e^(Pi*i) = -1 ???

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EnglshGentleman

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e^(Pi*i) = -1 ???
« on: February 08, 2010, 05:59:53 PM »
I have asked my Calc teachers at school as well as scoured the interwebz for a proof to this, but all my efforts have found no answer.

FES, can any of you people with your great wisdom show me a proof for this?

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optimisticcynic

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Re: e^(Pi*i) = -1 ???
« Reply #1 on: February 08, 2010, 08:38:41 PM »
I have asked my Calc teachers at school as well as scoured the interwebz for a proof to this, but all my efforts have found no answer.

FES, can any of you people with your great wisdom show me a proof for this?
have you done Taylor series yet?
You can't outrun death forever
But you can sure make the old bastard work for it.

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Parsifal

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Re: e^(Pi*i) = -1 ???
« Reply #2 on: February 08, 2010, 11:49:13 PM »
It comes out of Euler's formula. There are some proofs for Euler's formula here: http://en.wikipedia.org/wiki/Euler%27s_formula#Proofs
I'm going to side with the white supremacists.

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semperround

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Re: e^(Pi*i) = -1 ???
« Reply #3 on: February 09, 2010, 07:39:05 AM »
an vir

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EnglshGentleman

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Re: e^(Pi*i) = -1 ???
« Reply #4 on: February 09, 2010, 08:17:34 PM »
I have asked my Calc teachers at school as well as scoured the interwebz for a proof to this, but all my efforts have found no answer.

FES, can any of you people with your great wisdom show me a proof for this?
have you done Taylor series yet?

No :/ we arent that far yet. Looking at Euler's formula i guess i need to know it to understand it.

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #5 on: February 09, 2010, 09:25:28 PM »
Look at trigonometric form of a complex number. The best way is to picture a complex plane. The abscissa is the real axis and the ordinate is the imaginary axis. Every point in the plane corresponds to a single complex number. The Cartesian coordinates of that point (x, y) are the real and imaginary part of the complex number z. When we write:

z = x + I*y (1)

we say the complex number is in its algebraic form.

Now, instead of using Cartesian coordinates, one can use polar coordinates. The distance of the point from the center is called a modulus:

ρ = |z| = Sqrt[x2 + y2] (2)

The angle with the positive real axis is called the argument of a complex number (it ranges from 0 to 2*Pi, remember angles are measured in radians in math):

φ = arg(z) = arctan(y/x) (3)

If one knows the polar coordinates, one can go back to Cartesian coordinates by:

x = ρ*cos(φ)
y = ρ*sin(φ)   (4)

If you substitute (4) into (1), you get

z = ρ*[cos(φ) + I*sin(φ)] (5)

This is the trigonometric form of a complex number. Why is this useful. It turns out that when you multiply two complex numbers in this form, you get:

z1*z2 = ρ1*[cos(φ1) + I*sin(φ1)]*ρ2*[cos(φ2) + I*sin(φ2)]

z1*z2 = ρ12*{[cos(φ1)*cos(φ2) - sin(φ1)*sin(φ2)] + I*[sin(φ1)*cos(φ2) + cos(φ1)*sin(φ2)]}

If you refresh your addition formulas from trigonometry, you will notice that the real and imaginary parts in the parentheses are exactly:

cos(φ1 + φ2) and sin(φ1 + φ2)

So, we can write:

z1*z2 = ρ12*[cos(φ1 + φ2) + I*sin(φ1 + φ2)] (6)

Thus, the product of two complex numbers written in trigonometric form is again in trigonometric form. Furthermore, we can write:

|z1*z2| = |z1|*|z2| (7)

arg(z1*z2) = arg(z1) + arg(z2). (8)

Equation (7) says that the modulus of a product of complex numbers is a product of their moduli and equation (8) that the argument of a product is a sum of the arguments.

The last feature is very important. Let us look at complex numbers with |z| = 1. Geometrically, they lie on a circle with a unit radius and center in the origin in the complex plane. According to (5), they are given by the complex function of a real argument:

h(t) = cos(t) + I*sin(t) (9)

This function, according to (6) obeys the fundamental functional equation for exponential functions:

h(u + v) = h(u)*h(v) (10)

It is only left to find the basis of this exponential. In order to do this, let us find the derivative h'(t) from (9):

h'(t) = -sin(t) + I*cos(t),

where we used [cos(t)]' = -sin(t) and [sin(t)]' = cos(t). Now, because -1 = I2, we can take one I in front of both terms:

h'(t) = I*[cos(t) + I*sin(t)]

But, the term in the angle brackets is just h(t):

h'(t) = I*h(t). (10)

Every exponential function can be expressed as y(x) = exp(k*x). The derivative is then, according to the chain rule, y'(x) = k*exp(k*x) = k*y(x). Comparing this with (10), we see that k = I. So we can (formally) write:

h(t) = exp(I*t). (11)

Combining (5), (9) and (11), we can write the exponential form of a complex number as:

z = ρ*exp(I*φ). (12)

Now consider z = -1. For it x = -1, y = 0. If you plot it in the complex plane and use (2) and (3), you will find that ρ = 1 and φ = π. From here, we immediately arrive at:

-1 = exp(I*π). (13) Q.E.D.

