Prove the following using the definition of the limit? (Complex analysis)

Question

We want to prove the following:

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

I got this far but I don't know if it's right:

$|(z^2 + (1+i)z + 2)- i| < \epsilon$

$|z^2 + z + iz + 2 - i| < \epsilon$

$|(z - i) + (z^2 + iz + 2)| < \epsilon$

$|(z - i) + (z - i)(z + 2i)| < \epsilon$

$|(z - i)(1 + (z+2i))| < \epsilon$

$|(z - i)||(1 + (z+2i))| < \epsilon$

Now what do I do? Should I restrict z or z+2i ...? I have no clue.

Thank you.

Answer 1

Try to put $w=z-i$, so that $w \to 0$. Then $z^2=(w+i)^2=w^2+2iw-1$ and $(1+i)z= (1+i)(w+i)=w+i+iw-1$. Hence $$ z^2+(1+i)z+2=w^2+2iw-1+w+i+iw-1+2=w^2+(3i+1)w+i. $$ Now, pick $\epsilon>0$ and choose $\delta<\min \{\sqrt{\frac{\epsilon}{2}},\frac{1}{2}\frac{\epsilon}{|3i+1|}\}$ and deduce that $$ \left| z^2+(1+i)z+2 -i \right| = \left| w^2+(3i+1)w+i -i \right|= \left| w^2+(3i+1)w \right| $$ and $$ \left| w^2+(3i+1)w \right| \leq |w|^2+|3i+1||w|\leq\frac{1}{2}\left(\epsilon + \epsilon\right)=\epsilon $$ provided that $|w|<\delta$.

Answer 2

As $z$ tends to $i$, $z-i$ becomes as little as you please, so bound the other term.

Let's say that $\quad|z-i|<\delta<1$

It implies that $\quad|1+z+2i|=|1+3i+z-i|\le|1+3i|+|z-i|<4+\delta<5$

So you have

$$|(z - i)||(1 + (z+2i))| < 5\delta$$