Suppose $z$ is a complex number, $\operatorname{Arg}$ is the Argument of $z$, and $\arg$ is the principal argument of $z$.

Why $\operatorname{Arg}z=\arg z+2m\pi=\arg z+\operatorname{Arg}1$?

What does $1$ stands for in $\operatorname{Arg}1$?

  • $\begingroup$ 1 just refers to the real number 1. It lies on the Real axis, and so has argument $2m\pi$. $\endgroup$ – John Doe May 3 '17 at 17:19
  • $\begingroup$ Principal argument is the angle between 0 and less than 2 pi. Angles "repeat" when they go "full circle" or 360 degrees or 2 pi. So argument is always principal argument plus some multiple of 2 pi. The number 1=1+0i has an angle/principal argument of zero. $\endgroup$ – fleablood May 3 '17 at 17:25

This is because $1$ can be written as $1 = e^{i \theta}$, for any $\theta \in \{ 2 \pi m \mid m \in \mathbb{Z} \}$. So $\operatorname{Arg} 1 = \{ 2 \pi m \mid m \in \mathbb{Z} \}$. Of course, we drop the set notation, and just write $\operatorname{Arg} 1 = 2 \pi m$ for any $m \in \mathbb{Z}$.


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