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2fst4u

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Re: e^(Pi*i) = -1 ???
« Reply #6 on: February 09, 2010, 09:50:46 PM »
I would laugh my ass off if that's wrong.

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #7 on: February 09, 2010, 09:53:24 PM »
I would laugh my ass off if that's wrong.
you'll never know now, would you?  ;)

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Parsifal

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Re: e^(Pi*i) = -1 ???
« Reply #8 on: February 09, 2010, 09:56:19 PM »
I can usually follow parsec's working if I take the time to. Right now I just can't be bothered because I know the conclusion is true.
I'm going to side with the white supremacists.

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #9 on: February 09, 2010, 09:59:44 PM »
If they installed LaTeX support, it would've been easily readable. Also, this is not something unknown. I just wanted to use the "least sophisticated" method I know.

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crackpipe larry

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Re: e^(Pi*i) = -1 ???
« Reply #10 on: February 11, 2010, 03:49:20 AM »
Look at trigonometric form of a complex number. The best way is to picture a complex plane. The abscissa is the real axis and the ordinate is the imaginary axis. Every point in the plane corresponds to a single complex number. The Cartesian coordinates of that point (x, y) are the real and imaginary part of the complex number z. When we write:

z = x + I*y (1)

we say the complex number is in its algebraic form.

Now, instead of using Cartesian coordinates, one can use polar coordinates. The distance of the point from the center is called a modulus:

? = |z| = Sqrt[x2 + y2] (2)

The angle with the positive real axis is called the argument of a complex number (it ranges from 0 to 2*Pi, remember angles are measured in radians in math):

? = arg(z) = arctan(y/x) (3)

If one knows the polar coordinates, one can go back to Cartesian coordinates by:

x = ?*cos(?)
y = ?*sin(?)   (4)

If you substitute (4) into (1), you get

z = ?*[cos(?) + I*sin(?)] (5)

This is the trigonometric form of a complex number. Why is this useful. It turns out that when you multiply two complex numbers in this form, you get:

z1*z2 = ?1*[cos(?1) + I*sin(?1)]*?2*[cos(?2) + I*sin(?2)]

z1*z2 = ?1*?2*{[cos(?1)*cos(?2) - sin(?1)*sin(?2)] + I*[sin(?1)*cos(?2) + cos(?1)*sin(?2)]}

If you refresh your addition formulas from trigonometry, you will notice that the real and imaginary parts in the parentheses are exactly:

cos(?1 + ?2) and sin(?1 + ?2)

So, we can write:

z1*z2 = ?1*?2*[cos(?1 + ?2) + I*sin(?1 + ?2)] (6)

Thus, the product of two complex numbers written in trigonometric form is again in trigonometric form. Furthermore, we can write:

|z1*z2| = |z1|*|z2| (7)

arg(z1*z2) = arg(z1) + arg(z2). (8)

Equation (7) says that the modulus of a product of complex numbers is a product of their moduli and equation (8) that the argument of a product is a sum of the arguments.

The last feature is very important. Let us look at complex numbers with |z| = 1. Geometrically, they lie on a circle with a unit radius and center in the origin in the complex plane. According to (5), they are given by the complex function of a real argument:

h(t) = cos(t) + I*sin(t) (9)

This function, according to (6) obeys the fundamental functional equation for exponential functions:

h(u + v) = h(u)*h(v) (10)

It is only left to find the basis of this exponential. In order to do this, let us find the derivative h'(t) from (9):

h'(t) = -sin(t) + I*cos(t),

where we used [cos(t)]' = -sin(t) and [sin(t)]' = cos(t). Now, because -1 = I2, we can take one I in front of both terms:

h'(t) = I*[cos(t) + I*sin(t)]

But, the term in the angle brackets is just h(t):

h'(t) = I*h(t). (10)

Every exponential function can be expressed as y(x) = exp(k*x). The derivative is then, according to the chain rule, y'(x) = k*exp(k*x) = k*y(x). Comparing this with (10), we see that k = I. So we can (formally) write:

h(t) = exp(I*t). (11)

Combining (5), (9) and (11), we can write the exponential form of a complex number as:

z = ?*exp(I*?). (12)

Now consider z = -1. For it x = -1, y = 0. If you plot it in the complex plane and use (2) and (3), you will find that ? = 1 and ? = ?. From here, we immediately arrive at:

-1 = exp(I*?). (13) Q.E.D.


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Thermal Detonator

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Re: e^(Pi*i) = -1 ???
« Reply #11 on: February 11, 2010, 10:11:35 AM »
At the risk of sounding like I know nothing, what is i?
Gayer doesn't live in an atmosphere of vaporised mustard like you appear to, based on your latest photo.

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Thermal Detonator

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Re: e^(Pi*i) = -1 ???
« Reply #12 on: February 11, 2010, 10:21:17 AM »
At the risk of sounding like I know nothing, what is i?

What am i.

No, that's I. I meant i.
Gayer doesn't live in an atmosphere of vaporised mustard like you appear to, based on your latest photo.

