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?


1 Answer 1


More simply we can proceed as follows

  • 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 $\left(-\frac \pi 2, \frac \pi 2\right)$.

  • $\begingroup$ i still don't understand why the inequality reverses in direction. $\endgroup$
    – Derpp
    Oct 11, 2021 at 13:39

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