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My professor told me that in real analysis, e^z tends to infinity when z tends to infinity but in complex analysis, e^z does not tend to infinity when z tends to infinity. But I'm not understand this difference. Please help me.

$e^z$ has an essential singularity at $z=infty$, equivalently $e^{1/z}$
has an essential singularity at $z=0$.

If you have a function like $f(z)=z^3+z-1$, its behaviour as $ztoinfty$
is quite straightforward. If $(z_n)$ is a sequence of points with $z_ntoinfty$
(that means that $|z_n|toinfty$) then $g(z_n)toinfty$.

But $e^z$ isn't like that. If all we know about the sequence $(z_n)$
is that $z_ntoinfty$, then that tells us little about the sequence $e^{z_n}$.
For instance if $z_n=n$, then $e^{z_n}toinfty$ in the obvious manner. But
with $z_n=-n$ then $e^{z_n}to0$. With $z_n=2npi i$, then $e^{z_n}to1$
but with $z_n=ni$ then $e^{z_n}$ wanders around the unit circle but never
converging. One can cook up the $z_n$ to get any limit you like, and also
virtually all sorts of divergent behaviour too.

Infinity and poles in complex analysis make a lot more sense when you view them from the perspective of manifolds. I assume the definition of a complex manifold, chart, holomorphic function between manifolds.

If $U$ is an open set in $mathbb C$, $a in U$, and $f: U -{a} rightarrow mathbb C$ is a holomorphic function, then either $f$ either has a removable singularity, a pole, or an essential singularity at $a$. This depends, respectively, on whether the Laurent series expansion of $$f(z) = sumlimits_{n in mathbb Z} c_n (z-a)^n$$ has no nonzero negative terms, finitely many nonzero negative terms, or infinitely many such terms.

More generally, if $X$ is a complex manifold, $a in X$, and $f: X - {a} rightarrow mathbb C$ is holomorphic, it makes sense to talk about whether $f$ has a removable singularity, pole, or essential singularity there. One takes a chart $(U,varphi)$ containing $a$, and considers the holomorphic function $f circ varphi^{-1}$ in a punctured neighborhood of $varphi(a)$ in the above sense. This is independent of chart.

Essential singularities are very strange. If $f$ has an essential singularity at $a$, then in any punctured neighborhood of $a$, no matter how small, $f(z)$ takes all complex values with at one most exception. It follows that a sequence of complex numbers $z_n$ can be chosen to approach $a$ such that $limlimits_{n to infty} f(z_n)$ is any given complex number, with at most one exception. This is Picard's theorem.

Consider the exponential function $f(z) = e^z$ on $mathbb C$. We view $mathbb C$ as an open submanifold of the Riemann sphere $hat{mathbb C} = mathbb C cup {infty}$. Using a chart, we can view $f$ as a holomorphic function on a punctured neighborhood of $infty$, and ask whether $f$ has a removable singularity, pole, or essential singularity at $infty$.

We use the chart $(U,varphi)$ where $U = hat{mathbb C} - {0}$, and $varphi: U rightarrow mathbb C$ is the homeomorphism given by $varphi(z) = frac{1}{z}$ if $z neq infty$, and $varphi(infty) = 0$.

Then the question of what kind of singularity $e^z$ has at $infty$, is equivalent to what kind of singularity the holomorphic function $f circ varphi^{-1}: mathbb C -{0} rightarrow mathbb C$ has at $0$. For $z neq 0$, we have

$$f circ varphi^{-1}(z) = f(frac{1}{z}) = e^{frac{1}{z}} = sumlimits_{n leq 0} frac{z^n}{n!}$$

which is an essential singularity.

Therefore by Picard's theorem, for every complex number $lambda$ with at most one exception, there exists a sequence of complex numbers $z_n$ tending to infinity such that $limlimits_{n to infty} e^{z_n} = lambda$.