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 $\dfrac{3π}4$?

  • $\begingroup$ The capital generally means it's all possible angles whereas the lower case means it's restricted over some range of $2\pi$. $\endgroup$
    – Jared
    Apr 13, 2014 at 2:47
  • $\begingroup$ @Jared Isn't it exactly the other way round? $\endgroup$
    – user46234
    Apr 30, 2016 at 2:33

1 Answer 1


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).


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