Question

Harmonic conjugate of the function u=im e^z^2 is

Answer 1

If w=a+biw=a+bi then Rw=aℜw=a and Iw=bℑw=b, furthermore −iw=b−ai−iw=b−ai so R(−iw)=b=I(w)ℜ(−iw)=b=ℑ(w) and I(−iw)=R(w)ℑ(−iw)=ℜ(w). In particular if w=ez2w=ez2 then I(−iez2)=R(ez2)ℑ(−iez2)=ℜ(ez2); they are the same function. – anon May 20 '13 at 7:26

For real-valued functions u(x,y)u(x,y) and v(x,y)v(x,y), we say that vv is a harmonic conjugate of uu if u+ivu+iv is analytic. This is not a symmetric relation between uu and vv. For example, exsinyexsin⁡y is a harmonic conjugate of excosyexcos⁡y because excosy+iexsiny=ezexcos⁡y+iexsin⁡y=ez is analytic, but excosyexcos⁡y is not a harmonic conjugate of exsinyexsin⁡y because exsiny+iexcosyexsin⁡y+iexcos⁡y is not analytic (check the Cauchy-Riemann equations).

If vv is a harmonic conjugate of uu, meaning that the function f(z)=u+ivf(z)=u+iv is analytic, then −if(z)=v−iu−if(z)=v−iu, being a constant multiple of f(z)f(z), is also analytic; this shows that −u−u is a harmonic conjugate of vv, in other words, −Rf(z)−ℜf(z) is a harmonic conjugate of If(z)ℑf(z).

To repeat one more time: although If(z)ℑf(z) is a harmonic conjugate of Rf(z)ℜf(z) (assuming f(z)f(z) is analytic), it's −Rf(z)−ℜf(z), not Rf(z)ℜf(z), that is a harmonic conjugate of If(z)ℑf(z). This is true for any analytic function f(z)f(z), including your problem function f(z)=ez2f(z)=ez2.

shareciteimprove this answeredited May 20 '13 at 10:16answered May 20 '13 at 9:13user75900add a commentup vote1down vote

Let w=a+biw=a+bi. Note that Im (iw)=Im (−b+ai)=a=Re (w)Im (iw)=Im (−b+ai)=a=Re (w).

So yes, it is the same function up to a minus sign, which I don't know where it comes from - perhaps it comes from the manual's specific definition of a harmonic conjugate.