anonymous
  • anonymous
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 :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
Astrophysics
  • Astrophysics
You can use taylor series
zepdrix
  • zepdrix
Ivan have you taken any Calculus? You can generate the relationship using integration, it's pretty neat :) Yah Taylor Series is good too.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes - I just dont recall much of Taylors though. Working on it.. !
Astrophysics
  • Astrophysics
\[e^t = \sum_{n=0}^{\infty} \frac{ t^n }{ n! }, t \in \mathbb{R} \] 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
  • Astrophysics
Yes you will have to note A = a+bi, I think zep can explain it a better way without Taylor
zepdrix
  • zepdrix
If 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
  • zepdrix
Differentiate 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
  • zepdrix
Woops, 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
  • zepdrix
So 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
  • zepdrix
Woops woops, little mistake, my bad. the r is part of my z
zepdrix
  • zepdrix
\[\large\rm \frac{dz}{d \theta}= i (\color{orangered}{z})\]
anonymous
  • anonymous
I'm still here - just processing the information and looking everything up and cross referencing in my formula book ... !
zepdrix
  • zepdrix
Then 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
  • IrishBoy123
brill astro!! historically, Euler found this from comparing Taylor series
zepdrix
  • zepdrix
Exponentiating 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
  • Astrophysics
Thanks irishboy xD Nice one zep!
anonymous
  • anonymous
Okay. 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?
1 Attachment
anonymous
  • anonymous
Thank you so much! I finally got it, zapdrix! Thanks to everyone else as well for your input! :)
anonymous
  • anonymous
If 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
  • Astrophysics
Yeah it's just a constant, you can leave it alone, or just add it to your calculations after
anonymous
  • anonymous
Thanks
zepdrix
  • zepdrix
Yes, 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
  • zepdrix
Ah yes, good with the trig. Sorry I kinda skipped over that at the start >.<

Looking for something else?

Not the answer you are looking for? Search for more explanations.