Proving the Holomorphicity of $h(z)= \partial u / \partial y + i\partial u/ \partial x$ via $\nabla^2 \phi = 0$

223 Views Asked by At

In the book Function Theory of a Complex Variable I'm having trouble verifying my proof of proposition of $(1.)$

$(1.)$

Let $u$ be a $C^{2}$, real-valued harmonic function on an Open Set $\text{U} \subset \mathbb{C}$. Prove that $h(z)= \partial u / \partial y + i\partial u/ \partial x$ is holomorphic on $U$.

$$\text{Lemma}:$$ To formally begin attacking our proposition in $(1.)$ one can make the following observations in $(2.)$

$(2.)$

$\text{Remark}$:

Let $h :\, \, \text{U} \rightarrow \mathbb{C}$, one can note $h(z)$ is a harmonic function on our set $\text{U}$, following the observations made in $(3.)$

$(3)$

$$h(z)= \partial u / \partial y \, u(x,y)+ i\partial u/ \partial x\, iv(x,y)$$

$$h(z)= x +iv(y)$$

After considering Differability of $h(z)$, our function is Harmonic, since it satisfies Laplace's equation in $\text{Lemma (2)}$.

$$\text{Lemma (2)}$$

$$\nabla^2 \phi = 0$$

Applying Laplace's Equation it should be obvious to notation that:

$$\nabla \cdot \nabla f (h(z))$$

$$\nabla \cdot \nabla f (h(z)) = \partial ^{2}f/\partial x_{i}(x) +\partial ^{2}f/\partial y_{i}(iv(y))=0$$

From our prior developments it should be sound to reason that since $h(z)$ is harmonic it is also holomorphic.

1

There are 1 best solutions below

2
On BEST ANSWER

Hint: To show $h=u_y + i u_x$ is holomorphic, verify that the Cauchy-Riemann equations are satisfied. Here that means

$$(u_y)_x = (u_x)_y,\,\,(u_y)_y = -(u_x)_x.$$