$$f(z)= \frac{1}{Log(z)+i\cdot\frac{\pi}{2}}$$
Where the Log function is the principal Logartihm,
$$Log(z) = ln|r| +iArg(z)$$
$$ – \pi < Arg(z) \leq \pi$$
$\exists$ a singularity when,
$$Log(z)+i \cdot \frac{\pi}{2} = 0$$
$$Log(z) = -i \cdot \frac{\pi}{2}$$
$$\implies e^{Log(z)} = e^{-i \cdot \frac{\pi}{2}}$$
$$\implies z = -i$$
And we have an isolated singularity that is a pole of order 1 (simple).
What about when,
$$z=0$$
$$Log(z)$$
Or the jump discontinuity seen at:
$$M_1=\lim_{\theta \to \pi^{-}} ln|r|+i\theta = ln|r|+i\pi$$
and
$$M_2=\lim_{\theta \to \pi^{+}} ln|r|+i\theta = ln|r|-i\pi$$
Should these be considered when classifying and determining? If so, how?
Thanks.
Best Answer
1) Because $\log{z}$ is in the denominator, $\lim_{z \rightarrow 0} f(z) = 0$, regardless of the direction from which the limit is taken.
2) Since you are working with the principal branch of $\log{z}$, the question is moot: only the limit from below is relevant.