Calculating $\lim_n e^{-inz}$

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 $(1).$