# Why $\lim\limits_{z\to\infty}\frac{P(z)}{Q(z)}=\lim\limits_{y\to\infty}\frac{P(iy)}{Q(iy)}$?

While reading about the concept of $$L$$-stability I came across a result that said that for a rational (complex) function $$R(z)=P(z)/Q(z)$$ where $$P,Q$$ are polyomials we have $$\lim\limits_{z\to\infty}\frac{P(z)}{Q(z)}=\lim\limits_{y\to\infty}\frac{P(iy)}{Q(iy)},$$ where $$y$$ is real and $$i^2=-1$$. In words, the limit when moving on the real axis to infinity is the same as the limit when moving to infinity on the imaginary axis. I don't really know much about complex analysis but I was wondering why this is the case.

• Because the $\lim_{z\rightarrow \infty} R(z)=a/b$ where $a$ is the leading coefficient of $P$ and $b$ is the leading coefficient of $Q$ given that $deg(P)=deg(Q)$. It is not relevant in which direction you move to infinity as long as the absolute value blows up. Commented Mar 2, 2021 at 12:56

If $$\deg P(z)>\deg Q(z)$$, both limits are infinity. If $$\deg P(z)<\deg Q(z)$$, both limits are $$0$$. Otherwise, both limits are equal to$$\frac{\text{leading coefficient of }P(z)}{\text{leading coefficient of }Q(z)}.$$More generally, in each case there is the limit $$\lim_{z\to\infty}\frac{P(z)}{Q(z)}$$, and so the direction that you take as you go to $$\infty$$ doesn't matter.