0
$\begingroup$

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.

Memo answer My answer

$\endgroup$
1
  • 3
    $\begingroup$ 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. $\endgroup$
    – user376343
    Oct 30, 2020 at 19:09

1 Answer 1

-1
$\begingroup$

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 $z = 1 + 0i = 1$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – ArcLambda
    Oct 30, 2020 at 19:26
  • 1
    $\begingroup$ Why down-vote? @ $\endgroup$
    – Mikasa
    Jan 5 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.