Limit in $ \mathbb{C}$. Can the function be defined in $ z_{0} = i +1$ so that it remains continuous?

Question

Let $f \colon \mathbb{C} \backslash \left\{1+i\right\} \to \mathbb{C}$ $$ f(z) = \frac{z^2-2i}{z^2-2z+2}$$. Can the function be defined in $ z_{0} = i +1$ so that it remains continuous ?

I know I have to find the limit $ \lim_{ z \to 1 + i} f(z)$, if it exists, but I am having trouble finding it. I tried by writing $z$ as $x + iy$ but it got complicated. Any hint helps.

Answer 1

Note that $\bigl(\pm(1+i)\bigr)^2=2i$. Therefore\begin{align}f(z)&=\frac{z^2-2i}{z^2-2z+2}\\&=\frac{\bigl(z-(1+i)\bigr)(z+\bigl(1+i)\bigr)}{(z-1)^2+1}\\&=\frac{\bigl(z-(1+i)\bigl)\bigl(z+(1+i)\bigl)}{\bigl(z-(1+i)\bigl)\bigl(z-(1-i)\bigr)}\\&=\frac{z+1+i}{z-1+i}.\end{align}Therefore,$$\lim_{z\to1+i}f(z)=\frac{2+2i}{2i}=1-i$$and therefore, yes, you can extend $f$ to a continuous function near $1+i$.

However, there is a problem in the statement of your problem: $f$ is undefined at $1-i$ and you can't extend it to a continuous function near that point.

Answer 2

You can also apply L'Hopital rule:

$\lim _{z→1+i}\frac{z^2-2i}{z^2-2z+2}=\lim _{z→1+i}\frac{2z}{2z-2}=\frac{2i +2}{2i}=1-i$

That means at $z=1+i$ the function can be continuous.