Let E=\z:1<|z|<2 wedge dfracπ6<arg z<dfracπ3\ . Is E open…

Let $E=\left\{z:1<|z|<2 \ \wedge\ \dfrac\pi6<\arg z<\dfrac\pi3\right\}$. Is $E$ open, closed, both, or neither?

$E$ is open

Step 1 · Take an arbitrary point of $E$

Let $z_0\in E$, so $1<|z_0|<2$ and $\frac\pi6<\arg z_0<\frac\pi3$. Define the (all strictly positive) margins to each boundary:

$$ \delta_1=|z_0|-1, \quad \delta_2=2-|z_0|, \quad \delta_3=\arg z_0-\frac\pi6, \quad \delta_4=\frac\pi3-\arg z_0 $$

Step 2 · Choose a radius small enough to respect all four margins

$$ \varepsilon=\min\left\{\frac{\delta_1}2,\ \frac{\delta_2}2,\ |z_0|\sin\frac{\delta_3}2,\ |z_0|\sin\frac{\delta_4}2\right\}>0 $$

The modulus-margin terms $\frac{\delta_1}2,\frac{\delta_2}2$ directly bound how far $|z|$ can drift from $|z_0|$ for $z\in B(z_0,\varepsilon)$ (since $\big||z|-|z_0|\big|\le|z-z_0|<\varepsilon$). The argument-margin terms use that a chord of length $\varepsilon$ subtending angle $\phi$ at radius $|z_0|$ satisfies $\varepsilon\ge2|z_0|\sin\frac\phi2$, so keeping $\varepsilon$ below $|z_0|\sin\frac{\delta_3}2$ (resp. $\delta_4$) keeps $\arg z$ within $\delta_3$ (resp. $\delta_4$) of $\arg z_0$.

Step 3 · Conclude

Every $z\in B(z_0,\varepsilon)$ satisfies $1<|z|<2$ and $\frac\pi6<\arg z<\frac\pi3$, so $B(z_0,\varepsilon)\subseteq E$. Since $z_0$ was arbitrary:

$$ \boxed{E \text{ is open}} $$

$E$ is not closed

Step 4 · Exhibit a sequence in $E$ with a limit outside $E$

Let $z_n=\left(1+\frac1n\right)e^{i\pi/4}$ for $n\ge2$ (note $\frac\pi6<\frac\pi4<\frac\pi3$). Then $|z_n|=1+\frac1n\in(1,2)$ and $\arg z_n=\frac\pi4\in\left(\frac\pi6,\frac\pi3\right)$, so $z_n\in E$ for every $n\ge2$.

Step 5 · Take the limit

$$ \lim_{n\to\infty}z_n=e^{i\pi/4}, \qquad |e^{i\pi/4}|=1 $$

Since $|e^{i\pi/4}|=1$ fails the strict condition $|z|>1$, the limit point $e^{i\pi/4}\notin E$ — a limit point of $E$ that is not in $E$.

Step 6 · Conclude

$$ \boxed{E \text{ is not closed}} $$

Verification

Consistent with the general principle: a set defined entirely by strict inequalities on continuous functions ($|z|$ and $\arg z$, away from the branch cut) is always open, and — unless it happens to also be the whole space or empty — never closed, since its boundary (where some inequality becomes equality) consists of limit points excluded by the strict conditions.