Limit of a complex sequence So I wanted to calculate $$\lim_{n\rightarrow\infty}\frac{n^2}{(4+5i)n^2+(3+i)^n}$$
I thought that I could do it easier if I calculate $\lim_{n\rightarrow\infty}\frac{(3+i)^n}{n^2}$. First I write $\phi=\arctan(\frac{1}{3})$ so that $3+i=\sqrt{10}(\cos\phi+i\cdot\sin\phi)$. Now we have $\lim_{n\rightarrow\infty}\frac{(3+i)^n}{n^2}=\lim_{n\rightarrow\infty}\frac{(\sqrt{10}(\cos\phi+i\cdot\sin\phi))^n}{n^2}=\lim_{n\rightarrow\infty}\frac{10^{n/2}(\cos(n\cdot\phi)+i\cdot\sin(n\cdot\phi))}{n^2}$. Looking now at the limit of the absolute value of the real and imaginary part, we see both go to $\infty$. Knowing that we than have the complex number should go to $\pm\infty\pm i\infty$. Well adding there $4+5i$ doesn't change a lot. If we now look at $\frac{1}{\pm\infty\pm i\infty}$, can we say it equals to $0$? I am still a bit confused with the complex infinity, but in theory it should. Is there maybe a better proof of this limit?

The correct way of doing this is to show that
$$\lim_{n \to \infty} \left| \frac{n^2}{(4+5i)n^2 + (3+i)^n} \right| =0$$
Now, write
$$\frac{n^2}{(4+5i)n^2 + (3+i)^n} = \frac{1}{(4+5i) + (3+i)^n/n^2}$$
and using triangular inequality, $$|(4+5i) + (3+i)^n/n^2| \ge |(3+i)^n/n^2| - |4+5i| =$$ $$ =|3+i|^n/n^2 - |4+5i| = \frac{\sqrt{10}^n}{n^2} - \sqrt{41} \to + \infty$$
so that by comparison $\lim_{n \to \infty} |(4+5i) + (3+i)^n/n^2| = + \infty$ and you conclude.
