Consider the principal branch of logarithm in its polar form $$f(z)=\ln r+i\theta$$ If we take as domain $G=\mathbb{C}\setminus\{0\}$ then obviously $\theta$ is not a continuous function of $z$ as we approach the negative real axis. However, one can be tempted to verify the Cauchy-Riemann equations in their polar form and conclude (mistakenly) that the function is analytic since $$u=\ln r, v=\theta$$ and $$u_r=\frac{1}{r}, v_\theta=1$$ Which detail is the person overlooking while doing so?
Edit: I have seen proofs of why logarithm is analytic on $\mathbb{C}\setminus(-\infty,0]$ and it involves doing exactly what I did above; however, the proofs do not reflect why we need such a domain.
I believe that the problem is with multivalued functions.
In essence, the logarithm takes $z= re^{i\theta}\neq 0$ and gives the set $$\left\{\ln(r) +i\left( \theta + 2k\pi\right)\mid k\in \mathbb{Z}\right\}$$ as the exponential of any of them is $z$.
Once we have set up that $-\pi <\theta \leq \pi$ i.e. $z\in \mathbb{C} \setminus (-\infty, 0]$ we have a way to choose a member of the above list and make a (univalued) function that happens to be given by the expression $\ln(r) +i\theta$.
Let's see in an example how it works. I'll take the limit $\lim\limits_{z\to -1}\log(z)$ for two determinations of the logarithm.
First we fix the principal branch $-\pi <\theta \leq \pi$. Take $z = -1 + it$ then
$$\lim\limits_{t\to 0^+}\log(-1+it) = \lim\limits_{t\to 0^+} i\theta(t) = i\pi \, \text{ and } \lim\limits_{t\to 0^-}\log(-1+it) = -i\pi $$ Now if we select the branch $0 <\theta \leq 2\pi$ (defined on $\mathbb{C} \setminus [0,\infty)$) we have that $$ \lim\limits_{z\to -1}\log(z) = i\pi. $$
Both are given by the same expression but they differ when it comes to how $\theta$ behaves.