Limit of complex function if z doesn't approach zero

Question

$\lim_{z\to 3+i}Im(z) = Im(3+i) = 1$

I'm not sure if that works because I'm not sure if I actually covered all directions or not. How would I properly calculate this limit?

Answer 1

By definition $\forall \epsilon>0$ $\exists \delta>0$ shuch that for

$$|z-z_0|=|x-3+i(y-1)|=\sqrt{(x-3)^2+(y-1)^2}<\delta$$ $$\implies|\Im(z)-1|<\epsilon\iff|y-1|<\epsilon$$

then it suffices to take $\delta\le\epsilon$ such that the open ball $|z-z_0|<\delta$ is contained in the horizontal strip $|y-1|<\epsilon$.