## rvc one year ago separate into real and imaginary parts

1. rvc

$\sin^{-1} (e^{i\theta})$

2. rvc

$e^{i\theta}= \sin(x+iy)$

3. rvc

$\cos\theta+isin\theta=sinxcos(iy)+cosxsin(iy)$ $= sinxcoshy+icosxsinhy$

4. rvc

$\cos\theta=sinxcoshy~;~\sin\theta=cosxsinhy$

5. rvc

correct?

6. ribhu

hey

7. rvc

8. ribhu

can i help? if u want?

9. rvc

yep

10. rvc

11. ribhu

remember me?

12. dinamix

i think $\sin^{-1} (\sin(x+iy)) = x+iy$

13. dinamix

@rvc

14. dinamix

did u see my solution

15. rvc

no its sin inverse e^itheta

16. rvc

@ribhu if u want to help then i dont mind dont ask questions like that

17. dinamix

u didnt mean arcsin or what

18. rvc

$Let~\sin^{-1} e^{i\theta}=x+iy$

19. dinamix

cuz i know sin^-1 = arcsin

20. IrishBoy123

try starting with $sin z = \frac{e^z - e^{-z}}{2i}$ and note that $w = arcsin \ e^{i \theta} \implies sin(w) = \ e^{i \theta} =\frac{e^w - e^{-w}}{2i}$ ugly but potentially do-able

21. rvc

hmm what if i go with my steps

22. rvc

@hartnn pls help

23. IrishBoy123

you steps are technically correct, i believe. where are you taking it from here?

24. rvc

i need the value of x and y

25. IrishBoy123

well i found a reliable formula $arcsin z = -i \ ln (iz \pm \sqrt{1-z^2})$ and the answer i get from the method i propose actually looks very similar i do not see where you go next in you approach though i see no problems with it

26. IrishBoy123

following on: $$2i e^{i \theta } = e^w - e^{-w}$$ $$i = e^{\pi/2}$$ $$u = e^w$$ $$u - 2 e^{i(\theta + \pi/2)} - \frac{1}{u} = 0$$ $$u^2 - u.2 e^{i(\theta + \pi/2)} - 1 = 0$$ and apply quadratic formula => gets you that really ugly formula i posted above

27. IrishBoy123

the switch to $$e^{i(\theta + \pi /2 )}$$ is OTT and only complicates matters. you can just stuff it all straight into the quadratic mea culpa :-(

28. rvc

@hartnn @baru @IrishBoy123 can we get back to this question please..

29. baru

i dont even understand how to interpret that, sin( complex number) is a thing ?

30. baru

i think these are @ParthKohli 's hunting grounds

31. rvc

@Kainui

32. rvc

$$z= \frac{z+z^*}{2} + i \frac{z-z^*}{2i} = \Re(z)+i \Im(z)$$ That's probably the most important formula you can have, try it out on anything simple to test it.