), but the process of finding “square roots” applies only to real positive magnitudes. In reality, all complex numbers are in fact equivalent to two-dimensional vectors, that is, a + bi = (a,b). Multiplication and division are non-standard. Complex numbers cannot be ordered because true division is not possible using complex numbers. This is not because of the *Well Ordering Principle”, rather the Well Ordering Principle is an indirect result of the division process. Think about it as follows: division (a form of measure) arose from the concept of ratio and ratio is the comparison of two magnitudes or well defined numbers. The key word here is *comparison*, not well-ordering principle.
Complex theory is not required to prove any mathematical result since the trigonometric properties of sine and cosine are in fact the real reasons certain theorems are true. In some cases it is easier to use complex theory for such proofs but the same results can be proved without complex theory.
Euler’s identity e^(i * pi) = -1 (which is easy to show if one assumes i^2=-1) is nonsense. To wit, if one assumes this identity is true, then ln -1 = i * pi (where ln is the natural logarithm) which is obviously false. For starters, the exponential function is *always* positive given a well-defined number. Providing i*pi as input to the function is garbage in and what results is garbage out, that is, e^(i * pi) = -1.
Note that e^a=b ln b=a
Complex theory is the study of non-number (*) or partial-number concepts with properties used in the study of number theory and algebra.
Although sqrt(-1) is not a number, its role in complex theory is to help formulate properties or prove results in number theory.
Students are able to understand such an explanation rather than many hand-waving arguments that are unsound.
(*) It is just as easy to say that a polynomial of degree n has k real roots as opposed to saying a polynomial has n complex roots where a complex root is in fact not a number.
Group theorists fancy thinking of a group binary operation as a “squaring” operation if the operation is applied to the same element (**). However, this is errant thinking since binary operations are not the same, even though two groups may have the same fundamental properties.
Complex “numbers” are not only *not* numbers, they are *not* magnitudes also.
Some fallacious arguments involving complex numbers:
1. -1=(sqrt(-1))^2
2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
4. sqrt((-1)*(-1))=1
Academics will claim that step 3 is incorrect because sqrt(a)*sqrt(b)=sqrt(a*b) iff a and b are both greater than 0. Idiot academics (includes most mathematics professors) do not realize that when they claim sqrt(-4) = 2i, they have just contradicted this fact, that is,
sqrt(a)*sqrt(b) implies sqrt(a*b) where a, b > 0
and
sqrt(a*b) implies sqrt(a)*sqrt(b) where a, b > 0
Since sqrt(-4) = sqrt(4 x -1) is not equal to sqrt(4)*sqrt(-1) !!!
I claim that step 1 is also incorrect because the binary operation of squaring is valid only when performed on well-defined elements, that is, numbers.
In the set of complex numbers, the binary operation of *multiplication* has a dual nature: with real numbers a permutation or multiplicative principle is immediately evident, whereas with complex numbers this may or may not be the case. This chaotic organization is yet another reason why complex number theory is based on ill-defined concepts. In group theory, a cyclic group of order 4 can be represented using complex objects 1, i, -1, -i. A C4 group with complex objects is fundamentally similar to a C4 rotation group (0, tau/4, tau/2, tau) where the group operation is anticlockwise rotation by tau/4 and tau=2(pi) .
Is it human nature or just that modern academics are too stupid to realize these facts? How does one tell a student that the concept sqrt(-1) is a magnitude when there is no way of measuring its dimensions, and yet has learned throughout his studies (till that stage) that magnitudes can be measured – indeed, a magnitude becomes a number once it’s measurable. Just as well, inferential suspension of knowledge is a core part of the human learning process, for otherwise it would be impossible for any student to move beyond this ridiculously ill-defined concept!
Student: What does sqrt(-1) mean?
Teacher: Hmm. It’s a number and these are the operations that are valid…
No wonder it’s called *complex* theory – mathematicians are themselves confused. A better name would be Pseudo-number theory or the use of ill-defined concepts in arriving at some useful theory regarding well-defined numbers.
