Computing the limit $\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}$ For every $z\in\mathbb C\setminus\{0\}$ we define $\operatorname{Arg}(z)$ to be the principal argument of $z$, that is $\operatorname{Arg}(z)=\theta$, where $z=re^{i\theta }$ and $\theta\in(-\pi,\pi]$. Does the following limit exists?
$$
\lim_{z\to 0}\frac{\operatorname{Arg}(1+z)}{z}
$$
I guess that if it exists, then it must be $0$ since if $z\to0$ along the real line, then the limit is $0$.
 A: Given
$$f(z)=u(x,y)+iv(x,y)$$
$$z_0=x_0+iy_0$$
$$w_0=u_0+iv_0$$
Then
$$\lim_{z\to z_0}f(z)=w_0$$
If and only if
$$\lim_{(x,y)\to (x_0,y_0)}u(x,y)=u_0$$
And
$$\lim_{(x,y)\to (x_0,y_0)}v(x,y)=v_0$$

Let $z=x+iy$
$$\lim_{(x,y)\to (0,0)}\frac{x}{x^2+y^2}\arctan\left(\frac{y}{1+x}\right)$$
$$=\cos\phi\lim_{r\to 0^+}\frac1r\arctan\left(\frac{r\sin\phi}{1+r\cos\phi}\right)$$
$$=\sin\phi\cos\phi\lim_{r\to 0^+}\frac{1}{r^2+2r\cos\phi+1}$$
$$=\sin\phi\cos\phi$$
Since this limit is clearly dependent on $\phi$, we can conclude that
$$\lim_{(x,y)\to (0,0)}\frac{x}{x^2+y^2}\arctan\left(\frac{y}{1+x}\right)$$
$$\mbox{does not exist}$$
Therefore
$$\lim_{z\to 0}\frac{1}{z}\operatorname{Arg}\left(1+z\right)$$
$$\mbox{does not exist}$$
A: $$ \frac{\log (1+z)}{z} = \frac{\log |1+z|}{z}+ i \frac{\arg{(1+z)}}{z}$$
$$\lim_{z\to 0} \frac{\log(1+z)}{z}=\lim_{z\to 0}\frac{1}{1+z}=1.$$
Consider $z\to 0$ and $z=x$ ($z$ approaches zero along the real axis).
Then $$\lim_{z\to 0} \frac{\arg(1+z)}{z} = \Im \lim_{z\to 0} \frac{\log(1+z)}{z} =0.$$
Consider $z\to 0$ and $z=iy$ ($z$ approaches zero along the imaginary axis).
$$ \frac{\log (1+z)}{z} = \frac{\log |1+iy|}{iy}+ i \frac{\arg{(1+iy)}}{iy}$$
Then $$\lim_{y\to 0} \frac{\arg(1+iy)}{y} = 1$$
and  $$\lim_{z\to 0} \frac{ \arg(1+z)}{z} = -i \ne 0.$$
Since the limit obtained from different directions is different, the limit does not exist.
