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This is Exercise EP$10$ (g) from Fernadez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese)

How does one geometrically describes the set $S=\{z\in\mathbb{C}:|\operatorname{Arg}(z-i)|\lt\frac{\pi}{6}\}?$

I tried to solve this one as I did in others, but it did not work out as I wish. Let me show what I did so far.

$z=x+yi\in S \Longleftrightarrow |\operatorname{Arg}(z-i)|\lt\frac{\pi}{6}\Longleftrightarrow \frac{-\pi}{6}\lt \operatorname{Arg}(x+(y-1)i)\lt \frac{\pi}{6}.$

Unfortunately, I was not able to go further. I would appreciate any hint.

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The set of complex numbers whose argument is less than $\pi/6$ are those that lie below the real axis in quadrant III; or in quadrant IV, or in quadrant I below the line $k(\cos(\pi/6) + i\sin(\pi/6))$. The complex number $z-i$ lies one unit below the complex number $z$ on the complex plane. – Arturo Magidin Mar 4 '12 at 22:52
up vote 5 down vote accepted

$|\operatorname{Arg}(z)|<\pi/6$ is the set of nonzero complex numbers strictly between (to the right of) the rays emanating from the origin at angles $\pm\,\pi/6$. The $-i$ translates this set one unit in the $i$ direction.

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One can draw the following picture which may help to understand the solution.

enter image description here

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