What is the argument and module of the complex number sin(i+3) ?

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Answer 1 · 2017-03-06T15:46:28

See below.

Using de Moivre's identity

#sin x=(e^(ix)-e^(-ix))/(2i)# we have

#sin(i+3) = (e^(i(i+3))-e^(-i(i+3)))/(2i) = -1/2 i (e^(-1 + 3 i) - e^(1 - 3 i))=#

#=(1/(2 e) + 1/2 e) sin(3)+i (1/2 e-1/(2 e) ) cos(3)=#

#=sqrt[1 + e^4 - 2 e^2 cos(6)]/(2 e) e^(iphi)#

where

#phi = tan^-1((e^2-1) /(e^2+1)cot(3))#