(**) Claiming that i^2=-1 is equivalent to saying (stone)^2 = stick or (any object)^2= caterpillar because i is ill-defined. Indeed, i^2=-1 is exactly the same as claiming that [(0,1)]^2=-1 which is absurd.
An example of how an educator should explain complex theory:
Suppose that a quintic polynomial has only one pair of non-solutions (DO NOT call these roots because they are NOT numbers!), that is, two non-solutions containing i, then we can assume that three real roots exist.
The previous paragraph still fails the requirements for being well-defined, but it is more accurate than most of the rubbish you have heard muttered by your professor of mathematics.
Even better, an educator might say, a quintic polynomial has n real roots rather than arrive at conclusions about its solution based on non-attributes.
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“1. -1=(sqrt(-1))^2
2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
4. sqrt((-1)*(-1))=1
Academics will claim that step 3 is incorrect because sqrt(a)*sqrt(b)=sqrt(a*b) iff a and b are both greater than 0.”
Actually John…. The flaw is in step 4 where it is conveniently forgotten by you that the sqrt(1) (from step 3: sqrt((-1)*(-1))) also equals -1. I read through your websites for your “new calculus,” and I am just not impressed with people who claim every professor is an idiot and every mathematician in history is an idiot, to include people like Newton, Leibniz and Cauchy. following that up with ridiculous conspiracy theories is just icing on the cake.
Ron,
I think you are mistaken and Gabriel is correct. You would do well to check your logic. The flaw is really in Step 3 and not step 4 as you incorrectly assumed.
To be able to claim that sqrt(a)*sqrt(b)=sqrt((a)*(b)), it must be known beforehand that both a and b are real numbers.
Proposition 4 would be correct if nothing preceded it because (-1)*(-1) is a positive value and hence a square root is possible.
If e to the iπ equals -1, then any positive number to any imaginary number must be equal to a real, negative number.
So what is 4 to the i?
Why not also have a negative number to the power i is a real number?
@Diane Porter:
The example was given earlier in the article.
4^i = e^ln(4^i)
= e^i ln(4)
= cos(ln(4)) + i sin(ln(4))
= 0.183456975 + 0.98302774 i
Pi is a special case. Since cos π = -1 and sin π = 0:
e^iπ = cos π + i sin π
= -1 + i(0)
= -1.
Another special is e^(iπ/2) = cos π/2 + i sin π/2 = 0 + i(1) = i. Yet another is e^(3iπ/2) = -i (try this yourself to understand why).
e^ix has no imaginary part only when sin(x) = 0, which doesn’t leave many possibilities.
@James Hall
nothing wrong with this without any assumptions about a and b:
sqrt(a) * sqrt(b) = sqrt(a * b)
whether you know some property about a or b is irrelevant. sqrt is exponentiation by a half, and the usual exponentiation rules (per the post) still apply:
(a^0.5) * (b^0.5) = (a * b)^0.5
or are you saying that exponentiation does not distribute amongst the multiplicands?
@john
1. -1=(sqrt(-1))^2
2. (sqrt(-1))^2=sqrt(-1)*sqrt(-1)
3. sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
4. sqrt((-1)*(-1))=1
#1
sqrt(a) = +b or -b
such that (+b)^2 = a and (-b)^2 = a
since the result of sqrt is squared, and both values behave identically, this implies #1 is correct.
#2
since #1 is necessarily true numerically and algebraically, and since sqrt is dual-valued…
(sqrt(-1))^2=sqrt(-1)*sqrt(-1)
[from #1] -1 = sqrt(-1) * sqrt(-1)
stepping aside briefly, sqrt(-1), will have two solutions alternating sign. before you say this is illogical, consider that we have negative and positive infinity and that these are distinct, despite an infinity in your result being technically incorrect (the correct term is an undefined value).
since sqrt(-1) = some +b or -b, we have 4 possible solutions to #2
-1 = (-b) * (-b)
-1 = (-b) * (b)
-1 = (b) * (-b)
-1 = (b) * (b)
which reduces to two solutions by equivalence:
-1 = (-b) * (-b) = (b) * (b) = b^2
-1 = (-b) * (b) = (b) * (-b) = -(b^2)
so at best to your argument’s credit this is half wrong and half right. no one wins, but a solution does exist irrespective of what b is, so moving on…
#3
once more, the fault here is assuming sqrt yields only one valid solution.
sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))
using sqrt’s dual-valued nature, it is more correct to say:
sqrt((-1)*(-1)) = +(sqrt(-1)*sqrt(-1)) or -(sqrt(-1)*sqrt(-1))
simplifying:
sqrt(1) = +(-1) or -(-1) = 1 or -1
no fancy tricks here, just high school math.
