Prove that the following function tends to infinity as z approaches i (complex analysis)?

Question

Prove:

$\lim_{z\to i} [2 /({1 + z^2})] = \infty$

My attempt:

We want $M > 0$ such that if $0 < |z - i| < \delta$ then $|2/(1 + z^2)]>M$

Now

$|2/(1 + z^2)|>M$

$|2/((z-i)(z+i))|>M$

$|2/(M(z+i))|>|(z-i)|$

$(2/M)|1/(z+i)|>|(z-i)|$

Now what do I do? I know the left side will be our choice of $\delta$ but I'm not sure how to restrict $z+i$.

I was thinking ...

Restrict |z-i| < 1 so that

$|z+i| = |z-i +2i| \le |z-i|+|2i| < 1 + 2 = 3$

So we end up with

$(2/M)|1/(z+i)|>|(z-i)|$

$(2/M)|1/3|>|(z-i)|$

$2M/3 >|(z-i)|$

Is this correct?

Thank you.

Answer 1

Let $M>0$ be given. Suppose that $|z-i|<\min\{\frac{2}{2.5M},\frac{1}{2}\}$. Then, $$ |z+i|\leq|z-i|+|2i|<\frac{1}{2}+2=2.5 $$ implies $$ |1+z^2|=|z^2-i^2|=|z-i|\times|z+i|\leq\frac{2}{2.5M}\times2.5=\frac{2}{M}. $$ Can you take it from here?