What Happens to sin(i) for Complex i?

[HallsofIvy]

It occurred to me some time ago that this definition is ambiguous, since there are two solutions to that equation.

While this ambiguity does not seem to matter in practice, I'm still surprised that it never seems to get discussed.

God knows, I have discussed it on this forum often enough to sound as if I had a mania on the subject!

Not only are there, in the complex numbers, two square roots of i, but since the complex numbers do not form an "ordered field", there is no way to distinguish between "the positive root of -1" and "the negative root of -1"; there are no "positive" or "negative" complex numbers.

In fact, most really good textbooks on complex analysis are careful to define the complex numbers in terms of "pairs".

The set of complex numbers is the set of ordered pairs of real numbers with addition and multiplication defined by:
(a, b)+ (c, d)= (a+ c, b+ d) and (a,b)(c,d)= (ac- bd, ad+bc).

It is easy to see that the subset {(a, 0)}, of all pairs with second component 0, is isomorphic to the set of real numbers: (a,0)+ (b,0)= (a+b, 0+0)= (a+b, 0) and (a, 0)(b, 0)= (ab- 0(0),a(0)+ b(0))= (ab, 0) so we can, in that sense, think of the real numbers as a subfield of the complex numbers and identify (a, 0) with the real number a.

It is also easy to see that (0, 1)(0, 1)= (0(0)-1(1),0(1)+ 1(0))= (-1, 0)= -1 and that (0,-1)(0,-1)= (0(0)-(-1)(-1),0(-1)+ (-1)(0))= (-1, 0)= -1 so that (0, 1) and (0, -1) are both "square roots" of -1.

It is also easy to see that for any real number a, a(1,0)= (a, 0)(1, 0)=(a(1)-(0)(0), a(0)+ (0)(1))= (a, 0) and that b(0, 1)= (b, 0)(0,1)= (b(0)-(0)(1), b(1)+ (0)(1))= (0, b).

That is, for any complex number, (a, b), (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1). We have already identified (1, 0) with the real number 1. If we now define i to be (0, 1) we can write (a, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.

The difference now is that we have specifically identified i with (0, 1) while -i= (0, -1) so that they are distinguishable.

 

[tiny-tim]

rigorous definition

Hi HallsofIvy!

… In fact, that is not a definition at all because, again, in the real number system there is no such thing while in the complex number system there are two and that definition doesn't tell us which of the two is i.

ok, I'll be rigorous:

Define the commutative field {i} \bigoplus\,\Re with ii = - 1 …

it has two square roots of -1, and it is arbitrary which we call i.

Or, at least, it is arbitrary which we map onto i in the usual Argand plane etc.

(Similary for the non-commutative field {i} \bigoplus {j} \bigoplus {k} \bigoplus\,\Re with ii = - 1 ij = -ji etc … )

 

[kts123]

Yes, and, I would argue (and have here), in the same sense that any number exists!

I agree, and I used pi as an example because it's so easily observed to exist in nature.

Careful now! Strictly speaking you have just said that !

Pssfht. Well, someone might argue something ~similar~ using complex numbers. So there.

 

[Redbelly98]

God knows, I have discussed it on this forum often enough to sound as if I had a mania on the subject!

While I didn't realize this when I first posted, during the 2 days since I have become quite aware of it!

 

[Gear300]

It is useful in electrical engineering and signal processing and anytime you are doing a Fourier transform. Perhaps there are other applications also, but that is the field that I use it in. In the Fourier transform and signal processing complex numbers are used to represent phase information and i is a phase shift of 90º wrt 1 (i.e. a sin rather than a cos).

I see...interesting (I actually do not know much about the Fourier transformation as you can tell by my question). One additional question...has it ever occurred in any physical theory that i played a role in the meaning of the theory? If not, then much like other concepts, its meaning would occur more or less mathematically, right?

 

[HallsofIvy]

In the general theory of relativity the "metric" can be written as
ds^2= dt^2- dx^2- dy^2- dz^2[/itex] <br /> or as<br /> ds^2= -(dx^2+ dy^2+ dz^2+ d\tau^2)[/itex]&lt;br /&gt; &lt;br /&gt; where \tau= it&lt;br /&gt; &lt;br /&gt; The latter form makes it more like the &amp;quot;Euclidean&amp;quot; metric.

 

[Integral]

In the general theory of relativity the "metric" can be written as
ds^2= dt^2- dx^2- dy^2- dz^2[/itex] <br /> or as<br /> ds^2= -(dx^2+ dy^2+ dz^2+ d\tau^2)[/itex]&lt;br /&gt; &lt;br /&gt; where \tau= it&lt;br /&gt; &lt;br /&gt; The latter form makes it more like the &amp;quot;Euclidean&amp;quot; metric.

&lt;br /&gt; &lt;br /&gt; Halls, Do you have a sign error in that 2nd expression?