#4
but to appease all… ill address it directly…
sqrt(-1)*sqrt(-1) = sqrt((-1)*(-1)) = sqrt(1) = 1
sqrt(1^2) = 1 or -1
1^2 = 1
sqrt(1) = 1 or -1
therefore:
sqrt(-1)*sqrt(-1) = sqrt(1) = 1 or -1
i can soundly say that this does not disprove i or roots of negative numbers. what it does demonstrate is that using only part of the available solutions creates logical paradoxes.
Imaginary exponents only rotate a value, here pi, in an “orthogonal” or 90 degree direction from the number line, they do not increase its value; opposed to regular exponents which move a number left or right along the number line.
I would love. To look in the future and see children learning about the number i and how it was deemed an imaginary number. Thinking about the number i make sense to me just like negative numbers. However, at one point the idea of a number less than zero was quite absurd. I call i a number, and I don’t view it as a “imaginary” number.
Neal writes:
“For example, if you take the sequence it, t\to 0, since log(it) = t + ik\pi/2 (where k is some integer), (it)^i = e^[ilog(it)] = e^[i(t + ik\pi/2)] = e^[it]e^[-k\pi/2] limits to e^[-k\pi/2], not zero, as t goes to 0. So here we see that 0^i = e^[-k\pi/2].”
Wait a minute.
log(it) = log t + log i, not t + log i
log i = i*pi/2 + 2k*pi*i with k being any integer
So as t -> 0, log t -> -oo, and log i becomes irrelevant.
So if anything, 0^i = -oo
But you also get
(0^i)^i = 0^(-1) = 1/0
I don’t see anything reasonable or useful coming out of this.
A perhaps more interesting number is i^i. You get
i = e ^ ( (4k+1) * pi/2 * i )
i^i = e ^ ( -(4k+1) * pi/2 )
and for k = 0, i^i ~= 0.20788
You can prove that i^i is real without crunching the numbers:
-i ^ -i = -i ^ ( -1 * i ) = [ -i ^ -1 ] ^ i = i ^ i
This shows that the conjugate of i^i equals i^i, which means that it must be a real number. In fact, it’s a lot of real numbers!
[I’m assuming that (a^m)^n = a^(mn) for real and pure imaginary numbers. Seems to me that the exponent rules are legitimate for all non-zero complex numbers as long as you work in polar form. (Integral exponents should always work except for negative powers of zero.)]
In reply to John Gabriel:
First of all, complex numbers _are_ well defined: i^2 = -1. How is that not well defined?
They are indispensable in quantum mechanics. You simply cannot do quantum mechanics without them. The wave functions are in general complex. The wave function of a free particle is of the form e^(ikx), where k is the wave number and x is the position. Real numbers won’t do. Not even sines and cosines. The Schrodinger equation, the basic equation of QM, has an i in it. And the momentum operator has an i in it. There is no way around this.
And quantum mechanics may well be the most successful theory of all time. Certain quantities (the gyromagnetic ratio, e.g.) using QM are accurate to 10 parts in a billion, with the predicted value within the margin of error of the measured value. That’s pretty good!
Complex numbers are also immensely useful for electrical engineers and electromagnetism in general. Hardly ill-defined.
Complex theory is required to prove the fundamental theorem of algebra. It is also required to prove DeMoivre’s theorem. And it can be used to evaluate certain real definite integrals. See https://en.wikipedia.org/wiki/Complex_number for more.
Square roots apply to all complex numbers, real or imaginary.
Why is ln -1 = i*pi obviously false? Actually it’s multivalued, but i*pi is one of those values. A better statement would be
e^(i*pi) + 1 = 0
For real numbers e^x is always positive. But with the series you can easily generalize it to complex numbers. I don’t see a problem with this.
