Let v be the harmonic conjugate of u=Im(e^(z²)), then:
uᵪ=vᵧ, uᵧ=-vᵪ. And z can be written z=x+iy where x, y ∈ ℝ.
Therefore z²=x²-y²+2ixy.
So e^(z²)=e^(x²-y²+2ixy)=e^(x²-y²)e^(2ixy).
e^(2ixy)=cos(2xy)+isin(2xy), so u=Im(e^(z²))=e^(x²-y²)sin(2xy).
(1) vᵧ=uᵪ=2xe^(x²-y²)sin(2xy)+2ye^(x²-y²)cos(2xy);
vᵪ=-uᵧ=-(-2ye^(x²-y²)sin(2xy)+2xe^(x²-y²)cos(2xy)).
(2) vᵪ=-2xe^(x²-y²)cos(2xy)+2ye^(x²-y²)sin(2xy).
Now we need to integrate (1) and (2) and harmonise the common results.
Assume v=-e^(x²-y²)cos(2xy), then:
vᵪ=-2xe^(x²-y²)cos(2xy)+2ye^(x²-y²)sin(2xy)+g(y), same as (2), apart from the function g(y) yet to be found.
vᵧ=2ye^(x²-y²)cos(2xy)+2xe^(x²-y²)sin(2xy)+h(x), same as (1) apart from the undefined function h(x).
Because vᵪand vᵧare consistent with v=-e^(x²-y²)cos(2xy), h(x)=g(y)=0.
Therefore v is the harmonic conjugate of u.
But Re(e^(z²))=e^(x²-y²)cos(2xy)=-v, so the harmonic conjugate of u is:
-Re(e^(z²)).