The Grove

In mathematics and physics, many topics are named in honor of Swiss mathematician Leonhard Euler (1707–1783), who made many important discoveries and innovations. Many of these items named after Euler include their own unique function, equation, formula, identity, number (single or sequence), or other mathematical entity. Many of these entities have been given simple and ambiguous names such as Euler's function, Euler's equation, and Euler's formula. Euler's work touched upon so many fields that he is often the earliest written reference on a given matter. In an effort to avoid naming everything...

this one is just called "Euler's Formula". this is The Important One

Prince North Læraðr

What a chad :P

Deusovi

i mean... basically yeah i guess

but now you can do this:

e^i(a+b) = e^(ia) · e^(ib)

that's just a normal thing from exponentiation - do you see why that's true?

Prince North Læraðr

Yes

Deusovi

now use euler's formula on that and Magic will happen

Prince North Læraðr

Oooh okay

urm cos(a)+isin(a)*cos(b)+isin(b)

Deusovi

12:37 AM

parentheses!

don't forget them!

and apply it to both sides

@bobble the bold formula was us gathering reagents for the magic, and now the actual magic is happening

Prince North Læraðr

e ^ix = cos x + i sin x. Just typing that so I can see it to reference it

I'm confused on the magic :P

Do I have to expand the left side as well?

Deusovi

yep, do that

Prince North Læraðr

Ok, I'll try

cos(a+b)+isin(a+b)??

Deusovi

yep!

Gareth McCaughan

and the right hand side?

Deusovi

12:41 AM

(^)

Prince North Læraðr

cos(a+b)+isin(a+b)=(cos(a)+isin(a))*(cos(b)+isin(b))

Deusovi

parentheses on the right side!

you're missing them

Prince North Læraðr

What

Ohhh

Deusovi

(cos(a)+isin(a)) · (cos(b)+isin(b))

Prince North Læraðr

I just say what you were talking About

Forgot PEMDAS :P

Gareth McCaughan

12:42 AM

So, now, expand that out. What do you get?

Prince North Læraðr

Ack hold on. I'm messing myself up. Gonna do some substitutions

Deusovi

sounds good, take your time

Prince North Læraðr

cos(a+b)+isin(a+b)=cos(a)cos(b)+isin(a)cos(b)+icos(a)sin(b)-sin(a)sin(b)

Deusovi

now, when are two complex numbers equal?

that's a pretty general question

but what do you look at to check if two complex numbers are equal?

Prince North Læraðr

Oh darn complex number

Is it the what was that word... conjugate?

Deusovi

12:50 AM

nah, you're overthinking it

Prince North Læraðr

Hm

Oh do I just set it as an equation?

bobble

what's the general form of a complex number?

Deusovi

explain?

Prince North Læraðr

Like the general form is Ai+b

So isolate "i" to one side?

Deusovi

well, not to one side of the equation, but yes - break it down into its real and imaginary components

Gareth McCaughan

12:53 AM

(if z = x + iy then we say that x is the real part and iy is the imaginary part. Or component. Sometimes we say that y is the imaginary part instead, just to make life more interesting.)

Prince North Læraðr

Right

Gareth McCaughan

(two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal)

Prince North Læraðr

Oh. I did over think this

Deusovi

yes, the way i phrased it was probably unclear

Prince North Læraðr

So cos(a+b)+isin(a+b) is like the complex form

(or already in its "complex" form of a sort")

Deusovi

12:57 AM

"standard form" is the usual term

of course standard form has several meanings in different contexts, like most mathematical terms involving some synonym of "normal" or "regular"

Prince North Læraðr

Right. So what do I do with the right side?

Deusovi

the same thing!

group all the things without i together, and group all the things with i together, and then factor

to get (stuff) + i(other stuff)

Prince North Læraðr

Oh I see

cos(a)cos(b)-sin(a)sin(b)+i(sin(a)cos(b)+cos(a)sin(b))? That looks wrong. Nope, I think that's correct

Deusovi

looks right to me

now, as gareth said, two complex numbers are equal only when their real and imaginary components are both equal

Prince North Læraðr

Mhm

Deusovi

1:02 AM

( cos(a+b) ) + i( sin(a+b) ) = ( cos(a)cos(b)-sin(a)sin(b) )+i( sin(a)cos(b)+cos(a)sin(b) )

Um

i'm trying, one sec

No problem

there we go

this is the equation you had above, just with both converted to standard form (with some suggestive bolding)

Prince North Læraðr

Looks like the cosine and sine addition formulas to me with the i :P. But how do we prove it?

(Well from there)

Deusovi

1:04 AM

that is the proof

Prince North Læraðr

Ohhh

Deusovi

we already know those are equal

Prince North Læraðr

Ahhhhh

That is cool

Deusovi

because we started with two equal numbers - e^(i(a+b)) and e^ia · e^ib

Prince North Læraðr

Ohhhhh

Deusovi

1:04 AM

and everything else has just been us changing their forms -- same number, just rewriting and rewriting

Prince North Læraðr

Neat!

Deusovi

you can do... basically all trig identities this way

or at least a lot of them that you'll have to deal with

Prince North Læraðr

So like sin^2+cos^2=1?

Deusovi

sure

e^0 = e^(ia) · e^(-ia)

expand those out

Prince North Læraðr

hold on, I gotta call someone real quick

Gareth McCaughan

1:13 AM

PARENTHESES

Deusovi

you forgot the is

also what gareth said

Prince North Læraðr

(cos(a)+isin(a))*(cos(a)-isin(a)

Gareth McCaughan

right. Now expand that out and do what you need to do with i times i...

Prince North Læraðr

Which is (cos^2(a))-(-sin^2(a)) which is (cos^2(a))+(sin^2(a))

Is this also where people derive the "most beautiful" equation in math or whatnot? The e^(i)(pi)=1 or e^(i)(pi)-1=0?

Deusovi

yep

(personally i think it's overrated - the more general version is better)

Prince North Læraðr

1:18 AM

Hold on I gotta walk my dogs but I'm going to try expanding it out

Prince North Læraðr

1:38 AM

Ok so I was thinking about it and it's uh fairly straightforward

e^[(pi)(i)]=cos(pi)+isin(pi). Cosine of pi=-1, and sin of pi is 0. -1+0=-1. So e^[(pi)(i)]=-1, or e^[(pi)(i)]+1=0

Deusovi

1:54 AM

yep

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