# Inequality reversing with trigonometric function.

We are given that $$\arg(z-1) \leq \frac{3\pi}{4}$$, where $$z$$ is a complex number. When trying to shade this region, I let $$z=x+iy \implies$$ arg$$[(x-1)+i(y)] \leq \frac{3\pi}{4} \implies \tan^{-1}(\frac{y}{x-1}) \leq \frac{3\pi}{4}$$

I know $$\tan(\frac{3\pi}{4}) = -1,$$ and I had guessed that the inequality will need to be flipped to result in: $$\frac{y}{x-1} \geq -1$$, but I am not sure why. Can someone please explain?

• at first consider the region for $$w=z-1$$ such that $$\arg(w) \leq \frac{3\pi}4$$,
• then we can apply a translation for $$z=w+1$$.
Using $$\arctan$$ we need to pay attention since the function has range in $$\left(-\frac \pi 2, \frac \pi 2\right)$$.