## 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 :)

1. anonymous

2. Astrophysics

You can use taylor series

3. zepdrix

Ivan have you taken any Calculus? You can generate the relationship using integration, it's pretty neat :) Yah Taylor Series is good too.

4. anonymous

Yes - I just dont recall much of Taylors though. Working on it.. !

5. 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

6. Astrophysics

Yes you will have to note A = a+bi, I think zep can explain it a better way without Taylor

7. 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: $$\large\rm z=r(\cos\theta+i\sin\theta)$$

8. 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:

9. 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 $$\large\rm -1=i\cdot i$$

10. 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!$\large\rm \frac{dz}{d \theta}= i r(\color{orangered}{z})$

11. zepdrix

Woops woops, little mistake, my bad. the r is part of my z

12. zepdrix

$\large\rm \frac{dz}{d \theta}= i (\color{orangered}{z})$

13. anonymous

I'm still here - just processing the information and looking everything up and cross referencing in my formula book ... !

14. zepdrix

Then separation of variables and integrating gives us:$\large\rm \int\limits \frac{dz}{z}=\int\limits i d\theta$Leading to,$\large\rm \ln z=i \theta+c$

15. IrishBoy123

brill astro!! historically, Euler found this from comparing Taylor series

16. zepdrix

Exponentiating gives us:$\large\rm z=e^{i \theta+c}=e^{i \theta}e^c$ $\Large\rm z=e^c \color{orangered}{e^{i \theta}}=r(\color{orangered}{\cos \theta+i \sin \theta})$

17. Astrophysics

Thanks irishboy xD Nice one zep!

18. 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?

19. anonymous

Thank you so much! I finally got it, zapdrix! Thanks to everyone else as well for your input! :)

20. 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

21. Astrophysics

Yeah it's just a constant, you can leave it alone, or just add it to your calculations after

22. anonymous

Thanks

23. 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.

24. zepdrix

Ah yes, good with the trig. Sorry I kinda skipped over that at the start >.<