<s>[INST] Indicate on an Argand Diagram the region of the complex plane in which $ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $ 

Question: Indicate on an Argand Diagram the region of the complex plane in which $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$



I've tried this
Consider $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$
Split it into two inequalities : $ 0 \leq \arg (z+1)  $ and $\arg (z+1) \leq \frac{2\pi}{3} $
Then: 
$$ z+1 \Leftrightarrow x+iy+1 \Leftrightarrow (x+1)+i(y) $$
$$ \tan^{-1}\left(\frac{\Im(z)}{\Re(z)} \right) = \arg(z+1)$$
$$ \Leftrightarrow  \tan^{-1}\left(\frac{y}{x+1} \right) = \arg(z+1)$$
So for the first inequality 
$$ 0 \leq \arg (z+1)  $$
$$ 0 \leq \tan^{-1}\left(\frac{y}{x+1} \right) $$
$$ 0 \leq \frac{y}{x+1} $$
$$ 0 \leq y ~~ x \neq -1 $$ 
Second Inequality:
$$\tan^{-1}\left(\frac{y}{x+1} \right) \leq \frac{2\pi}{3} $$
$$ \frac{y}{x+1} \leq -\sqrt{3} $$
$$ y \leq  -\sqrt{3} (x+1) $$
So we have two conditions
$$ 0 \leq y ~~ x \neq -1 $$ 
and $$ y \leq  -\sqrt{3} (x+1) $$
How do I draw these?
 [/INST] 
You switched the sign of the inequality somewhere. It should be $y\ge-\sqrt3(x+1)$. So you plot the two lines $y=0$ and $y=-\sqrt3(x+1)$ and mark the sector between them as shown.
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