Prove $\lim_{z \rightarrow \infty}\left(1+\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right)$=1 Fix $c_0, c_1, \cdots, c_{d-1} \in \mathbb C$. Prove $$\lim_{z \rightarrow \infty}\left(1+\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right)=1$$
It seems stupid since I think we can just write it as
$$\begin{aligned}
\lim_{z \rightarrow \infty}\left(1+\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right) & =\lim_{z \rightarrow \infty}1+\lim_{z \rightarrow \infty}\frac{c_{d-1}}{z}+\lim_{z \rightarrow \infty}\frac{c_{d-2}}{z^2}+ \cdots + \lim_{z \rightarrow \infty}\frac{c_0}{z^d}\\ &=1+0+\cdots +0\\
& =1
\end{aligned}$$
But I am not sure whether I missed something here since it is a problem in the textbook...Thanks for any advise.
 A: You have to be careful here since a complex number has a real and an imaginary part. What should $z\to\infty$ say? Consider the following sequences:
$$
a_n=n+ni,~b_n=n-ni,~c_n=-n+ni,~d_n=-n-ni.
$$
Which of these sequences goes to infinity? But we can go further an consider a rotating sequence like $e_n=n\cos\left(n\frac\pi8\right)+n\sin\left(n\frac\pi8\right)$. Is here $e_n\to\infty$?
My point is, that for real limits, you can easily understand $\infty$ as extrem large because the reals are ordered. But there is no order on the complex plane, so $z\to\infty$ has to be used very, very carefully!
I'm actually sure that the textbook considers the limit $z\to\infty$ in sense of $|z|\to\infty$ and we have to use this to prove the statement.
Now let us prove

Fix $c_0, c_1, \cdots, c_{d-1} \in \mathbb C$, then
  $$\lim_{z \rightarrow \infty}\left(1+\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \ldots + \frac{c_0}{z^d}\right)=1.$$

First, we consider that is is sufficient to prove
$$
\lim_{z \rightarrow \infty}\left(\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \ldots + \frac{c_0}{z^d}\right)=0.
$$
Using the rules for existing limits, we can deduce the statement. 
Further, there should be a theorem or lemma in your textbook which says:

Lemma: If $\lim_{z\to a}|f(z)|=0$ then $\lim_{z\to a}f(z)=0$.

So we observe
$$
0\leq\left|\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right|
\leq \frac{|c_{d-1}|}{|z|}+\frac{|c_{d-2}|}{|z|^2}+ \ldots + \frac{|c_0|}{|z|^d}.
$$
This implies
$$
0\leq\lim_{|z|\to\infty}\left|\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right|
\leq \lim_{|z|\to\infty}\frac{|c_{d-1}|}{|z|}+\frac{|c_{d-2}|}{|z|^2}+ \ldots + \frac{|c_0|}{|z|^d}.
$$
On the RHS we can substitude $x=|z|$ and get a real limit, which is obviously $0$. We deduce
$$
\lim_{|z|\to\infty}\left|\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right|=0
$$
and using the above lemma we get further
$$
\lim_{|z|\to\infty}\left(\frac{c_{d-1}}{z}+\frac{c_{d-2}}{z^2}+ \cdots + \frac{c_0}{z^d}\right)=0
$$
and are done.
Complex limits can be similar to real limits, but you have to be more precise!
A: Your limit is equivalent to
$\lim_{z \to \infty} \sum_{k=1}^d \dfrac{c_{d-k}}{z^k}
= 0
$.
If
$|z| >1$,
then,
by the triangle inequality,
$\begin{array}\\
|\sum_{k=1}^d \dfrac{c_{d-k}}{z^k}|
&\le \sum_{k=1}^d |\dfrac{c_{d-k}}{z^k}|\\
&= \sum_{k=1}^d \dfrac{|c_{d-k}|}{|z^k|}\\
&\le \sum_{k=1}^d \dfrac{|c_{d-k}|}{|z|}
\qquad\text{since } |z^k| \ge |z|\\
&\le \dfrac1{|z|}\sum_{k=1}^d |c_{d-k}|\\
\end{array}
$
Therefore,
if
$|z|
\ge \max(1, \dfrac1{\epsilon}\sum_{k=1}^d |c_{d-k}|)
$,
$|\sum_{k=1}^d \dfrac{c_{d-k}}{z^k}|
\le \epsilon$.
