Calculating $\lim_n e^{-inz}$ In an exercise I have to prove that $f_n(z)=e^{-inz}$ converges uniformly for $\Re(z)>3$.
So I have to prove that:
$$\forall \varepsilon>0, \exists p \in \mathbb{N}:|e^{-inz}-f(z)|<\varepsilon\ \ \ \text{if } n\geq p$$
My question is, how can I find that $f(z)$?
I've tried calculating the $\lim_n e^{-inz}$ by doing the following:
if $z=x+iy$ then: $\lim_n e^{-inz}=\lim_n \frac{1}{e^{inz}}=\lim_n \frac{e^{ny}}{e^{inx}}$ but I could not continue from now on. How can I evaluate this limit?
 A: The given condition is based on the existance of the limit
$$L=\lim\limits_{n\to\infty} e^{-inz},\tag1$$ 
which exists only in the next cases.


*

*$(1.)\ \Im z<0,$ or

*$(2.)\ \Im z = 0\,\vee\, \Re z = \pi k,\quad k\in\mathbb Z.$
If $\Im z= y >0,$ then $\left|e^{-inz}\right| = e^{ny}$ is unbounded, if $n\to\infty.$
If $\Im z=0,\ \Re z = x\not\in\mathbb Z,$ then
$$\left|e^{-i(n+1)x} - e^{-inx}\right| = \left|e^{\large -i\frac{2n+1}2x}\right|\, \left|e^{\large \frac{1}2ix} - e^{\large -\frac{1}2ix}\right|= 2\left|\sin\dfrac x2\right|,$$
and the limiting transition is failed for $\varepsilon < \left|\sin\dfrac x2\right|.$
In the first case, $f(z) = f(x-i y)=0,\quad y>0,$
$$\left|e^{-inz}-f(z)\right| = e^{-ny}\left|e^{-inx}\right| = e^{-ny},$$
$$p(\varepsilon) = \lceil-\ln \varepsilon\rceil.$$
In the second case, $f(z)= f(-\pi i k) = 1,$
$$p(\varepsilon) = 1.$$
Uniformed convergency relates with the functions $f_n(z)$ for every bounded $n$ and does not relate with the limit $(1).$
A: Your question is how to find $f(z)$, so I'll attempt to address that.
I'm actually very nervous because no one seemed to mention that this is a geometric sequence, which makes my think I'm somehow deeply mistaken..... Here's what I think though.
--
Let $E$ denote the complex exponential function. Then, it is known that for any complex $x,y$, $E(x+y) = E(x)E(y)$.
From this, it follows that $E(nx) = E(x+ \cdots + x) = E(x)\cdots E(x) = (E(x))^n$ for any complex $x$. In other words, taking $x = -iz$,
$$
    E(-inz) = \left( E(-iz) \right)^n.
$$
Also, as $E(iz)E(-iz) = E(iz-iz) = E(0) = 1$, we have $E(-iz) = 1/E(iz)$. In other words,
$$
    E(-inz) = \left( \frac{1}{E(iz)} \right)^{n},
$$
which is a geometric sequence. Now geometric sequences in the complex plane behave pretty much just like on the real line (see attached picture, wherein the squares are $\left( \frac{\sqrt{99}+i}{9.9} \right)^n$ and the circles are $\left( \frac{\sqrt{99}+i}{10.1} \right)^n$).
$\left( \frac{\sqrt{99}+i}{9.9} \right)^n$ and the circles are $\left( \frac{\sqrt{99}+i}{10.1} \right)^n$" />
Specifically, for complex $x$, we have $x^n \to 0$ if $|x|<1$, and $(x^n)$ diverges otherwise (edit: unless $x=1$, in which case it still converges). Thus, the point-wise limit is zero when $\frac{1}{E(iz)}$ has modulus less than unity (or is equal to unity), and otherwise does not exist.
Now
$$
    \left| \frac{1}{E(iz)} \right|
=
    \frac{1}{|E(iz)|}
=
    \frac{1}{E\big(-\text{Im}(z)\big)}
=
    E\big(\text{Im}(z)\big),
$$
and so

the point-wise limit is zero whenever Im$(z)<0$, the point-wise limit is $1$ if $z$ is an integer multiple of $2 \pi$, and otherwise the point-wise limit does not exist.

