Limit of complex function by definition

Question

This is exercise from some book:

I have to prove by definition with $\epsilon $ and $ \delta$ the following limit:

$\lim_{z\to 1+i} \frac{1}{z-i} = 1$

What i've tried so far is that:

Let $\epsilon > 0$ we have to find $\delta > 0$ such that $0<|z-(i+1)|<\delta$ such that

$|\frac{1}{z-i}-1| <\epsilon$

$|\frac{z-(1+i)}{z-i}| <\epsilon$

$|\frac{1}{z-i}||z-(1+i)| <\epsilon$

And from here I've got stuck.

Can anyone help me with that?

Answer 1

Suppose that $|z-(1+i)|<\frac12$. Then$$|z-i|=|z-(1+i)+i|\geqslant\bigl||i|-|z-(1+i)|\bigr|>\frac12.$$But then$$\frac{|z-(1+i)|}{|z-i|}\leqslant2|z-(1+i)|.$$So, take $\delta=\min\left\{\frac12,\frac\varepsilon2\right\}$.

Answer 2

Hint: When $|z-(1+i)|<\frac12$ then $|z-i|>\frac12.$

Answer 3

$\displaystyle \lim_{z\to 1+i} \frac{1}{z-i} = 1$

Arguable alternative approach, that depends on what Complex Analysis theorems/results that you are permitted to use.

Preliminary Results

(1)
If $~~~\lim_{z \to z_0} f(z) = z_n~~~$ and $~~~\lim_{z \to z_0} g(z) = z_d \neq 0$
then $~~~\lim_{z \to z_0} \frac{f(z)}{g(z)} = \frac{z_n}{z_d}$.

(2)
For non-zero $~~w = (u + iv) \in \Bbb{C},~~$ you have that
$\displaystyle \frac{1}{w} = \frac{\overline{w}}{|w|^2} = \frac{u - iv}{u^2 + v^2}.$

Therefore, $~~\displaystyle \frac{1}{[1] + i[0]} = \frac{[1] - i[0]}{1^2 + 0^2} = [1] + i[0].$

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Take $~~\delta = \epsilon.$

Noting that $~~|z - (1 + i)| = |(z - i) - ([1] + i[0])|,~~$ you have that
$0 < |z - (1 + i)| < \delta ~~\implies ~~ |(z - i) - ([1] + i[0])| < \delta = \epsilon.$

Therefore,
$\displaystyle \lim_{z \to (1 + i)} (z - i) = [1] + i[0].$

Using (1) above, this implies that
$\displaystyle \lim_{z \to (1 + i)} \frac{1}{z - i} = \frac{1}{[1] + i[0])}.$

Using (2) above, this implies that
$\displaystyle \lim_{z \to (1 + i)} \frac{1}{z - i} = [1] + i[0].$