What Happens to sin(i) for Complex i?

[hadi amiri 4]

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

 

[Astronuc]

i is defined by i² = -1. It is referred to as the imaginary unit and arises from the fact that

\sqrt{-a}\,=\,\sqrt{-1}\sqrt{a}\,=\, {i}\sqrt{a}, where a is real.


See http://mathworld.wolfram.com/ImaginaryNumber.html, and
http://mathworld.wolfram.com/ImaginaryUnit.html

 

[hadi amiri 4]

I can not understand the existence of this

 

[HallsofIvy]

It is impossible to give an explanation that you will grasp if we do not know what mathematics you already know.

 

[nicksauce]

I can not understand the existence of this

What do you mean? It is defined to exist. All you have to do is to be able to grasp the concept of definition.

 

[hadi amiri 4]

pre unive math

 

[arildno]

pre unive math

Well, then you know about 2-D coordinates, don't you?

Instead of considering the quantities that lie on the real number line (i.e, the so-called "real" numbers!), let our basic "quantities" be points in the PLANE in stead.
We will call these quantities "complex numbers".

Now, each such "complex number" can be said to have two components, each of which is a standard real number, just like the coordinates for a point in the plane are given by, say, (x,y) where "x" is the "x-coordinate" and "y" the "y-coordinate"

Now, we want to DO SOMETHING with our new complex numbers, in particular, we want to be able to "multiply" two such numbers together. (Note that we here introduce a new meaning to the word "multiply" , namely as an operation between two points in the plane (i.e, our complex numbers!), rather than an operation between two numbers on the number line!)


Let two complex numbers have the shape (a,b) and (c,d) where a,b,c,d are ordinary real numbers.
Furthermore, let us introduce the short-hand symbol "i" for the point (0,1), that is i=(0,1), and in addition, we let the symbol ""1"" be defined as: "1"=(1,0), and, also, "-1"=(-1,0).


Now, we introduce a general multiplication rule:

(a,b)*(c,d)=(ac-bd,ad+bc)

Thus, let us multiply "i" with itself!

i*i=(0,1)*(0,1)=(0*0-1*1,0*1+1*0)=(-1,0)="-1"

That is, i*i equals the complex number we have called "-1"!


Furthermore, we will now look at those numbers that lie on just the x-axis, say numbers of the form (a,0) and (c,0).

What happens when we multiply two such numbers together?

Then we get:
(a,0)*(c,0)=(ac-0*0,a*0+0*a)=(ac,0)
THat is, we get a number ON the x-axis whose coordinate value equals exactly the product a*c, and we can therefore IDENTIFY the x-axis with our ordinary number line!


THe complex numbers constitute therefore a number PLANE, where we call the x-axis the "real axis", whereas the y-axis is called the "imaginary axis".

The number "i" lies therefore ON the "imaginary axis", has unit length, so we call it the "imaginary unit".




Thus, that i*i=-1 is something we are able to construct as long as we regard our NUMBERS as points in the plane, and in addition, have defined our "multiplication" in a smart manner as above.

 

[tiny-tim]

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

I can not understand the existence of this

Hi hadi amiri 4!

nicksauce is right … i is defined as a square-root of -1.

Mathematicians can define anything they like.

i does not exist.

Or, rather, it only exists in mathematicians' imaginations.

Why does its existence worry you?

 

[symbolipoint]

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

You already know that i is the imaginary unit, so you then know that i(/sqrt{-1}) is the same as i*i, which is -1.

 

[Redbelly98]

i is defined by i² = -1.

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.

 

[Werg22]

Hi hadi amiri 4!

nicksauce is right … i is defined as a square-root of -1.

Mathematicians can define anything they like.

i does not exist.

Or, rather, it only exists in mathematicians' imaginations.

Why does its existence worry you?

i exists just as much as the number 1. i is simply the element (0, 1) of the field formed by R^2 over addition and multiplication of complex numbers. That i^2 = -1 is simply due to the definition of multiplication for complex numbers. Also, to be clear, 1 in C is not the same as 1 in R. They are different objects; one is an ordered pair while the other is atomic, let alone the fact that they come from different fields. For convenience, we abbreviate (x, 0) as x and (0, y) as y*i. The pair (0, 1) exists just as much as the pair (1, 0).

 

[Gokul43201]

Well, then you know about 2-D coordinates, don't you?

Instead of considering the quantities that lie on the real number line (i.e, the so-called "real" numbers!), let our basic "quantities" be points in the PLANE in stead.
We will call these quantities "complex numbers".
...

Thus, that i*i=-1 is something we are able to construct as long as we regard our NUMBERS as points in the plane, and in addition, have defined our "multiplication" in a smart manner as above.

Nice! But I suspect a student's ability to appreciate this will depend strongly on whether or not they have been taught mathematics axiomatically.