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bowler

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Re: e^(Pi*i) = -1 ???
« Reply #13 on: February 11, 2010, 11:36:12 AM »

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Thermal Detonator

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Re: e^(Pi*i) = -1 ???
« Reply #14 on: February 11, 2010, 11:38:11 AM »
Gayer doesn't live in an atmosphere of vaporised mustard like you appear to, based on your latest photo.

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Its a Sphere

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Parsifal

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Re: e^(Pi*i) = -1 ???
« Reply #16 on: February 11, 2010, 01:02:41 PM »
At the risk of sounding like I know nothing

That never bothered you before.

Square root of -1

One of the square roots of -1. ;)
I'm going to side with the white supremacists.

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bowler

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Re: e^(Pi*i) = -1 ???
« Reply #17 on: February 11, 2010, 01:26:47 PM »
:p

Given the question i figured that wasn't important

At the risk of sounding like I know nothing

That never bothered you before.

Square root of -1

One of the square roots of -1. ;)

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #18 on: February 11, 2010, 02:07:38 PM »
At the risk of sounding like I know nothing, what is i?
If you didn't know that one, you probably shouldn't be reading this thread.

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #19 on: February 11, 2010, 03:25:59 PM »
At the risk of sounding like I know nothing, what is i?
If you didn't know that one, you probably shouldn't be reading this thread.

Yeah, if you don't know something, you are not allowed to learn.
No, if you don't know something that elementary, perhaps you should learn the fundamentals first before embarking on understanding more complicated stuff.
« Last Edit: February 11, 2010, 03:30:13 PM by parsec »

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EnglshGentleman

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Re: e^(Pi*i) = -1 ???
« Reply #20 on: February 11, 2010, 06:04:53 PM »
One of the square roots of -1. ;)

Well if you think of it like that there are an infinite amount of square roots of -1.  :-\

How so? I assume he meant it could be (+-)i  as the answer to the square root of -1

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #21 on: February 11, 2010, 07:06:55 PM »
One of the square roots of -1. ;)

Well if you think of it like that there are an infinite amount of square roots of -1.  :-\
This is not correct.

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EnglshGentleman

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Re: e^(Pi*i) = -1 ???
« Reply #22 on: February 11, 2010, 08:08:55 PM »
At the risk of sounding like I know nothing, what is i?

This is the first time you have ever suprised me.
The concept of "i" is taught during junior (sophomore is your in AP) math at my high school.

"i" appears whenever you take the root of a negative number. As far as I've learned anyways.

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bowler

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Re: e^(Pi*i) = -1 ???
« Reply #23 on: February 12, 2010, 04:09:14 AM »
One of the square roots of -1. ;)

Well if you think of it like that there are an infinite amount of square roots of -1.  :-\
This is not correct.

I was assuming there was only the 2.

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Thermal Detonator

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Re: e^(Pi*i) = -1 ???
« Reply #24 on: February 12, 2010, 10:04:55 AM »
At the risk of sounding like I know nothing, what is i?
If you didn't know that one, you probably shouldn't be reading this thread.

Yeah, if you don't know something, you are not allowed to learn.
No, if you don't know something that elementary, perhaps you should learn the fundamentals first before embarking on understanding more complicated stuff.

Given that i was the only component of the equation I was not familiar with, I didn't think it unreasonable to fill in the gap. I could talk about lots of things using abbreviations that Parsec wouldn't understand - that does not necessarily mean that he wouldn't understand the concepts if I told him what the abbreviations meant.

Though tbh if I talked to him about anything he couldn't bring equations into he'd shit his pants and cry.
Gayer doesn't live in an atmosphere of vaporised mustard like you appear to, based on your latest photo.

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #25 on: February 12, 2010, 11:01:40 AM »
One of the square roots of -1. ;)

Well if you think of it like that there are an infinite amount of square roots of -1.  :-\
This is not correct.

well if it's just +/- i, then that's every odd term of i.

i^1, i^3, i^5, i^7....i^(odd number here)

of course that all simplifies down to just i and -i, which, when squared, is -1.
So, then, your statement was incorrect.

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EnglshGentleman

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Re: e^(Pi*i) = -1 ???
« Reply #26 on: February 14, 2010, 08:44:56 PM »
So, then, your statement was incorrect.

Technically, no. But I thought that he DID say +/- i when he said 'only one of the square roots' so, yes, I am completely incorrect in that regard.

Technically No? Technically yes. You were incorrect in all senses.

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Parsifal

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Re: e^(Pi*i) = -1 ???
« Reply #27 on: February 17, 2010, 01:30:22 PM »
I'm pretty sure that i^5, i^7, i^9.... are all real numbers, well imaginary numbers, but definable.

Yes, but i = i5 = i9, just as i3 = i7 = i11.
I'm going to side with the white supremacists.

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bowler

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Re: e^(Pi*i) = -1 ???
« Reply #28 on: February 17, 2010, 02:54:07 PM »
Have you two considered stand up?

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parsec

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Re: e^(Pi*i) = -1 ???
« Reply #29 on: February 17, 2010, 02:56:00 PM »
I always stand.