Why is $\arg z$ continuous on the arc of the unit circle of those points with $|\arg z |< \frac{1}{2}\pi$

Question

I am looking for an intuitive/ simple argument why $\arg z$ is continuous on the arc of the unit circle of those points with $|\arg z| < \frac{1}{2}\pi$. Any help would be appreciated!

Answer 1

A function $f$ is continuous at a point $p$ if:

For every $\varepsilon > 0$, there is some $\delta > 0$, such that for all $x$ with $|x-p|< \delta$ we have $|f(x)-f(p)|<\varepsilon$

Consider $\left|\mathrm e^{\mathrm i x}-\mathrm e^{\mathrm i p}\right|=\left|\cos x + \mathrm i \sin x - \cos p - \mathrm i \sin p\right|=\left|(\cos x-\cos p)+\mathrm i(\sin x - \sin p)\right|$. Applying the definition of the modulus, we get \begin{eqnarray*} \left|\mathrm e^{\mathrm i x}-\mathrm e^{\mathrm i p}\right|^2 &=& (\cos x-\cos p)^2 + (\sin x - \sin p)^2 \\ \\ &=& 2-2\cos x\cos p - 2 \sin x \sin p \\ \\ &=& 2(1-\cos(x-p)) \end{eqnarray*} If $\left|\mathrm e^{\mathrm i x}-\mathrm e^{\mathrm i p}\right| < \varepsilon$, then $2(1-\cos(x-p)) < \varepsilon^2$, and so $\cos(x-p) > 1-\frac{1}{2}\varepsilon^2$. Hence, choose $$0 < \delta < \arccos\left(\frac{2-\varepsilon^2}{2}\right)$$