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

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rvc
 one year ago
Best ResponseYou've already chosen the best response.2\[\sin^{1} (e^{i\theta})\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.2\[e^{i\theta}= \sin(x+iy)\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos\theta+isin\theta=sinxcos(iy)+cosxsin(iy)\] \[= sinxcoshy+icosxsinhy\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos\theta=sinxcoshy~;~\sin\theta=cosxsinhy\]

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0i think \[\sin^{1} (\sin(x+iy)) = x+iy\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.2no its sin inverse e^itheta

rvc
 one year ago
Best ResponseYou've already chosen the best response.2@ribhu if u want to help then i dont mind dont ask questions like that

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0u didnt mean arcsin or what

rvc
 one year ago
Best ResponseYou've already chosen the best response.2\[Let~\sin^{1} e^{i\theta}=x+iy\]

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0cuz i know sin^1 = arcsin

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0try 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 doable

rvc
 one year ago
Best ResponseYou've already chosen the best response.2hmm what if i go with my steps

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you steps are technically correct, i believe. where are you taking it from here?

rvc
 one year ago
Best ResponseYou've already chosen the best response.2i need the value of x and y

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0well i found a reliable formula \[arcsin z = i \ ln (iz \pm \sqrt{1z^2})\] 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
 one year ago
Best ResponseYou've already chosen the best response.0following 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
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.2@hartnn @baru @IrishBoy123 can we get back to this question please..

baru
 one year ago
Best ResponseYou've already chosen the best response.0i dont even understand how to interpret that, sin( complex number) is a thing ?

baru
 one year ago
Best ResponseYou've already chosen the best response.0i think these are @ParthKohli 's hunting grounds

rvc
 one year ago
Best ResponseYou've already chosen the best response.2@hartnn please help whenever u r online...

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0$$z= \frac{z+z^*}{2} + i \frac{zz^*}{2i} = \Re(z)+i \Im(z)$$ That's probably the most important formula you can have, try it out on anything simple to test it.
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