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.

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
$$
|f(z)-f(2i)|\le 100 |z-2i|<100\delta\le 100\cdot\frac {\epsilon}{100}=\epsilon\ .
$$
