Not having clarity that what author is saying.

Question

The set of all points such that $\pi/2 < \operatorname{Arg} (z-2-3\iota)< \pi$. How to find those points who satisfy this inequality?

Answer 1

I expect that your "$\iota$" is just "i" the imaginary unit.

Write the complex number z as a+ bi. Then z- 2- 3i= (a- 2)+ (b- 3)i. We can write that in "polar form", $re^{i\theta}$ with $r= \sqrt{(a-2)^2+ (b- 3)^2}$ and "argument" $\theta= \arctan\left(\frac{b-3}{a- 3}\right)$. Saying that $\pi/2< \operatorname{Arg}(x- 2- 3i)< \pi$ is saying that $\pi/2< \arctan\left(\frac{b-3}{a- 3}\right)< \pi$, in the second quadrant.

Answer 2

The numbers with argument $\pi$ are the negative reals. So one boundary of your region is the set of $z$ with $$ (z-2-3i) = -t, \quad t>0 \\ z = -t+2+3i,\quad t>0 $$ which is a half line: staring at the point $2+3i$ with direction $-1$.

The numbers with argument $\pi/2$ are the imaginary axis above the origin. So the other boundary of your region is the set of $z$ with $$ (z-2-3i) = it,\quad t>0 \\ z = it+2+3i $$ which is again a half line: staring at the point $2+3i$ with direction $i$.

Your complete region is a sector with vertex 2+3i, and edges extending up and left from there.

Answer 3

Geometrically, it is the open second quadrant translated at the point with affix $2+3i$.