I'm reading a bit on complex numbers, but haven't deal with trigonometry a lot before, so here's my question; how do I calculate the argument of a complex number when the sin and cos of the argument aren't "recognizeable"?
I know I can take arccos.... but don't I need to take arccos and arcsin? What if they give me two different numbers?
if $z = a + ib$, $a,b \in \mathbb{R}$, then $$\theta = \begin{cases} \arctan \frac ba, & \text{if a > 0} \\ \pi + \arctan \frac ba, & \text{if a < 0} \end{cases}$$
Let $z=x+i y$ with $|z|=1$. Set $\arg(1)=0$, $\arg(-1)=\pi$, and if $y\neq 0$ then set $$ \arg(z)=(\mathrm{sign}\, y)\cdot (\arccos x). $$
Remark. Knowing how to compute the argument of a complex number having a unit modulus is sufficient. And there exists a unique function defined on the unit circle, let us call it $\arg$, such that. $$ \arg(\cos \theta+i \sin \theta)=\theta\quad \forall\theta\in(-\pi,\pi]. $$ This unique function is the one I reported.
Here is my answer to a similar question posed a few days ago:
One of the most important functions in analysis is the function $$\arg:\quad \dot{\mathbb R}^2\to{\mathbb R}/(2\pi),\quad{\rm resp.},\quad \dot{\mathbb C}\to{\mathbb R}/(2\pi),$$ written as $$\arg(x,y), \quad \arg(x+iy),\quad{\rm or}\quad \arg(z),$$ depending on context. It gives the angle you are talking about "up to multiples of $2\pi$". If you remove the negative $x$-axis (resp., negative real axis) from $\dot{\mathbb R}^2$ (resp., from $\dot{\mathbb C}$) you can single out the principal value of the argument, denoted by ${\rm Arg}(x,y)$, which is then a well defined continuous real-valued function on this restricted domain, taking values in $\ ]-\pi,\pi[\ $. One has $${\rm Arg}(x,y)=\arctan{y\over x}\qquad(x>0)$$ and similar formulas in other half planes.
Even though the values of $\arg$ are not "ordinary real numbers" the gradient of $\arg$ is a well defined vector field in $\dot{\mathbb R}^2$, and is given by $$\nabla\arg(x,y)=\left({-y\over x^2+y^2},\>{x\over x^2+y^2}\right)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$
