Find the locus of $\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3}$ I tried simplifying this first:
$$\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(-(z+(3-i)))\le \frac{5\pi}{3} \Leftrightarrow \\
\frac{\pi}{3}\le \arg(z+(3-i))+\pi \le \frac{5\pi}{3} \Leftrightarrow \\
???$$
My book does the following:
$$\frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} \Leftrightarrow \\
-\frac{2\pi}{3}\le \arg[z-(3+i)]\le \frac{2\pi}{3}$$
With the expression in this form I can easily find the locus, what i don't understand is how my book does this simplification. Can anyone explain it  to me?
 A: Here is how I was taught how to do this (I think it might be a slightly longer method):
$$ \frac{\pi}{3}\le \arg(-z+3+i)\le \frac{5\pi}{3} $$
Split this inequality into two inequalities:
$$ \frac{\pi}{3} \le \arg(-z+3+i) ~~\text{and}~~ \arg(-z+3+i) \le \frac{5\pi}{3}$$
Let's solve the first inequality first: Let $z=x+iy$
$$\begin{align*}
\frac{\pi}{3} &\le \arg(-z+3+i)\\
\frac{\pi}{3} &\le \arg(-x-iy+3+i) \\
\frac{\pi}{3} &\le \arg((3-x)+i(1-y))\\
\frac{\pi}{3} &\le \arctan \left( \frac{1-y}{3-x}\right)\\
 \sqrt{3} &\le \frac{1-y}{3-x} \\
\sqrt{3} (3-x) &\le 1-y \\
 3\sqrt{3} - \sqrt{3}x &\le 1-y\\
y &\le \sqrt{3}x + (1-3\sqrt{3})\\
\end{align*}$$
Second inequality:
$$\begin{align*}
\arg(-z+3+i) &\le \frac{5\pi}{3}\\
\arg(-x-iy+3+i) &\le \frac{5\pi}{3} \\
\arg((3-x)+i(1-y)) &\le  \frac{5\pi}{3} \\
\arctan \left( \frac{1-y}{3-x} \right) &\le  \frac{5\pi}{3}\\
 \frac{1-y}{3-x} &\le -\sqrt{3} \\
y &\ge -\sqrt{3}x+(3\sqrt{3}+1) \\
\end{align*}$$
So you have both of those regions with a hole at $x=3$
A: The answer is already given, in a rather nice and clear way, in the (first so far) comment below your question: observe that
$$\arg(-z+3+i)=\arg(-(z-(3+i))=e^{-\pi i}\arg(z-(3+i))$$
and you have thus "added" $\;-\pi\;$ to the argument...that's all!
