What is the proof behind the inexistence of $\arg(\theta)$ ? Is there anything besides the inability of $\cos\theta$ and $\sin \theta$ to be $0$ simultaneously?

  • 4
    $\begingroup$ For any $\theta \in [0, 2\pi)$, $0 = 0e^{i\theta}$ so the argument of zero is not well-defined. $\endgroup$ Jan 15, 2015 at 19:17
  • $\begingroup$ I didnt learn this formula yet. What is this? I've seen a similiar form for expressing cos in terms of $\theta and e and i though $\endgroup$
    – Rami Awar
    Jan 15, 2015 at 19:24

1 Answer 1


Every $z\in\mathbb C$ can be written as $z=re^{i\theta}$ where $r\geq0$ and $\theta\in[0,2\pi)$ (polar coordinates).

If $z\neq0$ then $r$ and $\theta$ (i.e. the argument of $z$) are both unique.

If $z=0$ then $r=0$ and is unique, but $\theta$ is not unique.

This because $0=0\cdot e^{i\theta}$ is true for every $\theta\in[0,2\pi)$.

In short: $\arg⁡(0)$ is not well-defined.


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