Show that $\log(x^2+y^2)$ is not a real part of analytic function in $z\ne 0$.

Question

Show that $$u(z)=\log(x^2+y^2)$$ is harmonic in $z\ne 0$ and that $u$ cannot be the real part of an analytic function in $z\ne 0$.

My attempt:

Direct calculation shows that $u_{xx}+u_{yy}=0$ in $z\ne 0$.

The solution in the book to show that $u$ cannot be a real part of an analyitc function in $z\ne0$ is that if such $f$ exists then $f(z)=\log(z)+c$ for all $\arg(z)\in(0,2\pi)$ but I don't see why that's true.

Answer 1

Consider the path $c(t) = \cos(t) + i\sin(t)$ tracing the unit circle, and consider the imaginary part $v$ of this analytic function, assuming it exists. Now using the Cauchy-Riemann equations you can find $\frac{d}{dt}v(c(t)) = 2$, so $v(c(2\pi)) = v(c(0)) + 4\pi$, but that can't be right since $c(2\pi) = c(0)$.

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To expand on why $\frac{d}{dt}v(c(t)) = 2$: $$ u_x(x,y) = \frac{2x}{x^2+y^2},~~ u_y(x,y) = \frac{2y}{x^2+y^2}.$$ So $$\frac{d}{dt}v(c(t)) = -\sin(t)v_x(c(t))+\cos(t)v_y(c(t))$$ $$=\cos(t) u_x(c(t)) + \sin(t)u_y(c(t))$$ $$=\cos(t)\cdot \frac{2\cos(t)}{\cos(t)^2+\sin(t)^2} + \sin(t)\cdot \frac{2\sin(t)}{\cos(t)^2+\sin(t)^2} = 2.$$