Confused about step in proof that $\lim\limits_{z\to i} z^2 = - 1$ I am reading a Complex Analysis text and I can't understand why they claim that $|z+i|$ can be bounded by $3$. They state, "If we choose $\delta$ smaller than $1$, then the factor $|z+i|$ on the right can be bounded by $3$.''
$$|z^2+1|=|z-i||z+i|<\delta|z+i|$$
But I don't understand why this is true. $|z+i|=|x+i(y+1)|=\sqrt{x^2+y^2+2y+1}=\sqrt{2+2y}.$ But if $y$ is really big, can't this be way bigger than $3$? I feel like a fool for not seeing what is supposed to be an obvious observation. I have been trying to figure this out for hours.
 A: $y$ cannot be "really big", because they assume $\underline{|z-i|<\delta}<1$, so
$$|z+i|=|z-i+2i|\le|z-i|+|2i|<\delta+2<3.$$
A: Alternative approach:
The problem can be visualized geometrically.
Draw a circle in the Complex plane, of radius $1$, centered at $(0 + i)$.  Specifying that $|z - i| \leq 1$ is geometrically equivalent to specifying that the point $z$ is in this circle.
Now, shift the circle up, one unit, so that each point $z$ in the circle is shifted to the point $(z + i)$.  In effect, you have a new circle of radius $(1)$, centered at $(0 + 2i)$.
So, any point $z$ that was in the original circle, will be shifted to the point $z + i$, that is in the new circle.

Now, consider the circle of radius $3$, centered at the origin.  This (larger) circle completely encompasses the circle of radius $1$ centered at $(0 + 2i).$
Therefore, any point inside the smaller circle, centered at $(0 + 2i)$ is also inside the larger circle, centered at the origin.  This implies that any point inside the smaller circle must have its distance to the origin be $\leq 3.$
