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Let $f$ be an analytic function. When is $\lvert f(x+iy) \rvert^2$ a harmonic function? So we know that if $f$ is analytic then $f = u(x,y) + iv(x,y)$ and $u$ and $y$ are both harmonic. So we can write $\lvert f(x+iy) \rvert^2 = u^2 + v^2 = h(x,y)$. In order for $h$ to be harmonic we need $h_{xx} + h_{yy} = 0$. So calculating $$ h_x = 2u\cdot u_x + 2v \cdot v_x, \quad h_y = 2u \cdot u_y + 2v \cdot v_y $$ and $$ h_{xx} = 2u_x^2 + 2v_x^2 + 2u \cdot u_{xx} + 2v \cdot v_{xx} \\ h_{yy} = 2u_y^2 + 2v_y^2 + 2u \cdot u_{yy} + 2v \cdot v_{yy} $$ So if we need $h_{xx} + h_{yy} = 0$ and knowing that $u$ and $v$ are harmonic we get the criteria $$ u_x^2 + u_y^2 + v_x^2 + v_y^2 = 0 $$ My question is that is this correct and can this criteria be improved some way?

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This can be seen quickly using Wirtinger derivatives. Note that $$ \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2} = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \overline{z}}$$ and $$\frac{\partial}{\partial z} \frac{\partial}{\partial \overline{z}} \lvert f(z)\rvert^2 = \lvert f’(z)\rvert^2.$$

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  • $\begingroup$ Well great but it doesn't really answer my question that is my thought process correct or not $\endgroup$ – Markus Punnar Dec 4 '19 at 18:50
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Note that for any constant $c \in \mathbb{C}$ and any holomorphic $f$ the difference $$\lvert f(z) \rvert^2 - \lvert f(z)-c \rvert^2$$ is harmonic. In particular if $\lvert f(z) \rvert^2$ is harmonic then so is $\lvert f(z)-c\rvert^2$. In that case $\lvert f(z)-f(z_0) \rvert^2$ is harmonic around $z_0$ and by the mean value property of harmonic functions $\lvert f(z)-f(z_0) \rvert^2$ must be identically zero around $z_0$.

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