 

[HallsofIvy]

No. I did when I first wrote it but that's why I put the "-" in front of it. I'm trying to remember if I posted it first, caught the error, and then edited. If so you might have seen in the few moments before my edit.

 

[hadi amiri 4]

what is the meaning i^i?

 

[kts123]

what is the meaning i^i?

There actually exists an infinite number of solutions to that problem.

When raising a real number to a complex value, we compute as follows:

a^z = e ^ ln(a)*z.

Where e ^ ln(a) = a, and z is a given complex number. In otherwords, when we raise a number "a"," to a complex power "z", the answer is e to the power of ln(a) times z. When a is complex, the same rule ultimately applies (though it complicates things, as you will shortly see.)

First convert to notion i into complex form: this can be written as 0 + 1i, which comes to just i. However, we can also write the complex value in an exponential form using e -- in this notation, 0+1i = e ^ {\pi*0.5*i}.

i^i = (e ^ {\pi*0.5*i})^(e ^ {\pi*0.5*i})

Which comes to (if memory serves me) e^ {[e ^ {(\pi*0.5*i)} * ln(i)]}, or e ^ {(0+1i)*ln(i)}. However, as you may notice, if we add 2pi to 0.5 (i.e. we rotate it exactly 360 degrees) the angle remains the same (technically speaking, at least, really the angles only behave the same.) For this reason, there is an infinite number of solutions to the problem, however they really just boil down to the two I mentioned.

Well, I think that's right, at least, there's a good chance I'm off my block!

 

[kts123]

Oh, I forgot to finish when I got to e ^ {ln(i)*i}.

Since e ^ 0.5*pi*i = i, then (obviously) ln(i) = 0.5*pi*i

Now we have:

e ^ {(\pi*0.5*i)*i}

In the exponent this yields: i*i (pi/2) = -1(pi/2) = -pi/2, so i^i = e ^ {-\pi/2}. Since we can add or subtract 2pi to pi/2 and not change anything, the possible solutions are: e ^ {(-\pi/2) +(2k\pi)}, where k = {1, 2, 3...n}

 

[Redbelly98]

has it ever occurred in any physical theory that i played a role in the meaning of the theory? If not, then much like other concepts, its meaning would occur more or less mathematically, right?

i appears explicitly in the Schrödinger Equation (time-dependent version).

<br /> i \hbar \frac{\partial \psi}{\partial t} = H \psi<br />

I'd have to think hard to put into words just how i "plays a role in the meaning" of quantum mechanics. But this is arguably the single most important equation in 20th-century physics, and it contains i.

 

[Gear300]

i appears explicitly in the Schrödinger Equation (time-dependent version).

<br /> i \hbar \frac{\partial \psi}{\partial t} = H \psi<br />

I'd have to think hard to put into words just how i "plays a role in the meaning" of quantum mechanics. But this is arguably the single most important equation in 20th-century physics, and it contains i.

Single most important...well that is interesting. I haven't studied quantum mechanics yet (only read concepts). Heh...it must be a he** of an equation then.

 

[hadi amiri 4]

What is the meaning Sin(i) do we have imaginary angel?

 

[HallsofIvy]

Well, I would say an imaginary angle. Angel's don't have a lot to do with mathematics!

You should be aware that the trig functions appear in a lot of mathematics that does not involve triangles or angles at all.

Since sin(x)= (e^{ix}- e^{-ix})/2i, sin(i)= (e^{i^2}- e^{-i^2})/2i= (e^{-1}- e^{1})/2i which is approximately -2.35/2i or 1.175i.

 

[tiny-tim]

hyperbolic sine

Hi hadi amiri 4!

What is the meaning Sin(i) do we have imaginary angel?

sin(i)= (e^{i^2}- e^{-i^2})/2i= (e^{-1}- e^{1})/2i

Or, more generally, sinh(x) = (e^x\ -\ e^{-x})/2

cosh(x) = (e^x\ +\ e^{-x})/2

and so cosh(ix) = cos(x), sinh(ix) = i sinx,

and, conversely, cos(ix) = cosh(x), sin(ix) = -i sinx.

sin(A + iB) = sinA cosiB + cosA siniB = sinA coshB - i cosA sinhB.

cosh and sinh are known as hyperbolic cosine and sine.

 

[DrGreg]

i appears explicitly in the Schrödinger Equation (time-dependent version).

<br /> i \hbar \frac{\partial \psi}{\partial t} = H \psi<br />

I'd have to think hard to put into words just how i "plays a role in the meaning" of quantum mechanics. But this is arguably the single most important equation in 20th-century physics, and it contains i.

Complex numbers are built into quantum mechanics at every level. Particles are represented by complex-valued functions e.g.

\psi(x,t) = e^{-a(x-ct)^2 + i\omega(x-ct)}​

might, under some circumstances, represent a photon of frequency \omega c/(2\pi).

