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anonymous
 one year ago
Explaination wanted > How can e^(i*smallomega*t) be equal to cos(smallomega*t) + i*sin(smallomega*t) ? A big cookie to whomever explains it so I understand the math behind the conversion. I believe it has something to do with Euler's  however I am unable to piece it together myself :)
anonymous
 one year ago
Explaination wanted > How can e^(i*smallomega*t) be equal to cos(smallomega*t) + i*sin(smallomega*t) ? A big cookie to whomever explains it so I understand the math behind the conversion. I believe it has something to do with Euler's  however I am unable to piece it together myself :)

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3You can use taylor series

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Ivan have you taken any Calculus? You can generate the relationship using integration, it's pretty neat :) Yah Taylor Series is good too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes  I just dont recall much of Taylors though. Working on it.. !

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3\[e^t = \sum_{n=0}^{\infty} \frac{ t^n }{ n! }, t \in \mathbb{R} \] so you can add an imaginary number \[e^{i t} = \sum_{n=0}^{\infty} \frac{ (i t)^n }{ n! } = \sum_{n=0}^{\infty} \frac{ i^nt^n }{ n! }\] ooh funny thing is if you don't put space in latex between the i and what ever letter, it makes it go away, truly imaginary xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Yes you will have to note A = a+bi, I think zep can explain it a better way without Taylor

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3If you have an complex number: \(\large\rm z=a+b i\) We can write it in terms of trig, yes? So we'll call the magnitude of this complex number \(\large\rm z=r\). And we can write it in this form: \(\large\rm z=r(\cos\theta+i\sin\theta)\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Differentiate this with respect to the argument, theta,\[\large\rm \frac{dz}{d \theta}=r(\sin \theta+i \cos \theta)\]And then I guess this is where the tricky step comes in. We're going to factor a 1 out of each term:

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Woops, I mean factor an \(\large\rm i\) out of each term :) \[\large\rm \frac{dz}{d \theta}=r i(i\sin \theta+\cos \theta)\]Recall that \(\large\rm 1=i\cdot i\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3So what we've shown is that:\[\large\rm \frac{dz}{d \theta}= i r(\color{orangered}{\cos \theta+i \sin \theta})\]Ooo look we've got our z showing up!\[\large\rm \frac{dz}{d \theta}= i r(\color{orangered}{z})\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Woops woops, little mistake, my bad. the r is part of my z

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm \frac{dz}{d \theta}= i (\color{orangered}{z})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still here  just processing the information and looking everything up and cross referencing in my formula book ... !

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Then separation of variables and integrating gives us:\[\large\rm \int\limits \frac{dz}{z}=\int\limits i d\theta\]Leading to,\[\large\rm \ln z=i \theta+c\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0brill astro!! historically, Euler found this from comparing Taylor series

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Exponentiating gives us:\[\large\rm z=e^{i \theta+c}=e^{i \theta}e^c\] \[\Large\rm z=e^c \color{orangered}{e^{i \theta}}=r(\color{orangered}{\cos \theta+i \sin \theta})\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Thanks irishboy xD Nice one zep!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. Working on the trig method. In order to understand (and recall old lessons) I am starting from scratch given from the examples above. Am I understanding this correctly thus far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! I finally got it, zapdrix! Thanks to everyone else as well for your input! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If anything  the only thing throwing me a bit off is the e^c and r. From the example e^(iw). Are those just constants we can.... throw away? ooh.. because they are the same we can just "remove" them! Right? No? @zepdrix @Astrophysics

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Yeah it's just a constant, you can leave it alone, or just add it to your calculations after

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Yes, good eye :) It turns out that the \(\large\rm e^c\) is the same \(\large\rm r\) that we started with. I was having trouble illustrating that though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Ah yes, good with the trig. Sorry I kinda skipped over that at the start >.<
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