Question

Can't we do the Particular solution by assuming sin(z) as imaginery part of e^iz,plz can u show how to do by that method

Answer 1

Yes , we can find particular solution by assuming sin(z) as imaginary part of $e^{iZ}$ .

let take a function Z with a variable t , so we write this function Z = cos t + i sin t

dz = (-sint + i cost )dt,

Taking i common from whole function , we get ,

dz = i(cost + i sint)dt = izdt

now seperating the variables , we get

-

$\dfrac{dz}{z}=idt$ ,

now integerating on both side of the function , we get -

$\int\dfrac{dz}{z}=\int idt$

ln z = it ,

now removing the function ln ,

Z=$e^{it}$ +c,

so this is a particular solution when we assume sin z as imaginary part of $e^{iz}.$