Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $g:W$ (Open in $\mathbb C$) $\to \mathbb C$ be analytic on $W$ & $g'(z)\neq 0$ $\forall$ $z\in W$. Show that, {$\Re z+\Im z:z\in$ $W$} is open in $\mathbb R$.

share|cite|improve this question
I'm not sure what the relevance of the function $g$ is. As far as I can tell, the question is: Let $W$ be an open subset of $\mathbb{C}$. Prove that $\{\Re z + \Im z : z\in W\}$ is open in $\mathbb{R}$. Is that right, or am I missing something? – froggie Dec 13 '12 at 1:47
up vote 0 down vote accepted

Let $A = \{\Re z + \Im z : z\in W\}$, and suppose $r\in A$. The problem is to show that there is an open interval around $r$ that is contained in $A$. Since $r\in A$, there is some $z\in W$ such that $r = \Re z + \Im z$. Now, since $W$ is open, there is an $\epsilon>0$ such that $B(z,\epsilon)\subseteq W$. In particular, if $t\in (-\epsilon,\epsilon)$, then the complex number $z_t:= z + t$ lies in $W$. But then $\Re z_t + \Im z_t = r + t\in A$. Thus $(r-\epsilon, r+\epsilon)\subseteq A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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