I think you mean e^(ix) = cos(x) + sin(x) i What I find useful about Laplace and Fourier transform is there is that applied to differential equations there pro

I think you mean e^(ix) = cos(x) + sin(x) i What I find useful about Laplace and Fourier transform is there is that applied to differential equations there produce algebraic equations, so is a very useful tool when you are working in problems that are modelled via differential equations. My introduction for someone totally new to this field is to consider y = cos(x) and y= sin(x) as functions whose second derivative satisfie y'' = -y, and the exponential y = e^(ax) as the solution of y'=ay then you can think of solving this kind of equations trying a solution of the form y=e^(kx) then you realize that the constant k is the solution of a first or second order algebraic equation that you know how to solve. Then if your equation has imaginary solutions you need to know what is the exponential of an imaginary number, exp(ik) = cos(k) + i sin(k), then you say ok but that is not a real function, answer the real part and the imaginary part of that are also solutions. Going this way you find that a complex number z=a+bi is the solution of a differential equation that often constitute a model a physical system. z=a+bi is the solution of the second order equation (z-a)^2+b^2 = 0 and this algebraic equation goes to a differential equation just by replacing z^2 with y'' (second derivative) z with y' (first derivative) and finally you have the solutions y=e^(ax)cos(bx) and y=e^(ax)sin(bx) in which the amplitude of the signal is controlled by the exponential so that if a is negative you have a signal that goes to zero and otherwise a signal that explode.