# Harmonic Conjugate

• Feb 11th 2010, 05:06 PM
seichan
Harmonic Conjugate
I have a problem with searching for the harmonic conjugate of the function Im(e^(z^2)). It would seem through all my work that this DNE, however, my homework program insists it should and does. Could you tell me if I am going wrong in my derivatives, etc?

Im(e^((x^2+y^2)+2xiy))=Im(e^(x^2+y^2)(cos(2xy)+isi n(2xy))= e^(x^2+y^2)*sin(2xy)

ux= 2xe^(x^2+y^2)*sin(2xy)+2ye^(x^2+y^2)cos(2xy)
uxx=2e^(x^2+y^2)*sin(2xy)+4x^2e^(x^2+y^2)*sin(2xy)-4y^2e^(x^2+y^2)sin(2xy)

uy= 2ye^(x^2+y^2)*sin(2xy)+2xe^(x^2+y^2)cos(2xy)
uyy=2e^(x^2+y^2)*sin(2xy)+4y^2*e^(x^2+y^2)*sin(2xy )-4x^2*e^(x^2+y^2)sin(2xy)

uyy+uxx=4e^(x^2+y^2)sin(2xy)

Thank you
• Feb 12th 2010, 12:03 AM
Opalg
Quote:

Originally Posted by seichan
I have a problem with searching for the harmonic conjugate of the function Im(e^(z^2)). It would seem through all my work that this DNE, however, my homework program insists it should and does. Could you tell me if I am going wrong in my derivatives, etc?

Im(e^((x^2+y^2)+2xiy))=Im(e^(x^2+y^2)(cos(2xy)+isi n(2xy))= e^(x^2+y^2)*sin(2xy)

ux= 2xe^(x^2+y^2)*sin(2xy)+2ye^(x^2+y^2)cos(2xy)
uxx=2e^(x^2+y^2)*sin(2xy)+4x^2e^(x^2+y^2)*sin(2xy)-4y^2e^(x^2+y^2)sin(2xy)

uy= 2ye^(x^2+y^2)*sin(2xy)+2xe^(x^2+y^2)cos(2xy)
uyy=2e^(x^2+y^2)*sin(2xy)+4y^2*e^(x^2+y^2)*sin(2xy )-4x^2*e^(x^2+y^2)sin(2xy)

uyy+uxx=4e^(x^2+y^2)sin(2xy)

Are you not allowed to use the fact that the harmonic conjugate of $\displaystyle \text{Im}(e^{z^2})$ is $\displaystyle \text{Re}(e^{z^2})$?

If you write z in polar form $\displaystyle z = re^{i\theta}$, then $\displaystyle z^2 = r^2e^{2i\theta} = r^2(\cos2\theta + i\sin2\theta)$, and $\displaystyle e^{z^2} = e^{r^2\cos2\theta}e^{ir^2\sin2\theta} = e^{r^2\cos2\theta}(\cos(r^2\sin2\theta) + i\sin(r^2\sin2\theta)).$ The real part of that is $\displaystyle e^{r^2\cos2\theta}(\cos(r^2\sin2\theta) = e^{r^2(\cos^\theta - \sin^2\theta)}(\cos(2r\cos\theta r\sin\theta)) = e^{x^2-y^2}\cos(2xy)$.

If you are not allowed to use the fact that the real and imaginary parts of an analytic function are harmonic conjugates of each other, then you will have to do the problem by working on the partial derivatives. But notice that the real part of $\displaystyle z^2$ is $\displaystyle x^2 \mathrel{\color{red}-} y^2$, not $\displaystyle x^2+y^2$.
• Feb 12th 2010, 12:31 PM
seichan
Ah! Of course! How did I overlook that? Thank you so much! *facepalm*