 

[Gear300]

Remember, math is a relatively arbitrary field. Something can be made valid so long as it serves a purpose that does not hold too many heavy contradictions. The number 1 is a concept and so is i. The number 0 is a concept and so is i (it can even be more mysterious than i).
Pre-university schools (and even universities can) have a crude way of teaching math...its very mechanical and rigid...it does not present any elegance and it pretends to put unnecessary boundaries between the several different categories. You could gain a better perspective from reading journals, etc... (how Gokul43201 said it).

 

[HallsofIvy]

Hi hadi amiri 4!

nicksauce is right … i is defined as a square-root of -1.

Mathematicians can define anything they like.

i does not exist.

Or, rather, it only exists in mathematicians' imaginations.

Why does its existence worry you?

Oh, tiny-tim, what were you thinking? "i" only exists in mathematician's imaginations in exactly the same way other abstract things, like the numbers "e", or "1/2", or "1" do. I understand that but the person who wrote the original post is certain to misunderstand it.

Oh, and as I am sure you have seen from my previous posts, I object to the bald statement "i is defined as a square root of -1". You did put "a" which a good thing but a definition has to specify it exactly. That does not say which of the two square roots of -1 is i.

 

[Dale]

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.

I hadn't thought of that, both +i and -i are solutions, but how can you say that one is the positive root and the other is the negative root? Hmmm, very interesting comment. I guess that is the ultimate source of conjugate symmetry.

 

[arildno]

Nice! But I suspect a student's ability to appreciate this will depend strongly on whether or not they have been taught mathematics axiomatically.

Agreed!
It would be interesting to see what the orinal poster makes out of it.

 

[matt grime]

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.

It is discussed. In one loose sense it is the starting point of Galois theory. In fact some mathematicians make a point of only using the symbol \sqrt{-1} to avoid having to make a choice of square root of -1.

 

[Redbelly98]

Thanks Matt, I just don't hang around with mathematician's enough to know what is going on. My exposure to math has been from a physics education earlier in life, and (these days) tutoring high school math.

Mark

 

[tiny-tim]

The most basic definition of i

Hi Werg22!

i exists just as much as the number 1. i is simply the element (0, 1) of the field formed by R^2 over addition and multiplication of complex numbers.

hmm … this is obviously a usage of the word "simply" that I haven't come across before …

Hi HallsofIvy!

"i" only exists in mathematician's imaginations in exactly the same way other abstract things, like the numbers "e", or "1/2", or "1" do. I understand that but the person who wrote the original post is certain to misunderstand it.

Sorry, but I think numbers like 1 2 3 … existed even in cavemen's imaginations.

And when it came to dividing food, so did numbers like 1/2.

e = 2.71828182889045etc is of course a much more recent invention of mathematicians … but is it an invention?

If you defined it (using that decimal series) it to a layman, surely the layman would reply "You haven't defined anything new … you've only given a name to something which I see no use for! I wish you well with it! But it was already there before you gave it that name … you didn't define it, you only named it!"

But i wasn't there before …

if so, where was it? ​

In that sense, hadi amiri 4 is quite right to question its existence!

Oh, and as I am sure you have seen from my previous posts

erm … sorry … what posts?

I object to the bald statement "i is defined as a square root of -1". You did put "a" which a good thing but a definition has to specify it exactly. That does not say which of the two square roots of -1 is i.

"i" can be defined in many ways.

Here's one definition: add a symbol "i" to the field R, define ii = -1, and then enlarge to a field.

Obviously, there are no other elements of the new field whose square is -1, except for -i.

What's wrong with that?​

All other definitions (the Argand plane construction, the pairs construction, etc) are just models of this definition.

And this definition … of i as a thing whose square is -1 … is the most basic.

 

[matt grime]

Sorry, but I think numbers like 1 2 3 … existed even in cavemen's imaginations.

So you don't think they exist in any real sense any more nor less than a symbol that squares to give -1?

Just because you can think of a 'real life' situation where things like 1,2,3 are useful, and describe some properties doesn't mean that these descriptors have a physical reality in this, or any (platonic) realm. My standard line is: the second I stub my toe on a '1', I'll believe it exists in a meaningful sense. The counter point is: when you stick your hand in the electric socket, it isn't the imaginary part of the current that kills you.

 

[tiny-tim]

… it doesn't really hurt …

My standard line is: the second I stub my toe on a '1', I'll believe it exists in a meaningful sense. The counter point is: when you stick your hand in the electric socket, it isn't the imaginary part of the current that kills you.

Hi matt!

But does your toe exist? or is it just a convenient name for an electromagnetic field with such a strong boundary that it repels what we choose to call the electromagnetic fields of other things?

 

[matt grime]

My toe certainly does exist, and you just described what it is. You didn't do that for the number 1. In fact all you will do is produce examples of 'one of something', such as one sheep, or one toe.

 

[Gear300]

Wait...to sidetrack just a little...what are the physical meanings i can hold (is it only applicable as a mathematical concept)?

