Complex Analysis: Epsilon - Delta proof Good morning, I tried to prove this limit but I don't know if I'm doing it the right way:
$$\lim_{z \rightarrow 2i} z^4 i + 3z^2 - 10i = -12 + 6i$$
This is how I approached it:
Per definition of limit: $\forall\epsilon>0,\exists\delta>0 / ||z-2i||<\delta\quad\land\quad||z^4i+3z^2-10i+12-6i||<\epsilon$
First, I factorized the second expresion:
$$||(z-2i)\cdot(z+2i)\cdot(iz^2+3-4i)||<\epsilon\tag{1}\label{eq1}$$
Now that I've factorized it, I tried to find a relation between $\quad\delta\quad$ and $\quad\epsilon$:
I tried to use triangular inequality in every factor in \eqref{eq1} in order to replace them with a function of $\delta$:
$$||(z+2i)||=||(z-2i)+4i||\rightarrow\text{By triangle inequality:}||(z-2i)+4i||\leq||(z-2i)||+||4i||\rightarrow||z+2i||\leq||z-2i||+4$$
$$||(iz^2+3-4i)||=||z^2+4-3i-8||\leq||z^2+4||+||-3i-8||\rightarrow||z^2-4-3i||\leq||z^2+4||+\sqrt{73}$$
Then I had the following expression:
$$\delta\cdot(\delta+4)\cdot(||z^2+4||+\sqrt{73})<\epsilon$$
Applying difference of squares and using the theorem of triangular inequality in the third factor I eventually had:
$$\delta\cdot(\delta+4)\cdot(\delta\cdot(\delta+4)+\sqrt73)<\epsilon\tag{2}\label{eq2}$$
Now, in \eqref{eq2} I have a relation between $\delta$ and $\epsilon$, but I don't think it's a good answer for a proof.
Second approach:
Other way I tried to solve it was to assume a value of $\delta$: 
$||z-2i||<\delta=3$ (I assumed $\delta=3$ to simplify operations), then by the theorem of triangular inequality I had: $\quad||z-2i||\leq||z||+||-2i||=||z||+2\rightarrow||z||<\delta-2\rightarrow||z||<1$, now replacing in \eqref{eq1}:
$$\delta\cdot||(z+2i)||\cdot||(iz^2+3-4i)||<\epsilon$$
By the triangle inequality I had eventually:
$$\delta\cdot(||z||+2)\cdot(||z||^2+5)<\epsilon\rightarrow\delta\cdot3\cdot6<\epsilon\rightarrow\delta<\frac{\epsilon}{18}$$
I  don't know if any of this aproaches are valid so please, I'd appreciate any help you can offer, thanks.
 A: In such cases it is ok to use generous estimations. First of all we note that for the function $f:\Bbb C\to\Bbb C$
$$
f(z)=z^4 i + 3z^2 - 10i 
$$
we have 
$$
f(2i)=(2i)^4\cdot i+3(2i)^2-10 i
=16i -12-10i
= -12 + 6i\ .
$$
Now we estimate for $|z|<3$...
$$
\begin{aligned}
|f(z)-f(2i)|
&=\Big|\ i(z^4-(2i)^4) + 3(z^2-(2i)^2)\ \Big|\\
&=\Big|\ i(z-2i)(z+2i)(z^2+(2i)^2)+3(z-2i)(z+2i)\ \Big|\\
&=|z-2i|\cdot 
\Big|\ i(z+2i)(z^2+(2i)^2)+3(z+2i)\ \Big|\\
&\le|z-2i|\cdot 
\Big[\ |i|\cdot |z+2i| \cdot |z^2+(2i)^2|+3|z+2i|\ \Big]\\
&\le|z-2i|\cdot 
\Big[\ |i|\cdot (|z|+ |2i|) \cdot (|z|^2+|2i|^2)+3(|z|+2|i|)\ \Big]\\
&\le|z-2i|\Big[\ (3+2)(9+4)+3(2+2)\ \Big]\\
&\le 100\cdot |z-2i|\ .
\end{aligned}
$$
(This could have been done in two steps, but let it be here so detailed.)
Now fix an $\epsilon>0$. 
Take $\delta$ to be the minimum under $1$ and $\epsilon/100$.
Consider now $z$ at distance $<\delta$ from $2i$. In particular $|z|\le |z-2i|+|2i|<\delta+2\le 3$, so the  above estimation holds, proving that
$$
|f(z)-f(2i)|\le 100 |z-2i|<100\delta\le 100\cdot\frac {\epsilon}{100}=\epsilon\ .
$$
