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Let $f(z)=Arg(z)$ denotes the principal argument of $z$. If $f(z)=u(x,y)+iv(x,y)$, then for $x\neq0$, we obtain that $u(x,y)=\arctan(y/x)+\{-\pi,0,\pi\}$ and $v(x,y)=0$, so $$ u_x(x,y)=-\frac{y}{x^2+y^2}\quad u_y(x,y)=\frac{x}{x^2+y^2} $$ and $v_x(x,y)=v_y(x,y)=0$. Hence, the Cauchy Riemann equations does not hold unless $(x,y)=(0,0)$. Therefore $f$ is not differentiable at every point with $x\neq0$.

Comment: we say that $f$ is differentiable at $z_0$ if the following limit exists $$ \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} $$

(a) What about differentiable in points with $x=0$ ?

(b) How to prove that $f(z)=Arg(z)$ is not differentiable without using the Cauchy Riemann equations?

boaz
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1 Answers1

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This is just a partial solution, but maybe it is of interest: Fix $z_0$ and consider the difference quotient

$$\frac{f(z_0+h)-f(z_0)}{h} = \frac{\text{Arg}(z_0+h)-\text{Arg}(z_0)}{h}$$

If we let $h \rightarrow 0$ along the line (or ray) $\text{Arg}(z) = \text{Arg}(z_0)$, this expression clearly tends to $0$. This at leasts shows that $f$ cannot be differentiable at any open set: If it were, the derivative would be zero and $f$ would be contant on that open set. Obviously $\text{Arg}(z)$ is not constant on any open set. So if $\text{Arg}(z)$ is differentiable somewhere, it will be in isolated points.

Thusle Gadelankz
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  • I would like to add here that since $z_0$ was choisen arbitrarily we can say that $\text{arg}'(z)\equiv 0$ on the domain of $\text{arg}$ function. $arg(z)$ is continuous on $\mathbb{C}\backslash{\Re(z)<0, \Im(z)=0}$ and therefore is constant there. Contradiction! – Levon Minasian Mar 15 '22 at 13:23