Complex Limits HW Help Prove the following limits by using $\epsilon$ and δ
1) Show $\lim_{z\to 2}z2 + iz = 4 + i2$.
2) Show $\lim_{z\to -i} 1/z = i$.
3) Show $\lim_{z\to4i}z/\overline{z}=-1$.
For 1) i'm stuck at $|z-2|< \delta$. Then I did $|z^2+iz-(4+iz)| = |z^2-4|=|z+2||z-2| <\delta |z+2|$. I'm not sure where to go next to find epsilon. I know I need to make a graph of some sort next but I'm not sure.
For 2) I'm stuck at $|z+i|< \delta$. Then I wrote $|(1/z) -i| = $?
For 3) I'm stuck at $|z-4i|< \delta$. Then I wrote $|(z/\overline{z}) - 1| = 0$ ?
 A: Some hints for proving that $\lim_{z\to z_0}f(z)=w$. (Say me if you need more help.) 
1) What's the definition of limit? Given an $\varepsilon>0$, that you can suposse sufficently small depending you necesities, you must show that there is some $\delta_\varepsilon>0$, which obviously depends on $\varepsilon$, such that if $|z-z_0|<\delta_\varepsilon$, then $|f(z)-w|<\varepsilon$. So, you are trying to solve
$$|z-z_0|<\delta_\varepsilon \Rightarrow |f(z)-w|<\varepsilon $$
2) How? What can you use? You can use the properties of the norm (triangular inequality and its alternative forms and, in this case, multiplicativity) and algebraic transformations of the function $f$. Sometimes, you will have to transform your epsilon in an $\varepsilon/C$ in order to obtain in the final step the desired epsilon.
I will work for you the second in order to illustrate this. Note that we want to prove that there is some $\delta_\varepsilon>0$ such that
$$|z+i|<\delta_\varepsilon \Rightarrow |1/z-i|<\varepsilon $$
But, by algebraic transformations,
$$1/z-i=\frac{1-iz}{z}=-i\frac{z+i}{z}$$
and now, using the properties of the norm,
$$\left|\frac{1}{z}-i\right|=\left|-i\frac{z+i}{z}\right|=\frac{|z+i|}{|z|}\text{.}$$
Therefore, we are reduce to prove that this $\delta_\varepsilon$ satifies
$$|z+i|<\delta_\varepsilon \Rightarrow \frac{|z+i|}{|z|}<\varepsilon \text{.}$$
However, since by triangular inequality $|z|\geq ||z+i|-1|> 1-\delta_\varepsilon$, after asuming $\delta_\varepsilon< 1$, our statement will be a consequence of showing that there is a $\delta_\varepsilon>0$ such that
$$\frac{\delta_\varepsilon}{1-\delta_\varepsilon-1}\leq \varepsilon$$
due to the fact that undr the asumption $|z+i|\delta_\varepsilon<1$, we have that
$$|1/z-i|<\frac{\delta_\varepsilon}{1-\delta_\varepsilon}\text{.}$$
Finally, one shows by usual transformations that $\frac{\delta_\varepsilon}{1-\delta_\varepsilon}$ iff $\delta_\varepsilon\leq \frac{\varepsilon}{1+\varepsilon}$. Thus$$\delta_\varepsilon=\frac{\varepsilon}{1+\varepsilon}$$
is enough for the desired purposes.
