Prove that the following holds for $z$.
$$\frac{\partial ^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2}=4\frac{\partial ^2}{\partial z \partial \bar z}$$
I'm having a little trouble showing the above statement.
I proceed by evaluating the right-hand side of the equation.
$$g(z)=\frac{\partial f}{\partial \bar z}=\frac 12 \times \left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)$$
$$\frac{\partial g}{\partial z}=\frac{\partial g}{\partial x}\times\frac{\partial x}{\partial z}+\frac{\partial g}{\partial x}\times\frac{\partial y}{\partial z}$$
we know that $\frac{\partial x}{\partial z}=\frac{\partial y}{\partial z}=\frac 12$
So, $$4\frac{\partial ^2 f}{\partial z \partial \bar z}= 4\times\frac 14 \left((f_{xx}+f_{yx}) + i(f_{xy}+f_{yy}) \right)$$
I'm not so sure how this equals the left-hand side which is $f_{xx}+f_{yy}$
Also, the above equation seems to imply that; If $f$ is analytic, then, $$\frac{\partial f}{\partial \bar z}=0$$ then the right-hand side is zero. Which says that $f$ is harmonic.
Is every analytic function harmonic?
Best Answer
You made a mistake in evaluating $\frac{\partial y}{\partial z}$, since
$$ \frac{\partial }{\partial z} = \frac 12 \left( \frac{\partial }{\partial x} - \sqrt{-1} \frac{\partial }{\partial y}\right),$$
you should have $\frac{\partial y}{\partial z} = -\frac 12 \sqrt{-1}$. With this you should get
$$4\frac{\partial ^2 f}{\partial z \partial \bar z}= 4\times\frac 14 \left((f_{xx}+f_{yy}) + i(f_{xy}-f_{yx}) \right) = f_{xx} + f_{yy}.$$