What does the "-" in front of the arccos do to its result, when wanting to compute the polar coordinates of a complex number?

Question

so I'm still training for my calculus exam but I stumbled upon a confusing thing again :( I think I kind of feel certain with computing polar coordinates now, but I just don't know if my train of thought with this one is right.

Soooo I have these three exercises, where I'm trying to come up with the respective $\arccos$ part that needs to be computed for $\varphi$ in $z=re^{i\varphi}$:

$$(-\frac{\sqrt{3}}{2}-\frac{1}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(-\frac{\sqrt{3}}{2})} = -(\frac{5\pi}{6}) = \frac{7\pi}{6}.$$

my understanding of the whole thing is that, for every negative minus sign, we move one quadrant to the left in the polar grid, to the corresponding value of $\cos$. I'm really really bad at explaining so I'll just draw what I mean:

the blue arrow is for the minus in front of the argument within $\arccos$, the red arrow for the $-$ in front of $\arccos$.

and for another exercise, it's pretty much the same:

$$(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(-\frac{\sqrt{2}}{2})} = -(\frac{3\pi}{4}) = \frac{5\pi}{4}.$$

with the same argumentation as above.

HOWEVER,, now, when I try to use this trick for the following exercise

$$(\frac{\sqrt{3}}{2}-\frac{1}{2}i) \Rightarrow r=1 \Rightarrow -\arccos{(\frac{\sqrt{3}}{2})} = -(\frac{\pi}{6}) = \frac{11\pi}{6}.$$

that is the sample solution. BUT, according to what I've gotten accustomed to, the arrow should only be moving from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$, not to $\frac{11\pi}{6}$.

like demonstrated here.

could someone please explain to me in simple terms what in particular is wrong with my way of thinking?... I would appreciate it so much ;-;

thank you guys in advance <3

Answer 1

Why are you not using \begin{align*} |-\sqrt{3}/2 - \mathrm{i}/2| &= 1 \text{,} \\ \Re(-\sqrt{3}/2 - \mathrm{i}/2) &= -\sqrt{3}/2 \text{, and } \\ \Im(-\sqrt{3}/2 - \mathrm{i}/2) &= -1/2 \text{, so} \\ \theta &= 7\pi/6 \text{.} \end{align*}

The first guarantees that we have extracted $r$, so have a point on the unit circle. If hte magnitude is not $1$ extract $r$, then proceed. The second and third pull out the coordinates of the point on that circle, $(x,y) = (-\sqrt{3}/2, -1/2)$ using the real part ($\Re$) and imaginary part ($\Im$) functions. The last is the angle in your unit circle diagram for the point that has those coordinates.