So I am trying to analyze the following:

Suppose $z=x+iy$. Let $u(x,y)=\ln(x^2+y^2)-\ln[(x+\frac{1}{2})^2+y^2]$. Does there exist single-valued analytic function whose real part is $u$ on the domain $D=\left\{ z \in \mathbb{C}: 1< |z|<2 \right\}$.

I am thinking the answer is none since in the first place, $x^2+y^2=|z|^2$ which is a case whose derivative may exist only at $z=0$ so as a consequence, $\ln(x^2+y^2)$ is not analytic and so is $u$. Is this incorrect? Do I have to utilize Cauchy-Riemann Eq. or harmonic conjugate/function?(Actually I tried but the math was a mess...)


1 Answer 1


$$ u(x, y) = \ln |z|^2 - \ln \left|z + \frac 12 \right|^2 = \frac 12 \cdot \ln \left| \frac{z}{z+\frac 12}\right| $$ is the real part of a holomorphic function in $D$ if and only if there is a holomorphic branch of $$ \ln \left( \frac{z}{z+\frac 12}\right) $$ in $D$, and that is the case if and only if $$ \frac 1z - \frac{1}{z + \frac 12} $$ has an antiderivative in $D$, which in turn is equivalent to $$ \int_\gamma \left(\frac 1z - \frac{1}{z + \frac 12}\right) \, dz = 0 $$ for all closed paths in $D$. But $$ \frac{1}{2 \pi i}\int_\gamma \left(\frac 1z - \frac{1}{z + \frac 12}\right)\, dz = I(\gamma, 0) - I(\gamma, \frac 12) $$ where $I$ is the winding number, and that difference is zero because $0$ and $1/2$ lie in the same component of $\Bbb C\setminus D$.

So $u$ is the real part of a holomorphic function in $D$.

Alternatively you can show that $$ z \mapsto \frac{z}{z+\frac 12} $$ maps $D$ to a subset of the right halfplane, where the principal branch $\operatorname{Log}$ of the logarithm is defined and holomorphic, so that $u$ is the real part of $$ \frac 12 \operatorname{Log}\left( \frac{z}{z+\frac 12}\right) \, . $$

  • $\begingroup$ I think I need more time to really understand this answer for I thought I will not use contour integrals yet. Will surely go back once I fully understood all necessary pieces of information. Thank you very much. $\endgroup$
    – Zai
    May 19, 2023 at 9:28
  • $\begingroup$ @Zai: You are welcome, and let me know if more details are needed. Note that the second solution (starting at “Alternatively ...”) does not use contour integrals. $\endgroup$
    – Martin R
    May 19, 2023 at 10:01

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