Solve $\lim\limits_{z\to 2i} 2x + y^2 = 4$ using the formal definition of limit First of all we need to remember that $z=x+iy$. Well, I'm confusing about a tip from the author where he says to consider $|z-z_{0}| < 1$. Then we need to prove $|x| < 1$ and $|y|<3$ . This is the part I'm not understanding! How can I prove that? Then he says to use this affirmation to conclude:
$$\begin{align}|f(z) - f(z_0)| &= |2x+y^2 - 4| \\&= |2x + (y-2)(y+2)| \\&\leq 2|x| + |y-2|(|y|+2)\\& \leq 5|x| + 5|y-2|\end{align}$$
and then $5|x| + 5|y-2| \leq 10|z-2i|$

$|a+ib|=\sqrt {a^{2}+b^{2}}$ if $a$ and $b$ are real numbers. Hence $|a+ib| \geq \sqrt {a^{2}}=|a|$ and $|a+ib| \geq \sqrt {b^{2}}=|b|$. Thus $|(x+iy)- (0+2i)| =|x+i(y-2)| <1$ implies that $|x| <1$ and $|y-2| <1$ so $|y| <1+2=3$.
[To finish the proof take $\delta =\frac {\epsilon} {10}$]
