I tried to solve it by taking two different paths like the way showing that $f(z)= Arg(z)$ is discontinuous at any point in the negative real axis, but the two paths I have taken gave me the same answer.. any other ideas please !!
Show that the function $f(z)=Arg(iz)$ is discontinuous at the point $z_0=i$.
complex numberscomplex-analysis
Best Answer
So, basically we're trying to show that $Arg(iz)$ is discontinuous at the point $i$ on the complex plane. Define a function $q$ given by:
$$q(t) = Arg(e^{i(\frac{1}{2}\pi + t)}), \quad t \in (- \frac{3\pi}{2}, \frac{\pi}{2}]$$
Observe that $e^{i(\frac{1}{2}\pi + t)}$ is the form of any point on the circle centered at $0$ with a radius of $1$.
Also, notice that
$$\lim_{t \to - \frac{3\pi}{2}^+} Arg(e^{i(\frac{1}{2}\pi + t)}) = -\pi$$
and
$$\lim_{t \to \frac{\pi}{2}^{-}} Arg(e^{i(\frac{1}{2}\pi + t)}) = \pi$$
What we have shown is that if you approach the point $i$ on the complex plane counter-clockwise around the circle parametrized by $e^{i(\frac{1}{2}\pi + t})$, our function outputs $\pi$. However, if on the same circle we approach $i$ clockwise instead, our function outputs $-\pi$. Since we obtain completely different outputs from $Arg(iz)$ as $z \to i$ (i.e. $\pi$ and $-\pi$) along the circle parametrized by $e^{i(\frac{1}{2}\pi + t)}$, we have to conclude that
$$\lim_{z \to i} Arg(iz)$$ does not exist. So, $Arg(iz)$ can't be continuous at $i$.