Find limlimits_(z→ i)dfrac1(z-i)^2

Find $\ \lim\limits_{z\to i}\dfrac1{(z-i)^2}$.

Step 1 · Parametrize the approach

Let $z=i+re^{i\theta}$, $r\to0^+$. Then:

$$ \frac1{(z-i)^2}=\frac1{r^2e^{2i\theta}}=\frac1{r^2}e^{-2i\theta} $$

Step 2 · Modulus is direction-independent

$$ \left|\frac1{(z-i)^2}\right|=\frac1{r^2}\to\infty \quad\text{for every }\theta $$

Step 3 · But the direction of blow-up depends on $\theta$

$$ \theta=0:\ \frac1{r^2}\to+\infty; \qquad \theta=\frac\pi2:\ -\frac1{r^2}\to-\infty; \qquad \theta=\frac\pi4:\ -\frac ir^2\to-i\infty $$

Different approach angles send the expression to infinity along entirely different rays.

Step 4 · Conclusion in $\mathbb C$ versus $\overline{\mathbb C}$

In the ordinary complex plane $\mathbb C$, there is no single finite value the expression approaches, so $\lim_{z\to i}\frac1{(z-i)^2}$ does not exist. But since $|{\cdot}|\to\infty$ uniformly regardless of direction, in the extended plane $\overline{\mathbb C}=\mathbb C\cup\{\infty\}$:

$$ \boxed{\lim_{z\to i}\frac1{(z-i)^2}=\infty \ \text{ in } \overline{\mathbb C}} $$

Classification

$z=i$ is a pole of order $2$ of $f(z)=\dfrac1{(z-i)^2}$ — the defining feature of a pole (as opposed to an essential singularity) is exactly that $f(z)\to\infty$ uniformly in the sense of $\overline{\mathbb C}$, regardless of the direction of approach.