<s>[INST] Conjugate Harmonic $\displaystyle u=\frac{1}{2}\log(x^2+y^2)$ is harmonic and find its harmonic conjugate function
$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$
$\displaystyle\frac{\partial u}{\partial y}=\frac{y}{x^2+y^2}$
$\displaystyle dv=\int\frac{-\partial u}{\partial y}dx  +\int\frac{\partial u}{\partial x}dy$
then
$\displaystyle v= \tan^{-1}\frac{y}{x} - \tan^{-1}\frac{x}{y} +c$
What did I do wrong here ?
 [/INST] $$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$$
One of the Cauchy-Riemann equations is:
$$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
so
$$\frac{\partial v}{\partial y}=\frac{x}{x^2+y^2}$$
$$v = \int \frac{x}{x^2+y^2} \, dy=\frac{1}{x} \int  \frac{1}{1+(\frac{y}{x})^2} \,dy= \tan^{-1} \frac{y}{x} + \text{const.} $$
Similarly,
$$v =- \int \frac{y}{x^2+y^2} \, dx=- \tan^{-1} \frac{x}{y} + \text{const.} $$
Since $\tan^{-1} \frac{y}{x} $ and $  - \tan^{-1} \frac{x}{y}$ differ by a constant $(|\tan^{-1} \frac{y}{x} +\tan^{-1} \frac{x}{y}| = \frac{\pi}{2})$, either answer is acceptable.
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