Question

separate into real and imaginary parts

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

\[\cos\theta=sinxcoshy~;~\sin\theta=cosxsinhy\]

Answer 5

correct?

Answer 6

hey

Answer 7

please help nn :)

Answer 8

can i help? if u want?

Answer 9

yep

Answer 10

please ans

Answer 11

remember me?

Answer 12

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

Answer 13

@rvc

Answer 14

did u see my solution

Answer 15

no its sin inverse e^itheta

Answer 16

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

Answer 17

u didnt mean arcsin or what

Answer 18

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

Answer 19

cuz i know sin^-1 = arcsin

Answer 20

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

Answer 21

hmm
what if i go with my steps

Answer 22

@hartnn pls help

Answer 23

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

Answer 24

i need the value of x and y

Answer 25

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

Answer 26

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

Answer 27

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 :-(

Answer 28

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

Answer 29

i dont even understand how to interpret that,

sin( complex number) is a thing ?

Answer 30

i think these are @ParthKohli 's hunting grounds

Answer 31

@Kainui

Answer 32

@hartnn please help whenever u r online...

Answer 33

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