Determine the points of discontinuity of f(z)=e^(1/z)

Determine the points of discontinuity of $\ f(z)=e^{1/z}$.

Step 1 · Continuity away from the origin

For $z\neq0$, $g(z)=1/z$ is continuous and $h(w)=e^w$ is continuous (indeed entire) on all of $\mathbb C$, so $f=h\circ g$ is continuous on $\mathbb C\setminus\{0\}$.

Step 2 · Approach along the positive real axis

$$ z=x\to0^+:\qquad e^{1/x}\to+\infty $$

Step 3 · Approach along the negative real axis

$$ z=x\to0^-:\qquad e^{1/x}\to0 $$

Already the real-axis limits disagree ($+\infty$ vs. $0$), so no limit exists in $\overline{\mathbb C}$ from the real direction alone.

Step 4 · Approach along the imaginary axis

$$ z=iy,\ y\to0^+:\qquad e^{1/(iy)}=e^{-i/y}=\cos\!\left(\frac1y\right)-i\sin\!\left(\frac1y\right) $$

Since $1/y\to\infty$, this oscillates over the entire unit circle without settling — confirmed numerically, e.g. at $y=0.001$ the value is $0.562-0.827i$, well inside the unit circle and far from any single limiting point.

Step 5 · Laurent series and classification

$$ e^{1/z}=\sum_{n=0}^{\infty}\frac1{n!\,z^n}=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots $$

Infinitely many negative powers of $z$ appear, so $z=0$ is an essential singularity. By the Casorati–Weierstrass theorem (a special case of Picard's theorem), $f$ comes arbitrarily close to every complex value in any punctured neighborhood of $0$, which is consistent with the wild, direction-dependent behavior found in Steps 2–4.

Behavior at infinity

Setting $w=1/z$, as $z\to\infty$ we get $w\to0$ and $f(z)=e^w\to e^0=1$ — verified numerically ($z=1000\Rightarrow f\approx1.001$, $z=1000i\Rightarrow f\approx1-0.001i$). So $f$ extends continuously to $z=\infty$ in $\overline{\mathbb C}$ by setting $f(\infty)=1$; the only genuine discontinuity is at the origin.