rvc
  • rvc
separate into real and imaginary parts
Mathematics
katieb
  • katieb
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rvc
  • rvc
\[\sin^{-1} (e^{i\theta})\]
rvc
  • rvc
\[e^{i\theta}= \sin(x+iy)\]
rvc
  • rvc
\[\cos\theta+isin\theta=sinxcos(iy)+cosxsin(iy)\] \[= sinxcoshy+icosxsinhy\]

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rvc
  • rvc
\[\cos\theta=sinxcoshy~;~\sin\theta=cosxsinhy\]
rvc
  • rvc
correct?
ribhu
  • ribhu
hey
rvc
  • rvc
please help nn :)
ribhu
  • ribhu
can i help? if u want?
rvc
  • rvc
yep
rvc
  • rvc
please ans
ribhu
  • ribhu
remember me?
dinamix
  • dinamix
i think \[\sin^{-1} (\sin(x+iy)) = x+iy\]
dinamix
  • dinamix
dinamix
  • dinamix
did u see my solution
rvc
  • rvc
no its sin inverse e^itheta
rvc
  • rvc
@ribhu if u want to help then i dont mind dont ask questions like that
dinamix
  • dinamix
u didnt mean arcsin or what
rvc
  • rvc
\[Let~\sin^{-1} e^{i\theta}=x+iy\]
dinamix
  • dinamix
cuz i know sin^-1 = arcsin
IrishBoy123
  • 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
rvc
  • rvc
hmm what if i go with my steps
rvc
  • rvc
@hartnn pls help
IrishBoy123
  • IrishBoy123
you steps are technically correct, i believe. where are you taking it from here?
rvc
  • rvc
i need the value of x and y
IrishBoy123
  • 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
IrishBoy123
  • 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
IrishBoy123
  • 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 :-(
rvc
  • rvc
@hartnn @baru @IrishBoy123 can we get back to this question please..
baru
  • baru
i dont even understand how to interpret that, sin( complex number) is a thing ?
baru
  • baru
i think these are @ParthKohli 's hunting grounds
rvc
  • rvc
rvc
  • rvc
@hartnn please help whenever u r online...
Kainui
  • Kainui
$$z= \frac{z+z^*}{2} + i \frac{z-z^*}{2i} = \Re(z)+i \Im(z)$$ That's probably the most important formula you can have, try it out on anything simple to test it.

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