The theory wouldn't make sense if you used real-valued functions instead.

 

[HallsofIvy]

I would also like to point out that the recognition that imaginary and complex numbers were really necessary historically came as a result of Cardano's method of solving cubic equations. There are cubic equations that can be shown to have only real solutions but to use Cardano's method to get those real solutions involves taking the square root of a negative numbers. (The imaginary parts cancel out in the end.)

 

[schroder]

can anyone tell me that what is i(\sqrt{-1})

what is the meaning i^i?

In case the concept of i is still not clear, may I offer another point of view? Never mind answering that, I will go ahead and offer it anyway.

As has arildno has already mentioned, it is helpful to visualize a plane formed by real numbers along the horizontal axis and complex numbers on the vertical axis. In considering this real/imaginary plane, i is just a unit vector with an angle of pi/2 radians, meaning it lays on the imaginary axis pointing straight up with a magnitude of 1. When you take i^2 and follow the rule for the multiplication of numbers expressed in polar coordinates which is to multiply the arguments and add the angles, the result is a unit vector with an angle of pi radians which lies on the real axis and is nothing more than the real number -1. There should be nothing ambiguous about that.

Now, when you wish to take i to the ith power, it is best to express i in its exponential form, which is e^(i*pi/2). This makes i to the ith power appear as (e^(i*pi/2))^i. Remembering that i^2 = -1, and following the rule for exponents, this becomes: e^(-pi/2).
This is a real number that results from the imaginary number e and the imaginary operator i.

If you study electrical engineering you will be constantly using the imaginary operator, only in that discipline it is known as j.

 

[HallsofIvy]

But now your distinction between "i" and "-i" depends upon which way is "up" and which way is "down", again an arbitrary choice.

 

[schroder]

But now your distinction between "i" and "-i" depends upon which way is "up" and which way is "down", again an arbitrary choice.

It would be arbitrary if I was the first person to adopt this notation, but it is by now conventional that i is up and –i is down. In Electrical Engineering, j is up and represents inductive reactance and –j is down for capacitive reactance so it seems a convention is already established.

 

[rbj]

It would be arbitrary if I was the first person to adopt this notation, but it is by now conventional that i is up and –i is down. In Electrical Engineering, j is up and represents inductive reactance and –j is down for capacitive reactance so it seems a convention is already established.

but it doesn't matter. whether you call it "j" or "-j", other than the fact that they are negatives of each other, every property of these two imaginary numbers are precisely the same; they both square to be -1. if all math (and physics and electrical engineering) literature was changed and every occurrence of i or j were replaced with -i or -j, every statement and every theorem would be just as valid as it was before. but you can't say that about replacing 1 with -1.

check out:

http://en.wikipedia.org/wiki/Imaginary_unit#i_and_.E2.88.92i

 

[schroder]

but it doesn't matter. whether you call it "j" or "-j", other than the fact that they are negatives of each other, every property of these two imaginary numbers are precisely the same; they both square to be -1. if all math (and physics and electrical engineering) literature was changed and every occurrence of i or j were replaced with -i or -j, every statement and every theorem would be just as valid as it was before. but you can't say that about replacing 1 with -1.

All of that is true. The only point that I was making is that all ambiguity is removed by making a choice, and staying with that as a convention which has already been made in applied science and engineering. Then only in the realm of the pure mathematician does any ambiguity reside but that’s what pure mathematicians thrive on! I do not reside in that realm.

 

[Redbelly98]

... only in the realm of the pure mathematician does any ambiguity reside but that’s what pure mathematicians thrive on! I do not reside in that realm.

Nor do I.

The realm of pure mathematics is a nice place to visit, but I wouldn't want to work there.

 

[tiny-tim]

… don't talk to me about reality …

The realm of pure mathematics is a nice place to visit, but I wouldn't want to work there.

Mathematician to physicist:

Reality is a nice place to visit … but I wouldn't want to live there! ​

 

[hadi amiri 4]

WE know that sin(x)\leq1 but sin(i) is something else an anyone explain

 

[tiny-tim]

WE know that sin(x)\leq1 but sin(i) is something else an anyone explain

Haven't we already gone over this :

Hi hadi amiri 4!

… and so cosh(ix) = cos(x), sinh(ix) = i sinx,

and, conversely, cos(ix) = cosh(x), sin(ix) = -i sinx.

sin(A + iB) = sinA cosiB + cosA siniB = sinA coshB - i cosA sinhB.

cosh and sinh are known as hyperbolic cosine and sine.

sin(i) = sin (i.1) = -i sinh(1) = -i(e - 1/e)/2

 

[HallsofIvy]

WE know that sin(x)\leq1 but sin(i) is something else an anyone explain

We know that sin(x)\le 1 for real x. sin(z) for complex z is not bounded- it looks like an exponential along the imaginary axis. In fact, there is a famous theorem that if an entire function is bounded, it is constant.