Prove that $\lim_{z \to i} |1-z|^2=2$

Question

Prove that $\lim_{z \to i} |1-z|^2=2$.

I let $\epsilon>0$ and $\delta$ = (blank, since I don't know what to choose yet)

When $|z - i| < \delta$, $||1 - z|^2 - 2| ≤ |(1 - z)^2 - 2| = |z^2 - 2z - 1|.$

And then I'm stuck. I'm not sure how I can extract out a $|z - i| < \delta$ in order to further the proof.

Answer 1

You could use the identification $\;\Bbb C\sim\Bbb R^2\;$ and then

$$\lim_{z\to i}|1-z|^2=\lim_{(x,y)\to(0,1)}||(1,0)-(x,y)||^2=\lim_{(x,y)\to (0,1)}\left[(1-x)^2+y^2\right]=1^2+1^2=2$$

Answer 2

You will have hard time bounding the distance this way. $\varepsilon-\delta$ tricks work well if you use more "homogenous" inequalities or you should decompose it to an atomic level (But thats just my two cents)...
Write $z=x+iy$, then: $$||1-z|^2-2| =|(1-x)^2 +y^2 -2|=|-2x+x^2+y^2-1| \tag1 $$ Recall we have $\delta^2 > |z-i|^2 =x^2+(y-1)^2 $, so obviously enough: $|x|<\delta$ and $|y-1| < \delta$

Returning to $(1)$ we have: $$|-2x+x^2+y^2-1|\leq|x||-2+x| + |y-1||y+1| $$

Which can be bounded in terms of delta (I leave it to you).