Prove that the set $\mathcal{U}=\left\{z\in\mathbb{C}\,\colon\,\Re{(z)}>0\right\}$ is open. Let $a\in \mathcal{U}$, we must show that there exists an $r>0$ such that the disk $$D(a, \,r) = \left\{z\in\mathbb{C}\,\colon\,\lvert z-a\rvert<r\right\}$$ is a subset of $\mathcal{U}$. Now take $0<r<\Re{(a)}$ and $z_0\in D(a, \,r)$ so that $\lvert z_0-a\rvert<r<\Re{(a)}$. We must show $\Re{(z_0)}>0$. Now recall

\begin{align} \Re{(a)}>\lvert z_0-a\rvert&\ge\Re{(z_0-a)}\\ &= \Re{(z_0)}-\Re{(a)} \end{align}

$$\implies \Re{(z_0)}< 2\Re{(a)}$$

Now I'm not really sure how to continue.

  • $\begingroup$ Let the real parts of $a$ and $z_0$ be $\alpha$ and $x$ respectively. First show that $|z_0 - a| < r \implies (x - \alpha)^2 < r^2 \iff -r < x - \alpha < r$. $\endgroup$ Oct 14, 2015 at 10:53
  • $\begingroup$ But you haven't shown that $x>0$... $\endgroup$ Oct 14, 2015 at 12:49
  • $\begingroup$ Just note that $|z_0-a|\geqslant\Re{(a-z_0)}=\Re{(a)}-\Re{(z_0)}.$ $\endgroup$
    – CIJ
    Oct 14, 2015 at 13:05

1 Answer 1


What you should do is consider the components of $w=z_0-a$. Since $|w|<r$ you would inevitably have that $|\Re w|<r$. Consequently $\Re z_0 = \Re w + \Re a$, and $|\Re w| < r = \Re a$ so $\Re z_0 > 0$.

The the last step is because $|\Re w|<r$ means that $-r< \Re w < +r$ (where the last inequality is of no interrest here) consequently we have by adding $r=\Re a$ $$0 = r-r = \Re a - r < \Re w + \Re a = \Re z$$

  • $\begingroup$ You have not shown $\Re{(z_0)}>0$. You have shown that $\Re{(z_0)} < 2\Re{(a)}$ exactly as I have. If $\Re{(z_0)} < 2\Re{(a)}$ then even though $r$ is positive, we have a less than sign and $\Re{(z_0)}$ could still be negative because a negative number is less than a positive number. $\endgroup$ Oct 14, 2015 at 12:47
  • $\begingroup$ @MichaelHowlard I think I have, note the absulute sign around $\Re w$. It means that $-r < \Re w < r$ and therefore $\Re a-r < \Re z_0 = \Re a + \Re w < \Re a + r$. $\endgroup$
    – skyking
    Oct 14, 2015 at 12:51

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