The problem is that you refuse to allow for legitimate generalization. Using your general argument we would be stuck with the natural numbers (or perhaps whole numbers). For example, how can you have 1/2 an object? If you divide an object by 2, you now have two objects! You would also say you can’t have a negative number of objects, so you would conclude that negative numbers are bogus. But they are a very useful generalization. Money (a balance) comes to mind, in addition to science.
Rational numbers allow division (except by 0). Generalize one step further to real numbers and you can measure what in physics we call “physical observables.”
There is no reason not to go one further step to complex numbers. They are quite legitimate. They allow for a measure of the wave function in quantum mechanics. It’s just a new type of “measure”, just like in the previous cases.
You need to be more precise about what you mean by “magnitudes”. The magnitude of a complex number normally means the (positive) square root of z*(conjugate of z). Or for z = x + iy, the magnitude is the (positive) square root of (x^2 + y^2)
True division of complex numbers is very real (sorry!).
(a+bi) / (c+di) = (a+bi) (c-di) / (c^2 + d^2)
What’s the problem? The denominator is a real number, and division by a real number is quite well defined.
The reason complex numbers are not ordered is because they fail to satisfy the criteria for an ordered field. It has nothing to do with division. One of the properties of an ordered field is that any number squared is a positive number. Well, i^2 is a negative number. Therefore, i is not a positive number, and therefore the field of complex numbers is not ordered. Also, a negative number squared is a positive number, and i^2 is a negative number. Therefore, i is not a negative number. Imaginary numbers are neither positive nor negative.
Now for the 1 thru 4 bit.
1. -1 = (sqrt(-1))^2
2. ( sqrt(-1) ) ^ 2 = sqrt(-1) * sqrt(-1)
3. sqrt(-1) * sqrt(-1) = sqrt( (-1) * (-1) )
4. sqrt( (-1) * (-1) ) = 1
First of all, use some spaces, for Chrissake. It’s hard enough interpreting text equations as it is.
Eq. 1 is already a problem, because the radical means take the positive square root. And as we’ve just seen, i is neither positive nor negative. It’s i. So sqrt(-1) is multivalued and can be i or -i. And once things are multivalued, you’re going to get into trouble. With positive real numbers there is a natural choice for square roots — the positive root. More generally you use the log-based definition except for a base of zero. Sticking with positive real numbers for the base, and real numbers for the exponent, you get single-valued functions and the exponent rules all work fine. (You can add zero to the mix if you’re careful. 0^(positive real number) = 0; 0^0 = 1.)
I see no problem with Eq. 2.
In Eq. 3, sqrt(a) * sqrt(b) = sqrt (a*b) if a and b are positive real numbers. It’s not iff; it’s just if. And since i is neither positive nor negative, this rule cannot be used anyway. You can choose values that come out right, but that’s about it.
The problem with complex numbers in exponentiation is that for y^x, you can multiply y by e^(2*k*pi*i). This doesn’t change the value of the base, but it does change the value of y^x, unless x is an integer.
The exponent rules work for complex numbers if you work in polar form (except for a base of zero, where the power must be a nonnegative real number).
MJ Young writes:
“Imaginary exponents only rotate a value, here pi, in an “orthogonal” or 90 degree direction from the number line, they do not increase its value; opposed to regular exponents which move a number left or right along the number line.”
This is not in general true. What _is_ true is that if you multiply a number by e^i*theta, then the complex number, treated as a vector, is rotated by the angle theta and its magnitude is unchanged.
Xerenarcy writes:
“. . . sqrt(-1), will have two solutions alternating sign.”
Well, it depends what you mean by sqrt.
Yes, there are two solutions of x^2 = 1:
-1 and 1
Fine. But conventionally sqrt is used to mean the positive root. (It certainly does in computer languages!) If you use ^(1/2) instead, convention says to go by the logarithmic definition y^x = e^(x ln y), which for x real and y positive real is always a positive number. In this case sqrt(1) = 1.
