Given that #sinh^-1x=ln{x+sqrt(x^2+1)}#, how do you determine #sinh^-1(2+j)# in the form #a + jb#?

Question

2277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-28T19:31:52

See below.

#sinhx_0=(e^(x_0)-e^(-x_0))/2=s_0#

or making #y=e^(x_0)#

#y-1/y=2s_0#. Solving for #y#

#y = s_0+sqrt(s_0^2+1) = e^(x_0)#

so

#x_0=log( s_0+sqrt(s_0^2+1))#. If now #x_0=2+i# then

#s_0 = (e^(2+i)-e^(-(2+i)))/2= (e^2 cdot e^i-e^(-2)cdot e^(-i))/2#
Using de Moivre's identity

#e^(iphi)=cosphi+isinphi# we get

#sinh(2+i)=1/2[(e^2-e^(-2))cos(1)+i(e^2+e^(-2))sin(1)]#

or

#sinh(2+i)=sinh(2)cos(1)+icosh(2)sin(1)#