# What kind of a point does $f(x)=\frac{e^{iz}}{z^2+1}$ have at i? [closed]

How should I approach a problem like this, given a function:

$$f(z)=\frac{e^{iz}}{z^2+1}$$

I need to find whether the limit of $$|\frac{e^{iz}}{z^2+1}|$$ at $$i$$ converges to some number, infinity, or if it does not converge to infinity or any number (IE.the real arguments converge to say 0 from negative side and infinity otherwise).

What would be the most useful things to try here?

## closed as unclear what you're asking by Did, Leucippus, user10354138, Rebellos, rtybaseDec 12 '18 at 9:34

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• You should use the fact that $f(z)$ you described is continuous at $i$. – NL1992 Dec 12 '18 at 0:03
• Plug in first, does $z=i$ look like a special point? – Michael Burr Dec 12 '18 at 0:04
• @MichaelBurr Sorry, a typo. – Dole Dec 12 '18 at 0:20
• Don't understand the problem here. Plugging in (or taking the limit) $z=i$ is the obvious thing to try right? – Winther Dec 12 '18 at 1:00
• @Winther Yes it goes to infinity, but does it go to somewhere else like zero as well approached from a different direction? To me it seems that it does given I think it might lead to a contradiction if i was a pole. – Dole Dec 12 '18 at 1:02

## 1 Answer

$$|f(z)|=\frac {|e^{iz}|} {|z+i| |z-i|} \to \infty$$ as $$z \to i$$.