What is the difference between $\arg(z)$ and $\operatorname{Arg}(z)$, where $z=a+bi$
By David Jones •
What is the difference between the $\arg(z)$ and the $\operatorname{Arg}(z)$, where $z$ is a complex number of the form $a+bi$, for example: $z = -2 - 2i$
The angle from the positive x-axis to the vector would be $5π/4$
Does that mean that the $\arg(z)=\dfrac{5π}4$?
If so, is $\operatorname{Arg}(z) = \dfracπ4, -\dfracπ4$, or $\dfrac{3π}4$?
$\endgroup$ 21 Answer
$\begingroup$It varies among authors, but: $-\pi < Arg(z) \leq \pi$ and $\arg(z) = Arg(z) + 2 \pi K$ for $K \in \mathbb{Z}$
To answer the example: $z = -2-2i \Rightarrow r=|z|=\sqrt{4+4}=2\sqrt{2} \Rightarrow z = 2\sqrt{2} (-\frac{2}{2\sqrt{2}}-\frac{2i}{2\sqrt{2}}) = 2\sqrt{2} (-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} i) $
$\Rightarrow \theta = \frac{5\pi}{4}-2\pi = -\frac{3\pi}{4}$ (this is done to get it in range).
$\endgroup$More in general
"Zoraya ter Beek, age 29, just died by assisted suicide in the Netherlands. She was physically healthy, but psychologically depressed. It's an abomination that an entire society would actively facilitate, even encourage, someone ending their own life because they had no hope. Th…"
