I have to prove that the complex-valued function $f(z)=e^\frac{1}{z^2}$ is uniformly continuous on the region of the complex plane: $0\lt\vert z\vert\leq R$ , $\vert\arg(z)\vert\leq\pi/6$.
I have tried using parametric equations for the given domain but I don't know how to relate them to any continuity definitions. Any help is appreciated.
It appears that the book is wrong.
Let $\displaystyle z_n = \frac{1}{2n}e^{i\pi/6}$ and $\displaystyle w_n = \frac{1}{n}e^{i\pi/6}$. We have $\displaystyle |z_n - w_n| = \frac{1}{2n}|e^{i\pi/6}| = \frac{1}{2n} \to 0$ as $n \to \infty$.
However, $z_n^{-2} = 4n^2e^{-i\pi/3}= 2n^2(1 - i\sqrt{3})$ and $w_n^{-2} = \frac{n^2}{2}e^{-i\pi/3}= \frac{n^2}{2}(1 - i\sqrt{3})$, and, hence, as $n \to \infty$,
$$|f(z_n) - f(w_n)| = \left|e^{2n^2}e^{-i2\sqrt{3}n^2} - e^{n^2/2}e^{-i\sqrt{3}n^2/2}\right|\geqslant e^{2n^2}\left|e^{-i2\sqrt{3}n^2}\right| - e^{n^2/2}\left|e^{-i\sqrt{3}n^2/2}\right|\\ = e^{2n^2}-e^{n^2/2}\to +\infty$$