So with this definition, Eq. 1 is invalid, as i is neither positive nor negative. I suppose you could define it so that that sqrt means to use the solution whose imaginary part is positive, but you still get into trouble with Eq. 3, as the exponent rule used is valid in general only when the arguments are positive real numbers. (Odd-power roots of negative real numbers are an exception, but then you are not using the logarithmic definition. And besides, 1/2 is not an odd root.)
For reference:
1. -1 = (sqrt(-1))^2
2. ( sqrt(-1) ) ^ 2 = sqrt(-1) * sqrt(-1)
3. sqrt(-1) * sqrt(-1) = sqrt( (-1) * (-1) )
4. sqrt( (-1) * (-1) ) = 1
I beg to differ on one point. You are ignoring the fact that the log of a complex number is in general multivalued.
Consequently, the exponent rule
\left(x^A\right)^B = x^{AB}
does not always work with complex numbers.
Consider the following equations:
(e ^ 5.5i) ^ 3.7 = ( 0.7087 – 0.7055i ) ^ 3.7 = -0.9704 – 0.2414i
= (e ^ -0.7832i) ^ 3.7 = e ^ (-2.898i)
But
e ^ (5.5i * 3.7) = e ^ 20.35i = 0.07029 + 0.9975i
= e ^ 1.5i
Not the same. The problem is the angle (aka argument or amplitude). In the first equation the angle for e^5.5i is 5.5 radians. Raising to the power 3.7 increases that angle to 20.35 radians. But the principal value of the angle of (0.7087 – 0.7055i) is -0.7832 radians.* This is then multiplied by the exponent 3.7, giving a total angle of -2.898 radians. In the second equation we have an angle of 20.35 radians. The two are not an even multiple of pi radians apart; therefore, we get different answers. But 20.35 – 1.5 = 6*pi, and even multiple of pi, hence e^ either gives the same answer.
So the rule \left(x^A\right)^B = x^{AB} only works if you do all your calculations in polar form and without converting intermediate results to use the principle angle or if your angles do not stray from the principal range.
* The range of the principal value of the angle on my HP 32S calculator is -pi < 0 <= pi.
Corrections to previous posts:
On July 29, 2017, at 7:16 pm I wrote in reply to John Gabriel:
“The magnitude of a complex number normally means the (positive) square root of z*(conjugate of z). ”
That’s wrong, of course. It should be, the magnitude of a complex number is
(z*z)^(1/2)
where z* is the complex conjugate of z.
(The magnitude here does come out to be positive number, as I’m using the logarithmic definition of y^x.)
In my post of July 30, 2017, at 8:59 am I wrote:
“The range of the principal value of the angle on my HP 32S calculator is -pi < 0 <= pi."
That should, of course, be
-pi < theta <= pi
where theta is the angle of the complex number in the complex plane.
@Xerenarcy
sqrt(a) * sqrt(b) = sqrt(a*b)
holds _in general_ only when a and b are positive real numbers.
sqrt(-1) * sqrt(-1) = sqrt(1)
sqrt(-1) = i
So,
sqrt(-1) * sqrt(-1) = i * i = -1 != 1
sqrt normally means the positive square root. Since i is neither positive nor negative, this definition of sqrt is meaningless in this context, which is why we got the nonsensical result.
Since i and -i both satisfy x^2 = -1, you can choose values to make the rule true if you loosen up the definition of sqrt:
sqrt(-1) * sqrt(-1) = -i * i = 1 = 1
sqrt[ (-1) * (-1) ] = sqrt(1) = 1
But you have to choose different values for the same thing, not to mention having to choose among possible values in the first place!
Even roots of negative numbers are always going to cause trouble like this.
Roots in a sense are not really powers. They are solutions of certain equations, but expressed using rational exponents. Since the equations often have multiple solutions, you get into trouble with the exponent laws.
The exponent laws work fine when using the logarithmic definition of powers, y^x = e ^ (x * ln y) and limiting y to positive real numbers and x to real numbers. This avoids the problem multiple values for the same expression. When you get to complex numbers, the exponent laws only work _in general_ when you use the polar form or when your exponents are integers (except for negative powers of zero, of course.)