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I denote with $\phi(x,y)$ a hamornic function. I would like to show that $\phi$ is the real part of any analytic function $f(z)$ of the form $$f(z)=2\phi\left(\frac{z+1}{2},\frac{z-1}{2i}\right)-\phi(1,0)+ic$$ where c is a real constant and provided that the RHS exists.

My idea was to consider $$g(z,\bar{z})=\phi\left(\frac{1}{2}(z+\bar{z}),\frac{1}{2i}(z-\bar{z})\right)+i\psi\left(\frac{1}{2}(z+\bar{z}),\frac{1}{2i}(z-\bar{z})\right)$$

and set this equal to $f(z)=g(z,\bar{z})$ with $\bar{z}=1$, but is this helpful?

Remark: In the end I would like to find an analytic function whose real part is $\tan^{-1}y/x$, my lecture notes states that the obvious function is $-i\log z$ but I do not see why this is so obvious.

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Search Cauchy-Riemann equations. –  Martín-Blas Pérez Pinilla Jan 26 at 23:59
    
@Martín-BlasPérezPinilla, that would show that the real part of an analytic function was harmonic, but not the converse without a little work. I'm sure the poster is aware of the cauchy-riemann equations. –  JSempliner Jan 27 at 0:01

2 Answers 2

up vote 2 down vote accepted

Unless I'm missing something, your question makes little sense to me:

  • $\phi\left(\frac{z+1}{2},\frac{z-1}{2i}\right)$ is not defined since $\phi(x,y)$ only exists for $x$ and $y$ real
  • Your candidate function $f$ has constant imaginary part, which is impossible for a non constant holomorphic function.

It's a classical fact or exercise that a harmonic function is (at least locally) the real part of a holomorphic function, and that this holomorphic function is unique up to a (purely imaginary) additive contant (and this function looks nothing like what you wrote). You should try to do this exercise (use the Cauchy-Riemann equations), it would be instructive.

As for your example, I can explain it to you: recall the principal value logarithm is $\log z = \ln |z| + i \operatorname{Arg}(z)$ (for $z \notin \mathbb{R}_-$), where $\operatorname{Arg}(z)$ is the principal value argument of $z$, i.e. in $(-\pi, \pi]$. In your problem, I'm assuming you are working in the half-plane $H = \{z\in \mathbb{C}, ~\operatorname{Re}(z)>0\}$. For $z \in H$, it's not hard to check that $\operatorname{Arg}(z) = \tan^{-1}(y/x)$, where $z = x+iy$. It follows that $\operatorname{Re}(-i\log(z)) = \operatorname{Arg}(z) = \tan^{-1}(y/x)$.

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Thanks for your answer, do you also have an idea how to find an analytic fct whose real part is $\frac{x}{x^2+y^2}$? –  Alexander Feb 10 at 12:43
    
yes: take $1/z$ –  Seub Feb 10 at 17:47

Your first phrase should read "... is the real part of some analytic function...".

As Seub said, your first formula does not makes sense. Your second formula makes sense, but the RHS depends only of $z$ ($\bar{z}$ is the conjugate of $z$!), so $\bar{z}=1$ is absurd.

The usual solution is take $g=\phi_x-i\phi_y$ (the presumptive derivative of $f=\phi+i\psi$), check analiticity with Cauchy-Riemann and integrate (calculate a primitive). Warning: the topology of the domain is important, as your example proves.

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