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The details provided are that the function is analytic and that its real part along the line $y=c x$ is constant. What conclusions can I draw from here? I think that this imposes $\frac{\partial u}{\partial x}=0 $ and $\frac{\partial u}{\partial y}=0 $ along the line, however I am unsure how to take it from here.

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  • $\begingroup$ Pretty poor answers, considering there's a bounty on here $\endgroup$ Nov 10 '15 at 7:48
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Consider the function $g(z) := \mathrm i (f((1+\mathrm ic)z)-f(0))$. Clearly $g$ is analytic iff $f$ is analytic. Moreover, $(1+\mathrm ic)z$ has the property that $y=cx$ iff $z\in\mathbb R$. Finally, for real $z$, clearly $g(z)$ has a zero imaginary part, that is, it is real.

So $g(x)$ can be any analytic function that maps real values to real values.

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Look at $f(z)=z^2=x^2-y^2+2ixy$. We have $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$. Take $c=1$. Along the line $y=x$, we have $u=0$, but the partial derivatives of $u$ are not $0$ on the line.

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