5
$\begingroup$

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 $\dfrac{3π}4$?

$\endgroup$
2
  • $\begingroup$ 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

5
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.