Question

I know that
e^(ix)=cosx+isinx
but why does it? I'm bored of memorising things, can someone explain it so I can understand why?

Answer 1

is this de Moivre's theorem in complex numbers ?
You ask "why" which is often a tricky q to ask in maths/science. After all i is the square root of minus one which only "exists" in the mind of the imaginary part of a a complex number.
Actually, thinking about it, is this an equation or and identity ?????

Answer 2

Identity I think
demoivre's yes! although i think demoivres is to powers of n
I just want to know why this specific example or in general if you can explain, is the same as this

I want to start understanding math instead becoming a person who only memorises

Answer 3

Yeahh, pretty stumped with this as well, so is this what the equation is saying?\[\large e^{\pi \sqrt{-1}} = -1\]

Answer 4

that is eulers identity

I guess I am talking about demoivres formula

Answer 5

ah okee

Answer 6

Define a complex value using polar coordinates,\[\large\rm z=\cos x+i \sin x\]Differentiate,\[\large\rm z'=-\sin x+i \cos x\]This step is a little tricky,
factor an i out of each term,\[\large\rm z'=i(i \sin x+\cos x)=i(\cos x+i \sin x)\]Oh but cosx+isinx is what we started with!
So we've determined this,\[\large\rm z'=i z\]Let's integrate and see what happens,\[\large\rm \frac{dz}{dx}=iz\qquad\to\qquad \int\limits\frac{dz}{z}=\int\limits i~dx\]Giving us,\[\large\rm \ln z=ix\]Exponentiating,\[\large\rm z=e^{ix}\]So we've shown that our complex number has two equivalent representations,
cosx+isinx but also e^{ix}.

Answer 7

That's one silly way to establish the relationship.
You can also look at the uhhh Taylor Series for sine and cosine,
and together they will relate to e^x Series in an interesting way :D

Answer 8

I think starting at Z=cosx+isinx, which is the result we want to get to, is not a good way to start to explain something

hmm the taylor series is definitely an interesting approach! Never even thought of that, could use e^x and let x=ix I suppose! thank you Zepdrix :)

how do we know that integration works as naturally with imaginary numbers, as it does other numbers?

Answer 9

Also take the series of e^{ix}
\[
1+i x-\frac{x^2}{2}-\frac{i x^3}{6}+\frac{x^4}{24}+\frac{i
x^5}{120}-\frac{x^6}{720}-\frac{i x^7}{5040}+\frac{x^8}{40320}+\frac{i
x^9}{362880}-\frac{x^{10}}{3628800}+O\left(x^{11}\right)
\]
The even powers will give you cos(x)
and i(odd powers)= i sin(x)

Answer 10

Ahh that's a good question :D

I'd have to look at my Complex Analysis book to remember how they justify integration techniques. I can't seem to remember.

Answer 11

Ooo, any chance you can point me to the book online?

Answer 12

What do you mean by complex integration? Do you mean contour integrals?

Answer 13

yes that is what I mean