Consider $u(x,y)=\dfrac{\text{log}(x^2+y^2)}{2}$ on $\Omega=\{0<r<|z|<R\}.$ Show there is no holomorphic function on $\Omega$ whose real part is $u.$

My attempt:

I understand that $u$ is real part of $\text{log}(z)$ and $\text{log}(z)$ is not well defined on $\Omega.$ How do I use this fact and identity theorem to show there isn't any holomorphic function on $\Omega$ whose real part is $u \ ?$

Thank you.


2 Answers 2


My preferred approach (real-analysis mostly)

Suppose that $u+iv$ is holomorphic in $\Omega$. By the Cauchy-Riemann equations, $$ \nabla v(x,y) = (-u_y,u_x) = (-y,x)/(x^2+y^2) $$ Integrate $\nabla v$ along the circle $x^2+y^2=\rho^2$ for some $\rho\in (r,R)$. You will find that the integral is strictly positive (actually, equal to $2\pi$) which contradicts the fundamental theorem of calculus for line integrals.

Approach via identity theorem.

Suppose such a holomorphic function $f$ exists. Consider $f(z)-\log z$, which is defined and holomorphic in the slit annulus $\Omega\setminus (-R,-r)$. Observe that the real part of this function is zero: hence, the function is identically equal to some imaginary constant. Contrast this with the fact that $f$ is continuous on the negative real axis while the principal branch of logarithm is not.


They idea behind my answer is similar to the one of Weapon of Choice.

If $f=u+iv$ was such a holomorphic function, then Cauchy-Riemann equations provide $$ f'(z)=u_x(z)+iv_x(z)=u_x(z)-iu_y(z)=\frac{x-iy}{x^2+y^2}=\frac{1}{z}. $$ But if $f$ is holomorphic in a region $\Omega$, then $\int_\gamma f'(z)\,dz=0$, for every closed path in $\Omega$. To make it clearer, let $\gamma: [a,b]\to\Omega$, with $\gamma(a)=\gamma(b)$. Then $$ \int_\gamma f'(z)\,dz=\int_a^bf'(\gamma(t))\, \gamma'(t)\,dt= f(\gamma(t))\,\big|_{\,a}^{\,b} = f(\gamma(b))-f(\gamma(a))=0. $$

In our case, for $r<\varrho<R$, $$ \int_{\lvert z\rvert=\varrho}f'(z)\,dz=\int_{\lvert z\rvert=\varrho}\frac{dz}{z}=2\pi i\ne 0. $$


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