Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be analytic. In the natural way, let $f = u + vi$ for $u,v : \mathbb{C} \rightarrow \mathbb{R}$. Let $z \in \mathbb{C}$.
Suppose that $u$ and $v$ satisfy Laplace's equation at $z$ (which they do since $f$ is analytic) so that
$$ \Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $$
and
$$ \Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 $$
Question 1: Does this mean that the real and complex parts of $f''(z)$ are equal to each other in magnitude?
EDIT: This is why I think that the Laplace Equations holding for $u$ and $v$ implies that $f''(z)$ has real and complex components which are equal in magnitude to each other.
Since $f$ is analytic, we have that the Cauchy-Riemann Equations hold for $u$ and $v$. In particular, we have that
$$ \frac{\partial u}{\partial x} = \frac{\partial v }{\partial y} $$
and
$$ \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} $$
This then means that $f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y}$.
Then $\color{red}{f''(z) = \frac{\partial^2 u}{\partial x^2} - i \frac{\partial^2 u}{\partial y^2}}$.
Yet since $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ via Laplace, we have that $f''(z)$ satisfies that its real and complex parts are equal in magnitude to each other.
Is this correct?
Question 2: If no to the above answer, what is a geometric interpretation of this statement?