Complex Limit $\lim_{z \rightarrow i} \frac{z^3 + i}{|z| - 1}$ I wish to determine whether the limit $L = \lim_{z \rightarrow i} \frac{z^3 + i}{|z| - 1}$ exists. Noticing it to be of the form $0/0$, I separate the expression into its real and imaginary parts:
$$L = \lim_{(r, \theta) \rightarrow (1, \pi/2)} \frac{(re^{i\theta})^3 +i}{r-1} = \lim_{(r, \theta) \rightarrow (1, \pi/2)} \frac{r^3(\cos(3\theta) + i\sin(3\theta)) + i}{r-1} = \lim_{(r, \theta) \rightarrow (1, \pi/2)} \left(\frac{r^3\cos(3\theta)}{r-1} + i\frac{r^3\sin(3\theta) + 1}{r-1}\right)$$, but I am still stuck with the $0/0$ indeterminate form. How might I evaluate this limit?
 A: If $t\in\Bbb R$,$$\operatorname{Re}\left(\frac{(t+i)^3+i}{|t+i|-1}\right)=\operatorname{Re}\left(\frac{t^3+3it^2-3t}{\sqrt{t^2+1}-1}\right)=\frac{t^3-3t}{\sqrt{t^2+1}-1},$$and$$\lim_{t\to0^+}\frac{t^3-3t}{\sqrt{t^2+1}-1}=\lim_{t\to0^+}(t^2-3)\frac t{\sqrt{t^2+1}-1}=-\infty.$$
A: If the limit existed, it would need to be the same via any path of $z$ to $i$.
Using the path $z=it,\;t\in(0,1)$, and using the difference of two cubes formula, we find (as $t\to 1$): $$f(z)=f(it)=(-i)(t^2+t+1) \to -3i.$$
But using the path $z=i+t,\;t>0,$ and introducing the factor $1+\sqrt{t^2+1},$ we find (as $t\to 0$):
$$|f(z)|=|f(i+t)|= \left| \frac{(1+\sqrt{t^2+1})(t^2+3it-3)}{t} \right| \to \infty.$$
A: I'd rather translate to complex variable $h:=z-i\to0$:
$$\frac{z^3 + i}{|z|-1}=\frac{(i+h)^3+i}{|i+h|-1}=\frac{-3h+o(h)}{\sqrt{a^2+(1+b)^2}-1}$$
hence there is no limit as $(a,b)\to(0,0)$. This is because there are sequences with various limits, depending on the relative speed of convergence of $a$ and $b$. E.g.:
For $z_n=i+\frac1n$,
$$\frac{z_n^3+i}{|z_n|-1}\sim\frac{-3/n}{1/(2n^2)}=-6n\to-\infty.$$
For $z_n=i+\frac{1+i}n$,
$$\frac{z_n^3+i}{|z_n|-1}\sim\frac{-3(1+i)/n}{1/n}=-3(1+i).$$
