If $\ln(z)=\ln|z|+i\arg(z)$, and $\arg(z) \in (-\pi,\pi]$ i tried to check is $\ln(z)$ analytic function.
I wrote $\ln(z)=\ln|z|+i(\arctan(\frac{y}{x})+c)$ where $c$ is some of numbers $ 0, \pm \pi$. So real part of this function is $u(x,y)=\ln\sqrt{x^2+y^2}$ and imaginary part is $v(x,y)=\arctan(\frac{y}{x})+c$
If I check $u_x=v_y=\frac{x}{x^2+y^2}$ and $u_y=-v_x$ and these partial derivatives are continues as a function over $\mathbb{R}^2\setminus0$. So, it implies that $\ln(z)$ is R differentiable and Cauchy Rimann conditions are true, so it implies $\ln(z)$ is differentiable.
The problem is I know that $\ln(z)$ is not continuous on $x<0, y=0$, so it can't be differentiable. Where is error in above analysis.