# Determine whether the region $\text{Im}(\frac{z+1}{z-1}) \leq 3$ is open, closed or neither.

Determine whether the region $$\text{Im}\left(\frac{z+1}{z-1}\right) \leq 3$$ is open, closed or neither.

My attempt and the memo for the question is below.

My answer is somewhat similar to the memo's except that I didn't have the $$z \neq 1+0i$$ condition. Is this additional $$z \neq 1+0i$$ part correct? While I was working on the solution it didn't seem immediately evident to me, that that condition was necessary. If the memo is correct, do you have any tips on how I can spot such conditions in similar questions in the future? Thank you.

• Your answer looks right except the point z=1 ... it has to be excluded and the set is no more closed (it is surely not open). Your mistake accured when you multiplied the equation by a potential 0 &ädenominator). To avoid similar mistakes, I suggest to start posting "conditions" or, alternatively, not to forget write them when necessary. Oct 30, 2020 at 19:09

See what happens when you substitute $$z= 1+ 0i = 1\;$$ into your inequality.
Then $$\text{Im}\left(\frac{z+1}{z-1}\right) = \text{Im}\left(\frac 20\right).$$
You can't do that, because $$\frac 20$$ is undefined; so we must forbid $$z = 1 + 0i = 1$$.
• Thank you, for the answer. So just to clarify, when determining regions in the complex plane, I would also have to forbid points in the same way if I was determining the regions of, for example $Im(\frac{1}{z-2})\leq3$ or $Re(\frac{z}{2z-1})\leq1$ etc? Oct 30, 2020 at 19:26