The limit as $z \to \infty$ of $e^z+z$, does it exist? There was a bit of confusion in a homework exercise from complex function theory. The problem was to classify the isolated singularities of
\begin{equation}
e^{\frac{1}{1-z}} + \frac{1}{1-z}.
\end{equation}
The isolated singularity being of course $z_0=1$. One can easily show that $z_0$ is an essential singularity by looking at the Laurent-series of $e^{\frac{1}{1-z}}$ and adding $1/(1-z)$ to get that the above function is equal to
$$
1 + \frac{2}{1-z}+ \frac{1}{2!(1-z)^2} + \frac{1}{3!(1-z)^3}+ \ldots
$$
However, a good friend of mine had a different approach. He tried to prove that it was a pole as follows: $z_0$ is a pole if $\lim\limits_{z \to z_0}f(z) = \infty$. Next we get that
\begin{align}
\lim_{z\to 1} e^{\frac{1}{1-z}} + \frac{1}{1-z} = \lim_{w \to 0} e^{\frac{1}{w}} + \frac{1}{w}.
\end{align}
Next, for complex functions it is true that
$$
\lim_{w \to 0} f(w) = \lim_{w \to \infty} f\left(\frac{1}{w}\right) \implies \lim_{w \to 0} e^{\frac{1}{w}} + \frac{1}{w} = \lim_{w \to \infty} e^{w} + w.
$$
He then concluded that, since $e^w + w \to \infty$ as $w \to \infty$, $\lim_{z\to 1} e^{\frac{1}{1-z}} + \frac{1}{1-z} = \infty$, and hence $z_0$ is a pole. These are two conflicting conclusions, hence one of the proofs must be wrong. I believe the essential singularity proof is correct, and that the pole proof might be incorrect if
$e^w + w \not\to \infty$ as $w \to \infty$. What path can $w$ take to infinity to have this expression not be infinity? Many thanks!
 A: Your friend is wrong. In fact the first solution shows, by Picard's Theorem, that in every region of then type $\{w: |w| >R\}$ the function  $e^{w}+w$ takes all values except possibly one (infinitely many times!). In particular, the image of $\{w: |w|>R\}$ intersects the unit disk for every $R$.  So it is wrong to say that $|e^{w}+w| \to \infty$ as $|w| \to \infty$. The same argument also shows that $\lim_{w \to \infty} (e^{w}+w)$ does not exist.
A: Your friend is wrong. Without using Picard's Great Theorem, we may directly prove that for every $k\ge1$, there exists some $x\in\mathbb R$ and $(2k-1)\pi<y<2k\pi$ such that $w=x+iy$ is a solution to $e^w+w=0$. First, rewrite the equation as the system of equations
$$\begin{align}
e^x\cos y&=-x,\\
e^x\sin y&=-y.
\end{align}$$
Therefore $x=y\cot y$ and the equation $e^w+w=0$ is equivalent to
$$e^{y\cot y}\sin y=-y.\tag{1}$$
When $y=(2k-1)\pi+\delta$ or $y=2k\pi-\delta$, we have $\sin y=-\sin\delta$ and hence
$$\begin{align}
e^{y\cot y}\sin y
&=\exp\left(\frac{-y\cos y}{\delta}\frac{\delta}{\sin\delta}\right)\frac{\sin\delta}{\delta}(-\delta).
\end{align}$$
Therefore, when $k\ge1$,
$$\begin{align}
\lim_{y\searrow(2k-1)\pi}e^{y\cot y}\sin y
&=\lim_{\substack{\delta\to0^+\\y=(2k-1)\pi+\delta}}\exp\left(\frac{y}{\delta}\right)(-\delta)=-\infty,\\
\lim_{y\nearrow2k\pi}e^{y\cot y}\sin y
&=\lim_{\substack{\delta\to0^+\\y=2k\pi-\delta}}\exp\left(\frac{-y}{\delta}\right)(-\delta)=0.\\
\end{align}$$
Hence $(1)$ is solvable on $(2k-1)\pi<y<2k\pi$ for every $k\ge1$, by Intermediate Value Theorem.
