# Thread: Complex Limit z \to i

1. ## Complex Limit z \to i

$\displaystyle \displaystyle\lim_{z\to i}\frac{z^2-1}{z^2+1}=\frac{(z+1)(z-1)}{(z+i)(z-i)}$

This limit is giving me a little trouble.

I tried factoring but I wasn't able to go anywhere from there.

Anyone have any suggestions?

Thanks.

2. May be this helps

$\displaystyle \displaystyle { \frac{z^2+1-2}{z^2+1}=1-\frac{2}{z^2+1} }$

3. I am not sure how that is going to help though. I will still have 2/0.

4. Originally Posted by dwsmith
I am not sure how that is going to help though. I will still have 2/0.
Maybe it tells you the limit doesn't exist...

5. It is infinity. I just don't know how to show it.

6. Suppose it were $\displaystyle \displaytype \lim_{x\to 0}\frac{1}{x}$. How would you show that goes to infinity?
For z close to i, $\displaystyle z^2- 1$ will be close to -2 and $\displaystyle z+ i$ will be close to 2i so
$\displaystyle \frac{z^2- 1}{(z+i)(z-i)}$ will be close to $\displaystyle \frac{i}{2i(z- i)}= \frac{1}{2}\frac{1}{z- i}$. You want to show that, given any M, there exist r such that if $\displaystyle |z- i|< r$, then $\displaystyle \frac{1}{2}\left|\frac{1}{z-i}\right|< M$. That should be pretty straight forward.