Show that the function f(z)=dfrac1z is continuous in the…

Show that the function $\ f(z)=\dfrac1z\ $ is continuous in the region $|z|>0$ (i.e. $\mathbb C\setminus\{0\}$), but is not bounded there.

Part 1 · Continuity

Fix $z_0\neq0$. Choose $\delta_1=\dfrac{|z_0|}2$; for $|z-z_0|<\delta_1$, the reverse triangle inequality gives $$ |z|\ge|z_0|-|z-z_0|>|z_0|-\frac{|z_0|}2=\frac{|z_0|}2 $$ Then $$ \left|\frac1z-\frac1{z_0}\right|=\frac{|z-z_0|}{|z||z_0|}<\frac{|z-z_0|}{\tfrac{|z_0|}2\cdot|z_0|}=\frac{2|z-z_0|}{|z_0|^2} $$ Given $\varepsilon>0$, choose $$ \delta=\min\left\{\frac{|z_0|}2,\ \frac{\varepsilon|z_0|^2}2\right\} $$ Then $|z-z_0|<\delta$ forces $\left|\dfrac1z-\dfrac1{z_0}\right|<\varepsilon$, so $$ \lim_{z\to z_0}\frac1z=\frac1{z_0}\qquad\text{for every }z_0\neq0 $$

Part 2 · Unboundedness

Take $z_n=\dfrac1n$ for $n=1,2,3,\ldots$, each nonzero. Then $$ f(z_n)=\frac1{1/n}=n\to\infty $$ Since no finite $M$ bounds $|f(z_n)|=n$ for all $n$, there is no $M$ with $|f(z)|\le M$ for all $z\in\mathbb C\setminus\{0\}$.

Why this doesn't contradict the extreme value theorem

A continuous function on a compact set is automatically bounded. But $\mathbb C\setminus\{0\}$ is not compact — it is neither closed (the excluded point $0$ is a limit point) nor bounded (it extends to infinity) — so the theorem simply does not apply here.

Consistently, as $z\to0$: $|f(z)|=1/|z|\to\infty$; and as $|z|\to\infty$: $|f(z)|=1/|z|\to0$ — unbounded near the missing point, decaying at infinity.