- #1
thoughtgaze
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Can anyone point me to some material on applying the Fourier transform to the case of an analytic function of one complex variable?
I've tried to generalize it myself, but I want to see if I'm overlooking some important things. I've started by writing the analytic function with
u + iv where u and v satisfy the cauchy riemann equations. I'm tempted to start by saying that to take the Fourier transform of u + iv, simply take the Fourier transform of u and v in the usual way for a function of two variables.
Usually you would have something like u(x,y) → μ(k1, k2) and similarly for v(x,y) → γ(k1, k2)
However, applying the cauchy riemann equations necessarily sets k2 = i k1
and
which implies that the Fourier transform of the full analytic function is simply
μ(k1, k2) =μ(k1)δ(k2 - ik1) = iγ(k1, k2) = iγ(k1)δ(k2 - ik1)
(the δ(k2 - ik1) is just the dirac delta function, which I'm hoping is okay to use even with a complex argument.)
u + iv = ∫dk[2μ(k)]eik(x +iy) = ∫dkdx'dy'2e-i(k(x' - x +i(y'-y))) u(x,y)
μ(k) = ∫dxdy e-ik(x +iy) u(x,y)
I've tried to generalize it myself, but I want to see if I'm overlooking some important things. I've started by writing the analytic function with
u + iv where u and v satisfy the cauchy riemann equations. I'm tempted to start by saying that to take the Fourier transform of u + iv, simply take the Fourier transform of u and v in the usual way for a function of two variables.
Usually you would have something like u(x,y) → μ(k1, k2) and similarly for v(x,y) → γ(k1, k2)
However, applying the cauchy riemann equations necessarily sets k2 = i k1
and
which implies that the Fourier transform of the full analytic function is simply
μ(k1, k2) =μ(k1)δ(k2 - ik1) = iγ(k1, k2) = iγ(k1)δ(k2 - ik1)
(the δ(k2 - ik1) is just the dirac delta function, which I'm hoping is okay to use even with a complex argument.)
u + iv = ∫dk[2μ(k)]eik(x +iy) = ∫dkdx'dy'2e-i(k(x' - x +i(y'-y))) u(x,y)
μ(k) = ∫dxdy e-ik(x +iy) u(x,y)