# {$\Re z+\Im z:z\in$ $W$} is open in $\mathbb R$.

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 $\mathbb R$.

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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

## 1 Answer

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 $(r-\epsilon, r+\epsilon)\subseteq A$.

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