 

[hadi amiri 4]

we know that
\sqrt{a}\sqrt{b}=\sqrt{ab}
so
\sqrt{-1}\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}
=1
and i^2=-1
finally 1=-1

 

[Kaimyn]

we know that
\sqrt{a}\sqrt{b}=\sqrt{ab}
so
\sqrt{-1}\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}
=1
and i^2=-1
finally 1=-1

Do you see the errors you made in there? For one thing, \sqrt{a}\sqrt{b}=\sqrt{ab} works for real numbers, but not always complex/imaginary numbers. Also, \sqrt{1} has two answers, -1 and 1, just as \sqrt{25} has the two roots -5 and 5 and \sqrt{-9} has (-3)i, 3i, 3(-i) and (-3)(-i). You could even go to say that \sqrt{-9} has the root -3i^{5}.

But really, \sqrt{-1}\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}=-1

 

[epkid08]

we know that
\sqrt{a}\sqrt{b}=\sqrt{ab}
so
\sqrt{-1}\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}
=1
and i^2=-1
finally 1=-1

Only in your imaginary world

 

[HallsofIvy]

Hi Werg22!


hmm … this is obviously a usage of the word "simply" that I haven't come across before …

Hi HallsofIvy!


Sorry, but I think numbers like 1 2 3 … existed even in cavemen's imaginations.

You assume that cavement couldn't be mathematiciains? Do you write Geico commercials?
I will confess that when I wrote that I really meant that all numbers exist in the imagination. I only said a mathematician's imagination because you did and I wanted to agree with you as much as possible. Did you really intend to say that only professional mathematicians can conceive of the number i?

And when it came to dividing food, so did numbers like 1/2.

1/2 of an apple is an object. 1/2 of a loaf of bread is an object. "1/2" is none of those things.

e = 2.71828182889045etc is of course a much more recent invention of mathematicians … but is it an invention?
If you defined it (using that decimal series) it to a layman, surely the layman would reply "You haven't defined anything new … you've only given a name to something which I see no use for! I wish you well with it! But it was already there before you gave it that name … you didn't define it, you only named it!"

But i wasn't there before …

if so, where was it? ​

In that sense, hadi amiri 4 is quite right to question its existence!


erm … sorry … what posts?


"i" can be defined in many ways.

Here's one definition: add a symbol "i" to the field R, define ii = -1, and then enlarge to a field.

Obviously, there are no other elements of the new field whose square is -1, except for -i.

What's wrong with that?​

All other definitions (the Argand plane construction, the pairs construction, etc) are just models of this definition.

And this definition … of i as a thing whose square is -1 … is the most basic.

No, it is not. 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.

 

[Dale]

Wait...to sidetrack just a little...what are the physical meanings i can hold (is it only applicable as a mathematical concept)?

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).

 

[kts123]

i is a mathematical entity equatable to pi.

i exists in the sense that pi exists -- that is, under special situations, it is a value that arrises naturally. A cartesian coordinate system does not "exist," but it most certainly behaves as if it does when we use trignometry to measure apparent size, paralax, or a number of other things (and in these contexts, pi also appears.)

i is easily demonstrated to exist in the same sense as pi via observing the "addition" of forces. Pi let's us represent angle, for example Pi/4 rads over "level" yields maximum distance when firing a cannon, and complex values allow us to "add" forces (as well as other useful applications.) Anyone claiming that i does not "exist" must also concede that pi does not "exist." (In fact, after studying complex analysis, I would have to say that complex numbers are the "true" numbers, making coherence of ambigous concepts such as "limits at infinity," "rotation," and other topologies.)

I'm not an expert though, and I have no room to mention topology considering I have almost no experience in the field. Still, I feel that complex numbers are extremely real -- perhaps more so than even integers (well, that's going a bit far, but still.)

 

[HallsofIvy]

i is a mathematical entity equatable to pi.

Careful now! Strictly speaking you have just said that e= \pi!

i exists in the sense that pi exists -- that is, under special situations, it is a value that arrises naturally.

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

A cartesian coordinate system does not "exist," but it most certainly behaves as if it does when we use trignometry to measure apparent size, paralax, or a number of other things (and in these contexts, pi also appears.)

i is easily demonstrated to exist in the same sense as pi via observing the "addition" of forces. Pi let's us represent angle, for example Pi/4 rads over "level" yields maximum distance when firing a cannon, and complex values allow us to "add" forces (as well as other useful applications.) Anyone claiming that i does not "exist" must also concede that pi does not "exist." (In fact, after studying complex analysis, I would have to say that complex numbers are the "true" numbers, making coherence of ambigous concepts such as "limits at infinity," "rotation," and other topologies.)

I'm not an expert though, and I have no room to mention topology considering I have almost no experience in the field. Still, I feel that complex numbers are extremely real -- perhaps more so than even integers (well, that's going a bit far, but still.)

 

[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 Schrodinger 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 Schrodinger 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 Schrodinger 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.